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Chapter 2 Structure Of Atom
The wide range of chemical behaviors observed among different elements is a direct consequence of the distinct internal structures of their atoms. Understanding the arrangement of particles within an atom is key to explaining their properties and interactions.
The idea that matter is composed of indivisible building blocks, atoms, dates back to ancient Indian and Greek philosophers (around 400 B.C.). They believed that repeatedly dividing matter would eventually lead to particles that could not be cut further, coining the term 'atom' from the Greek word 'a-tomio', meaning 'uncut-able'. However, these were purely philosophical ideas lacking experimental evidence.
Scientific investigation into the nature of atoms was rekindled in the 19th century. John Dalton's atomic theory (1808) provided a scientific foundation for these ideas, describing atoms as the fundamental, indivisible particles of matter. Dalton's theory successfully explained key laws of chemical combination, such as the Law of Conservation of Mass and the Law of Definite Proportions. However, it could not account for observations like the electrical charging of materials when rubbed, suggesting that atoms were not, in fact, the ultimate, indivisible particles.
Experiments conducted towards the end of the 19th and beginning of the 20th centuries revealed that atoms are composed of even smaller, sub-atomic particles. This unit explores the discovery and properties of these particles and the models developed to describe atomic structure.
Discovery Of Sub-Atomic Particles
Experiments involving electrical discharge through gases provided crucial insights into the composition of atoms.
Recall that particles with the same electrical charge repel each other, while particles with opposite charges attract each other. This fundamental principle helps understand the behavior of charged particles in electric and magnetic fields.
Discovery Of Electron
In the 1830s, Michael Faraday's experiments on electrolysis showed that passing electricity through electrolyte solutions caused chemical reactions at the electrodes, involving the deposition or liberation of matter. His findings suggested that electricity had a particulate nature.
Later, in the mid-1850s, scientists like Faraday studied electrical discharge in partially evacuated tubes, known as cathode ray discharge tubes.
A typical cathode ray tube is a glass tube with two metal electrodes sealed inside. Electrical discharge occurs at very low gas pressures and high voltages. When a sufficiently high voltage is applied, a stream of invisible particles flows from the negative electrode (cathode) to the positive electrode (anode). These were named cathode rays.
To visualise these rays, a hole was made in the anode, and the glass tube behind it was coated with a phosphorescent material like zinc sulfide. When the cathode rays passed through the hole and struck the coating, a bright spot appeared.
Observations about cathode rays:
- They originate from the cathode and travel towards the anode.
- They are invisible, but their presence is detected by materials that fluoresce or phosphoresce when struck by them (like a TV screen).
- In the absence of external electric or magnetic fields, they travel in straight lines.
- When electric or magnetic fields are applied, they behave like streams of negatively charged particles. The deflection direction is consistent with a negative charge.
- The properties of cathode rays are independent of the material of the electrodes or the type of gas in the tube.
This led to the conclusion that these negatively charged particles, named electrons, are fundamental constituents of all atoms.
Charge To Mass Ratio Of Electron
In 1897, British physicist J.J. Thomson quantified the properties of electrons by measuring the ratio of their electrical charge ($e$) to their mass ($m_e$). He used a cathode ray tube with perpendicular electric and magnetic fields applied to the electron beam's path.
By applying either field, Thomson observed the deflection of the electrons. Applying both fields and adjusting their strengths, he could balance the deflecting forces so the electrons passed through undeflected (hitting point B). The amount of deflection was found to depend on:
- The magnitude of the charge ($e$) on the particle: Higher charge leads to greater deflection by the fields.
- The mass ($m_e$) of the particle: Lighter particles are deflected more easily.
- The strength of the electric or magnetic field: Stronger fields cause greater deflection.
Using precise measurements of field strengths and deflections, Thomson determined the charge-to-mass ratio ($e/m_e$) for the electron:
$\frac{e}{m_e} = 1.758820 \times 10^{11} \text{ C kg}^{-1}$
Since electrons are negatively charged, the charge is denoted as $-e$.
Charge On The Electron
While Thomson found the ratio $e/m_e$, he did not determine the individual values of $e$ or $m_e$. American physicist R.A. Millikan devised the famous oil drop experiment (1906-1914) to determine the charge of a single electron.
In this experiment, tiny oil droplets were introduced into a chamber containing two charged plates. Some droplets acquired electrical charge by colliding with ions produced by X-rays. By observing the motion of these charged droplets under the influence of gravity and an adjustable electric field, Millikan could determine the charge on each droplet. He found that the charge on any droplet ($q$) was always an integral multiple ($n$) of a fundamental unit of charge ($e$), i.e., $q = n \times e$.
Millikan's experiments yielded a value for the electronic charge: $-1.6 \times 10^{-19}$ C. The currently accepted value is $-1.602176 \times 10^{-19}$ C.
Knowing the charge ($e$) and Thomson's charge-to-mass ratio ($e/m_e$), the mass of the electron ($m_e$) could be calculated:
$m_e = \frac{e}{(e/m_e)} = \frac{1.602176 \times 10^{-19} \text{ C}}{1.758820 \times 10^{11} \text{ C kg}^{-1}} \approx 9.1094 \times 10^{-31} \text{ kg}$.
Discovery Of Protons And Neutrons
Following the discovery of electrons, experiments with modified cathode ray tubes led to the discovery of positively charged particles. When a perforated cathode was used, streams of positively charged particles were observed moving in the direction opposite to the cathode rays, passing through the holes in the cathode. These were called canal rays or anode rays.
Characteristics of these positively charged particles:
- Unlike cathode rays, their mass depends on the type of gas in the discharge tube. These particles are essentially the positively charged ions formed from the gas molecules.
- Their charge-to-mass ratio also varies depending on the gas.
- Some particles carry a charge that is a multiple of the fundamental unit of charge, $e$.
- They are deflected by electric and magnetic fields in the opposite direction compared to electrons, confirming their positive charge.
The smallest and lightest positive particle was observed when hydrogen gas was used. This particle was named the proton and was characterised in 1919. A proton carries a positive charge equal in magnitude but opposite in sign to the electron's charge ($+1.602176 \times 10^{-19}$ C).
Later, it became apparent that the total mass of an atom could not be explained by only protons and electrons. This suggested the presence of electrically neutral particles in the nucleus. In 1932, James Chadwick discovered these particles by bombarding a thin sheet of beryllium with alpha ($\alpha$) particles. Neutral particles with mass slightly greater than protons were emitted. These were named neutrons.
Key properties of the three fundamental sub-atomic particles are summarised below:
Name | Symbol | Absolute charge/C | Relative charge | Mass/kg | Mass/u | Approx. Mass/u |
Electron | e | $-1.602176 \times 10^{-19}$ | $-1$ | $9.109382 \times 10^{-31}$ | $0.00054$ | $0$ |
Proton | p | $+1.602176 \times 10^{-19}$ | $+1$ | $1.6726216 \times 10^{-27}$ | $1.00727$ | $1$ |
Neutron | n | $0$ | $0$ | $1.674927 \times 10^{-27}$ | $1.00867$ | $1$ |
Besides these discoveries, other forms of radiation were identified. X-rays, discovered by Wilhelm Röntgen in 1895, are high-energy electromagnetic radiation produced when electrons hit a metal target. Radioactivity, discovered by Henri Becquerel and further studied by Marie and Pierre Curie and Rutherford, is the spontaneous emission of radiation by certain elements. This radiation consists of alpha ($\alpha$) particles (helium nuclei, $+2$ charge), beta ($\beta$) particles (electrons, $-1$ charge), and gamma ($\gamma$) rays (high-energy electromagnetic radiation, no charge).
Atomic Models
The experimental evidence for sub-atomic particles contradicted Dalton's model of an indivisible atom. Scientists faced the challenge of developing models that could explain the arrangement of electrons, protons, and neutrons within the atom and account for atomic properties.
Key questions atomic models needed to address:
- How is the atom stable?
- How do physical and chemical properties relate to atomic structure?
- How do atoms combine to form molecules?
- What is the origin of electromagnetic radiation absorbed or emitted by atoms (atomic spectra)?
Several models were proposed, but two early ones, by J.J. Thomson and Ernest Rutherford, were particularly significant.
Thomson Model Of Atom
Following his work on electrons, J.J. Thomson proposed the first atomic model in 1898, often called the "plum pudding" model.
According to this model:
- The atom is a sphere with a radius of approximately $10^{-10}$ m.
- The positive charge is spread uniformly throughout this sphere.
- The electrons are embedded within this positive sphere, like plums in a pudding or seeds in a watermelon, arranged to achieve the most stable electrostatic configuration.
- The total positive charge equals the total negative charge, making the atom electrically neutral.
An important assumption was that the mass of the atom was evenly distributed throughout the sphere. While this model explained the atom's overall neutrality, it could not be reconciled with the results of later experiments, particularly Rutherford's scattering experiment.
Rutherford’s Nuclear Model Of Atom
Ernest Rutherford, a former student of Thomson, conducted a landmark experiment in 1909 with his students Hans Geiger and Ernest Marsden, known as the alpha ($\alpha$)-particle scattering experiment.
A beam of high-energy $\alpha$-particles (which are positively charged helium nuclei, $\text{He}^{2+}$) was directed at a very thin gold foil (around 100 nm thick). A circular screen coated with zinc sulfide (a fluorescent material) was placed around the foil to detect the scattered $\alpha$-particles, as each strike produced a flash of light.
Based on Thomson's model, which proposed a uniform distribution of positive charge and mass, Rutherford expected the $\alpha$-particles to pass through the foil with minimal deflection. However, the results were surprising:
- Most $\alpha$-particles passed straight through the foil without any deflection.
- A small fraction of the $\alpha$-particles were deflected by noticeable, small angles.
- A very few $\alpha$-particles (about 1 in 20,000) were deflected back by large angles, some even exceeding 90$^\circ$, as if they had hit something dense and bounced off.
Based on these observations, Rutherford drew revolutionary conclusions about atomic structure:
- Since most $\alpha$-particles passed through undeflected, most of the space within an atom must be empty.
- The small angle deflections indicated a significant repulsive force, suggesting that the atom's positive charge is not spread out (as in Thomson's model) but is concentrated in a very small, dense region at the center.
- The rare large-angle deflections and backward scattering meant that $\alpha$-particles sometimes hit this dense positive core directly. Calculations showed this central region, which he named the nucleus, occupies an extremely small volume compared to the atom's total volume. The radius of the atom is about $10^{-10}$ m, while the nucleus's radius is about $10^{-15}$ m. To appreciate the scale, if an atom were the size of a large sports stadium, the nucleus would be like a small pebble in the center.
Based on these conclusions, Rutherford proposed the Nuclear Model of the Atom:
- Most of the atom's mass and all its positive charge are concentrated in a tiny, dense central region called the nucleus.
- The electrons, which are negatively charged and have negligible mass, revolve rapidly around the nucleus in well-defined circular paths called orbits.
- The electrostatic force of attraction between the positively charged nucleus and the negatively charged electrons holds the atom together, similar to how planets orbit the sun due to gravitational force (hence, often called the planetary model).
Atomic Number And Mass Number
Rutherford's model established the nucleus as the location of the positive charge, carried by protons. For an atom to be electrically neutral, the number of protons in the nucleus must equal the number of electrons orbiting it.
The number of protons in the nucleus of an atom is defined as its atomic number ($Z$). This number uniquely identifies an element. In a neutral atom, the number of electrons equals the atomic number.
$Z = \text{Number of protons in the nucleus}$
$Z = \text{Number of electrons in a neutral atom}$
For example, Hydrogen has 1 proton ($Z=1$), Sodium has 11 protons ($Z=11$). A neutral Hydrogen atom has 1 electron, and a neutral Sodium atom has 11 electrons.
The mass of the nucleus comes from both protons and neutrons. Protons and neutrons collectively are called nucleons. The total number of nucleons (protons + neutrons) is the mass number ($A$) of the atom.
$A = \text{Number of protons (Z)} + \text{Number of neutrons (n)}$
An atom of an element is typically represented using the notation $^A_Z$X, where X is the element symbol, A is the mass number, and Z is the atomic number.
Isobars And Isotopes
Atoms can have variations in their composition, leading to different types:
- Isobars are atoms of different elements that have the same mass number ($A$) but different atomic numbers ($Z$). This means they have different numbers of protons and thus different chemical properties, but the total number of nucleons is the same. Example: Carbon-14 ($^{14}_6$C) and Nitrogen-14 ($^{14}_7$N).
- Isotopes are atoms of the same element (same atomic number, $Z$) but different mass numbers ($A$). This difference in mass number arises from having a different number of neutrons in the nucleus. Since isotopes of an element have the same number of protons and electrons, they exhibit very similar chemical properties.
Examples of isotopes:
- Hydrogen has three common isotopes:
- Protium ($^1_1$H): 1 proton, 0 neutrons ($A=1$, $Z=1$). Most abundant ($\approx 99.985\%$).
- Deuterium ($^2_1$D): 1 proton, 1 neutron ($A=2$, $Z=1$). (approx. 0.015%)
- Tritium ($^3_1$T): 1 proton, 2 neutrons ($A=3$, $Z=1$). Radioactive, found in trace amounts.
- Carbon has common isotopes $^{12}_6$C, $^{13}_6$C, $^{14}_6$C (6 protons, 6, 7, and 8 neutrons respectively). Carbon-12 is the standard for atomic mass.
- Chlorine has isotopes $^{35}_{17}$Cl and $^{37}_{17}$Cl (17 protons, 18 and 20 neutrons respectively).
Since chemical behaviour is primarily determined by the number of electrons (equal to the number of protons, Z), isotopes of an element show almost identical chemical reactivity. Differences are mainly in physical properties influenced by mass (e.g., density, melting/boiling points, rates of reaction, especially for lighter elements like Hydrogen).
Problem 2.1. Calculate the number of protons, neutrons and electrons in $^{80}_{35}$Br.
Answer:
The symbol for the atom is $^{80}_{35}$Br.
From the notation $_Z^A$X, we can identify the atomic number (Z) and mass number (A).
Atomic number ($Z$) = The subscript on the left = 35.
Mass number ($A$) = The superscript on the left = 80.
The species is given as a neutral atom (no charge indicated).
Number of protons = Atomic number ($Z$) = 35.
In a neutral atom, number of electrons = number of protons = 35.
Number of neutrons = Mass number ($A$) - Atomic number ($Z$).
Number of neutrons = 80 - 35 = 45.
So, $^{80}_{35}$Br has 35 protons, 45 neutrons, and 35 electrons.
Problem 2.2. The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Answer:
Given:
- Number of electrons = 18
- Number of protons = 16
- Number of neutrons = 16
The atomic number ($Z$) of an element is defined by the number of protons. Here, the number of protons is 16. The element with atomic number 16 is Sulfur (S).
The mass number ($A$) is the total number of protons and neutrons.
Mass number ($A$) = Number of protons + Number of neutrons = 16 + 16 = 32.
To determine if the species is neutral or an ion, compare the number of protons and electrons. In a neutral species, the number of protons equals the number of electrons.
Here, number of protons (16) is not equal to the number of electrons (18). The species carries an electrical charge.
The charge is determined by the difference between the number of protons (positive charges) and electrons (negative charges).
Net charge = (Number of protons $\times$ charge of proton) + (Number of electrons $\times$ charge of electron)
Relative charge = Number of protons - Number of electrons = 16 - 18 = -2.
Since there are more electrons than protons, the species is a negatively charged ion (anion) with a charge of -2.
The symbol for an ion is written as $^A_Z$X$^{charge}$.
Substituting the values: $A=32$, $Z=16$, Element symbol is S, Charge is -2.
The proper symbol for the species is $^{32}_{16}$S$^{2-}$.
Drawbacks Of Rutherford Model
Despite being a significant advancement, Rutherford's nuclear model faced challenges based on classical physics:
- Instability of the atom: According to classical electromagnetic theory (Maxwell's equations), an accelerating charged particle (like an electron orbiting the nucleus) should continuously emit electromagnetic radiation. This emission would cause the electron to lose energy, slow down, and spiral inwards, eventually collapsing into the nucleus. Calculations showed this process should take a fraction of a second ($\approx 10^{-8}$ s), yet atoms are stable.
- Cannot explain the distribution and energies of electrons: Rutherford's model did not specify how electrons are arranged around the nucleus or what their specific energy levels might be. It only described them as orbiting the nucleus like planets.
These limitations highlighted the need for a new atomic model that could explain both the stability of atoms and their characteristic spectra.
Developments Leading To The Bohr’s Model Of Atom
The limitations of Rutherford's model paved the way for new theoretical developments. Niels Bohr's model of the hydrogen atom was significantly influenced by insights gained from the study of how radiation interacts with matter. Two major concepts were particularly important:
- The dual nature of electromagnetic radiation (behaving as both waves and particles).
- Experimental data on atomic spectra.
Wave Nature Of Electromagnetic Radiation
By the mid-19th century, physicists were actively studying thermal radiation emitted by heated objects. James Clerk Maxwell unified the theories of electricity and magnetism in the 1870s, proposing that accelerating electric charges produce and transmit alternating electric and magnetic fields. These disturbances propagate as electromagnetic waves or electromagnetic radiation.
Light, the most familiar form of radiation, was shown by Maxwell to be an electromagnetic wave, consisting of oscillating electric and magnetic fields.
Properties of electromagnetic waves:
- The oscillating electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction the wave travels.
- They do not require a medium to propagate and can travel through a vacuum (unlike sound waves).
- There are many types of electromagnetic radiation, differing in their wavelengths ($\lambda$) and frequencies ($\nu$). This range of wavelengths/frequencies constitutes the electromagnetic spectrum.
Different regions of the spectrum have different names, such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. The visible light spectrum is a small portion that our eyes can detect.
Electromagnetic radiation is characterised by:
- Frequency ($\nu$): The number of waves passing a point per second. SI unit is Hertz (Hz or s$^{-1}$).
- Wavelength ($\lambda$): The distance between two consecutive crests or troughs. SI unit is meter (m), though often smaller units like nanometer (nm) or angstrom (Å) are used.
- Velocity (c): In a vacuum, all electromagnetic radiation travels at the speed of light, $c = 3.0 \times 10^8$ m s$^{-1}$ (approx.).
These properties are related by the equation:
$c = \nu \lambda$
Another quantity used, especially in spectroscopy, is the wavenumber ($\bar{\nu}$), defined as the number of wavelengths per unit length. It's the reciprocal of wavelength:
$\bar{\nu} = \frac{1}{\lambda}$
SI unit for wavenumber is m$^{-1}$, but cm$^{-1}$ is also commonly used.
Problem 2.3. The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?
Answer:
Given frequency ($\nu$) = 1,368 kHz.
First, convert kHz to Hz: $1 \text{ kHz} = 10^3 \text{ Hz}$.
So, $\nu = 1368 \times 10^3 \text{ Hz} = 1.368 \times 10^6 \text{ Hz}$.
The speed of electromagnetic radiation (radio waves) in vacuum is the speed of light, $c = 3.0 \times 10^8$ m s$^{-1}$.
The relationship between wavelength, frequency, and speed of light is $c = \nu \lambda$.
Rearranging to find wavelength ($\lambda$): $\lambda = \frac{c}{\nu}$.
Substitute the given values:
$\lambda = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{1.368 \times 10^6 \text{ Hz}} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{1.368 \times 10^6 \text{ s}^{-1}}$
$\lambda \approx 219.3 \text{ m}$.
Electromagnetic radiation with wavelengths around hundreds of meters belongs to the radio wave part of the electromagnetic spectrum. Radio broadcasting typically uses this range.
Problem 2.4. The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1nm = 10–9 m)
Answer:
Given wavelength range: Violet $\lambda_{\text{violet}} = 400$ nm, Red $\lambda_{\text{red}} = 750$ nm.
Conversion: $1 \text{ nm} = 10^{-9} \text{ m}$.
So, $\lambda_{\text{violet}} = 400 \times 10^{-9} \text{ m} = 4.00 \times 10^{-7} \text{ m}$.
And, $\lambda_{\text{red}} = 750 \times 10^{-9} \text{ m} = 7.50 \times 10^{-7} \text{ m}$.
Using the relationship $c = \nu \lambda$, we can find the frequency $\nu = \frac{c}{\lambda}$.
The speed of light $c = 3.0 \times 10^8$ m s$^{-1}$.
Frequency of violet light:
$\nu_{\text{violet}} = \frac{c}{\lambda_{\text{violet}}} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{4.00 \times 10^{-7} \text{ m}} = \frac{3.0}{4.00} \times 10^{(8 - (-7))} \text{ s}^{-1} = 0.75 \times 10^{15} \text{ Hz} = 7.5 \times 10^{14} \text{ Hz}$.
Frequency of red light:
$\nu_{\text{red}} = \frac{c}{\lambda_{\text{red}}} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{7.50 \times 10^{-7} \text{ m}} = \frac{3.0}{7.50} \times 10^{(8 - (-7))} \text{ s}^{-1} = 0.40 \times 10^{15} \text{ Hz} = 4.0 \times 10^{14} \text{ Hz}$.
The frequency range of the visible spectrum is from $4.0 \times 10^{14}$ Hz (red) to $7.5 \times 10^{14}$ Hz (violet).
Problem 2.5. Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.
Answer:
Given wavelength ($\lambda$) = 5800 Å.
Conversion: $1 \text{ Å} = 10^{-10} \text{ m}$.
So, $\lambda = 5800 \times 10^{-10} \text{ m} = 5.800 \times 10^{-7} \text{ m}$.
(a) Calculation of wavenumber ($\bar{\nu}$):
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.800 \times 10^{-7} \text{ m}} = \frac{1}{5.800} \times 10^7 \text{ m}^{-1} \approx 0.1724 \times 10^7 \text{ m}^{-1} = 1.724 \times 10^6 \text{ m}^{-1}$.
Commonly used unit is cm$^{-1}$. $1 \text{ m} = 100 \text{ cm}$, so $1 \text{ m}^{-1} = (100 \text{ cm})^{-1} = \frac{1}{100} \text{ cm}^{-1} = 10^{-2} \text{ cm}^{-1}$.
Wait, $1 \text{ m}^{-1} = 1 \text{ m}^{-1} \times \frac{1 \text{ m}}{100 \text{ cm}} \times \frac{1 \text{ m}}{100 \text{ cm}}$... this is wrong. Let's do the conversion properly:
$\lambda = 5800 \text{ Å} = 5800 \times 10^{-8} \text{ cm} = 5.800 \times 10^{-5} \text{ cm}$.
$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.800 \times 10^{-5} \text{ cm}} = \frac{1}{5.800} \times 10^5 \text{ cm}^{-1} \approx 0.1724 \times 10^5 \text{ cm}^{-1} = 1.724 \times 10^4 \text{ cm}^{-1}$.
(b) Calculation of frequency ($\nu$):
Using $c = \nu \lambda$, so $\nu = \frac{c}{\lambda}$.
Speed of light $c = 3.0 \times 10^8$ m s$^{-1}$.
Using wavelength in meters: $\lambda = 5.800 \times 10^{-7} \text{ m}$.
$\nu = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{5.800 \times 10^{-7} \text{ m}} = \frac{3.0}{5.800} \times 10^{(8 - (-7))} \text{ s}^{-1} \approx 0.5172 \times 10^{15} \text{ s}^{-1} = 5.172 \times 10^{14} \text{ Hz}$.
Particle Nature Of Electromagnetic Radiation: Planck’s Quantum Theory
While wave nature explains phenomena like diffraction and interference, some experimental observations were not explained by classical wave theory. These included:
- Black-body radiation: The pattern of radiation emitted by ideal heated objects.
- Photoelectric effect: The ejection of electrons from a metal surface by light.
- The variation of heat capacity of solids with temperature.
- The line spectra of atoms (emission of light at specific wavelengths).
These phenomena suggested that energy is not continuously variable but can only be taken up or released in discrete amounts.
Max Planck provided a key explanation for black-body radiation in 1900. A hot object emits radiation over a range of wavelengths, and the intensity distribution changes with temperature. An ideal emitter/absorber is called a black body.
Classical physics failed to predict this intensity distribution accurately. Planck proposed that energy is emitted or absorbed by atoms in discrete bundles, not continuously. He called the smallest such bundle a quantum. The energy ($E$) of a quantum of radiation is directly proportional to its frequency ($\nu$).
$E = h\nu$
where $h$ is Planck's constant, with a value of $6.626 \times 10^{-34}$ J s.
This idea of energy quantisation is like climbing a staircase – you can only stand on the steps, not in between. Energy can only take specific values, $0, h\nu, 2h\nu, 3h\nu, \dots, nh\nu$.
Photoelectric Effect
In 1887, Heinrich Hertz observed that shining light on certain metal surfaces (like potassium, rubidium, caesium) caused electrons to be ejected. This is the photoelectric effect.
Key observations from the photoelectric effect:
- Electron ejection is instantaneous when light strikes the surface; there's no time delay.
- The number of ejected electrons is proportional to the intensity (brightness) of the light.
- For each metal, there is a minimum frequency, called the threshold frequency ($\nu_0$), below which no electrons are ejected, regardless of the light intensity.
- If the light frequency ($\nu$) is greater than the threshold frequency ($\nu_0$), the ejected electrons have kinetic energy. This kinetic energy increases linearly with the frequency of the light used, but is independent of the intensity of the light.
Classical physics predicted that electron ejection and kinetic energy should depend on light intensity, contradicting observations (especially the threshold frequency). Albert Einstein explained the photoelectric effect in 1905 using Planck's quantum theory.
Einstein proposed that light consists of a stream of particles called photons, each with energy $E = h\nu$. When a photon hits an electron in the metal, it transfers its energy to the electron. If the photon's energy ($h\nu$) is greater than the minimum energy required to remove an electron from the metal surface (called the work function, $W_0 = h\nu_0$), the electron is ejected. The excess energy becomes the kinetic energy of the ejected electron.
According to the law of conservation of energy:
Energy of incoming photon = Work function + Kinetic energy of ejected electron
$h\nu = W_0 + \frac{1}{2}m_e v^2$
or
$h\nu = h\nu_0 + \frac{1}{2}m_e v^2$
A more intense light beam has more photons, so it ejects more electrons, but the energy of each photon (and thus the maximum kinetic energy of each electron) depends only on the frequency, not the intensity.
Metal | Li | Na | K | Mg | Cu | Ag |
$W_0$ / eV | 2.42 | 2.3 | 2.25 | 3.7 | 4.8 | 4.3 |
Dual Behaviour of Electromagnetic Radiation
The successful explanation of phenomena like black-body radiation and the photoelectric effect using the particle nature of light, alongside phenomena like interference and diffraction explained by wave nature, led scientists to accept that light (and all electromagnetic radiation) exhibits dual behaviour – it can behave as both a wave and a stream of particles, depending on the experiment.
Problem 2.6. Calculate energy of one mole of photons of radiation whose frequency is 5 ×10$^{14}$ Hz.
Answer:
Given frequency ($\nu$) = $5 \times 10^{14}$ Hz (or s$^{-1}$).
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
The energy of a single photon is given by $E = h\nu$.
$E_{\text{photon}} = (6.626 \times 10^{-34} \text{ J s}) \times (5 \times 10^{14} \text{ s}^{-1})$
$E_{\text{photon}} = (6.626 \times 5) \times 10^{(-34 + 14)} \text{ J} = 33.13 \times 10^{-20} \text{ J} = 3.313 \times 10^{-19} \text{ J}$.
One mole of photons contains Avogadro's number ($N_A$) of photons.
Avogadro's constant ($N_A$) = $6.022 \times 10^{23}$ mol$^{-1}$.
Energy of one mole of photons = $E_{\text{photon}} \times N_A$
= $(3.313 \times 10^{-19} \text{ J}) \times (6.022 \times 10^{23} \text{ mol}^{-1})$
= $(3.313 \times 6.022) \times 10^{(-19 + 23)} \text{ J mol}^{-1}$
= $19.95 \times 10^4 \text{ J mol}^{-1} = 1.995 \times 10^5 \text{ J mol}^{-1}$.
Converting to kJ: $1.995 \times 10^5 \text{ J mol}^{-1} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 199.5 \text{ kJ mol}^{-1}$.
So, the energy of one mole of photons is approximately 199.5 kJ mol$^{-1}$.
Problem 2.7. A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Answer:
Power of the bulb = 100 watt = 100 J/s (Energy emitted per second).
Wavelength ($\lambda$) of light = 400 nm.
Conversion: $1 \text{ nm} = 10^{-9} \text{ m}$. So, $\lambda = 400 \times 10^{-9} \text{ m} = 4.00 \times 10^{-7} \text{ m}$.
Speed of light ($c$) = $3.0 \times 10^8$ m s$^{-1}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
The energy of a single photon is given by $E = h\nu = \frac{hc}{\lambda}$.
$E_{\text{photon}} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{4.00 \times 10^{-7} \text{ m}}$
$E_{\text{photon}} = \frac{6.626 \times 3.0}{4.00} \times 10^{(-34 + 8 - (-7))} \text{ J} = \frac{19.878}{4.00} \times 10^{-19} \text{ J}$
$E_{\text{photon}} \approx 4.9695 \times 10^{-19} \text{ J}$.
The number of photons emitted per second is the total energy emitted per second (power) divided by the energy of a single photon.
Number of photons per second = $\frac{\text{Total energy per second}}{\text{Energy per photon}}$
= $\frac{100 \text{ J s}^{-1}}{4.9695 \times 10^{-19} \text{ J/photon}}$
= $\frac{100}{4.9695} \times 10^{19}$ photons s$^{-1}$
$\approx 20.12 \times 10^{19}$ photons s$^{-1} = 2.012 \times 10^{20}$ photons s$^{-1}$.
The bulb emits approximately $2.012 \times 10^{20}$ photons per second.
Problem 2.8. When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 ×10$^5$ J mol$^{-1}$. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Answer:
Given wavelength ($\lambda$) = 300 nm = $300 \times 10^{-9} \text{ m} = 3.00 \times 10^{-7} \text{ m}$.
Kinetic energy (KE) of emitted electrons = $1.68 \times 10^5$ J mol$^{-1}$.
Speed of light ($c$) = $3.0 \times 10^8$ m s$^{-1}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
(i) Calculate the energy of the incoming photon.
$E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{3.00 \times 10^{-7} \text{ m}}$
$E_{\text{photon}} = \frac{6.626 \times 3.0}{3.00} \times 10^{(-34 + 8 - (-7))} \text{ J} = 6.626 \times 10^{-19} \text{ J}$.
This is the energy of one photon. The kinetic energy is given per mole of electrons. To compare, find the energy of one mole of photons:
Energy of 1 mole of photons = $E_{\text{photon}} \times N_A = (6.626 \times 10^{-19} \text{ J/photon}) \times (6.022 \times 10^{23} \text{ photons/mol})$
= $(6.626 \times 6.022) \times 10^{(-19 + 23)} \text{ J mol}^{-1} \approx 39.90 \times 10^4 \text{ J mol}^{-1} = 3.990 \times 10^5 \text{ J mol}^{-1}$.
The minimum energy needed to remove an electron from sodium is the work function ($W_0$), which corresponds to the energy required to remove one mole of electrons.
Using Einstein's photoelectric equation (per mole):
Energy of 1 mole photons = Work function (per mole) + Kinetic energy (per mole)
$3.990 \times 10^5 \text{ J mol}^{-1} = W_0 (\text{mol}) + 1.68 \times 10^5 \text{ J mol}^{-1}$.
$W_0 (\text{mol}) = (3.990 - 1.68) \times 10^5 \text{ J mol}^{-1} = 2.31 \times 10^5 \text{ J mol}^{-1}$.
This is the minimum energy needed to remove one mole of electrons from sodium (the work function per mole).
(ii) What is the maximum wavelength that will cause a photoelectron to be emitted?
Photoelectric effect occurs only if the photon energy is greater than or equal to the work function ($h\nu \ge W_0$). The maximum wavelength ($\lambda_{\text{max}}$ or threshold wavelength, $\lambda_0$) corresponds to the threshold frequency ($\nu_0$), where the photon energy equals the work function ($h\nu_0 = W_0$).
First, find the work function per single electron:
$W_0 = \frac{2.31 \times 10^5 \text{ J mol}^{-1}}{6.022 \times 10^{23} \text{ mol}^{-1}} \approx 3.836 \times 10^{-19} \text{ J}$.
Now, use $W_0 = h\nu_0 = \frac{hc}{\lambda_0}$ to find $\lambda_0$ (the maximum wavelength).
$\lambda_0 = \frac{hc}{W_0} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{3.836 \times 10^{-19} \text{ J}}$
$\lambda_0 = \frac{6.626 \times 3.0}{3.836} \times 10^{(-34 + 8 - (-19))} \text{ m} = \frac{19.878}{3.836} \times 10^{-7} \text{ m}$
$\lambda_0 \approx 5.182 \times 10^{-7} \text{ m}$.
Converting to nanometers: $5.182 \times 10^{-7} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 518.2 \text{ nm}$.
The minimum energy to remove one mole of electrons is $2.31 \times 10^5$ J mol$^{-1}$. The maximum wavelength that will cause photoemission is approximately 518.2 nm.
Problem 2.9. The threshold frequency n0 for a metal is 7.0 ×10$^{14}$ s$^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency n =1.0 ×10$^{15}$ s$^{-1}$ hits the metal.
Answer:
Given threshold frequency ($\nu_0$) = $7.0 \times 10^{14}$ s$^{-1}$.
Given frequency of incident radiation ($\nu$) = $1.0 \times 10^{15}$ s$^{-1}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
Using Einstein's photoelectric equation:
Kinetic energy (KE) of emitted electron = $h\nu - h\nu_0 = h(\nu - \nu_0)$.
Note that the incident frequency ($\nu = 1.0 \times 10^{15} \text{ s}^{-1} = 10.0 \times 10^{14} \text{ s}^{-1}$) is indeed greater than the threshold frequency ($\nu_0 = 7.0 \times 10^{14} \text{ s}^{-1}$), so photoemission will occur.
KE = $(6.626 \times 10^{-34} \text{ J s}) \times (1.0 \times 10^{15} \text{ s}^{-1} - 7.0 \times 10^{14} \text{ s}^{-1})$
KE = $(6.626 \times 10^{-34} \text{ J s}) \times (10.0 \times 10^{14} \text{ s}^{-1} - 7.0 \times 10^{14} \text{ s}^{-1})$
KE = $(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^{14} \text{ s}^{-1})$
KE = $(6.626 \times 3.0) \times 10^{(-34 + 14)} \text{ J} = 19.878 \times 10^{-20} \text{ J} = 1.988 \times 10^{-19} \text{ J}$.
The kinetic energy of the emitted electron is approximately $1.988 \times 10^{-19}$ J.
Evidence For The Quantized Electronic Energy Levels: Atomic Spectra
When white light passes through a prism, it splits into its constituent colors (spectrum) because the different wavelengths bend at different angles. This is a continuous spectrum because all wavelengths in the visible range are present and merge into each other.
However, when atoms in the gas phase are heated or subjected to electrical discharge, they emit light only at specific, discrete wavelengths. This produces a line emission spectrum, seen as bright lines on a dark background.
Conversely, if white light is passed through a gas of the same atoms, the atoms absorb light at those specific wavelengths they would normally emit. The resulting spectrum is a line absorption spectrum, showing dark lines at the same positions where the emission spectrum had bright lines (it's like a photographic negative).
The study of these spectra is called spectroscopy. Atomic line spectra are unique to each element, serving as a 'fingerprint' for identification. The discovery of elements like rubidium and caesium was made using spectroscopy.
Line Spectrum of Hydrogen
The hydrogen atom has the simplest line spectrum. When hydrogen gas is excited, it emits light at discrete frequencies. The hydrogen spectrum consists of several series of lines in different regions of the electromagnetic spectrum.
Johann Balmer discovered a formula in 1885 that described the wavelengths of the visible lines (Balmer series):
$\bar{\nu} = R_H (\frac{1}{2^2} - \frac{1}{n^2})$, where $n = 3, 4, 5, \dots$
Here, $\bar{\nu}$ is the wavenumber and $R_H$ is a constant (Rydberg constant for Hydrogen).
Johannes Rydberg later generalised this, finding that all series in the hydrogen spectrum fit the formula:
$\bar{\nu} = R_H (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
where $n_1 = 1, 2, 3, \dots$ and $n_2 = n_1 + 1, n_1 + 2, \dots$.
The value of the Rydberg constant $R_H$ is approximately $109,677$ cm$^{-1}$.
Different values of $n_1$ correspond to different series:
Series | $n_1$ | $n_2$ | Spectral Region |
---|---|---|---|
Lyman | 1 | 2, 3, 4, ... | Ultraviolet |
Balmer | 2 | 3, 4, 5, ... | Visible |
Paschen | 3 | 4, 5, 6, ... | Infrared |
Bracket | 4 | 5, 6, 7, ... | Infrared |
Pfund | 5 | 6, 7, 8, ... | Infrared |
Humphreys | 6 | 7, 8, 9, ... | Infrared |
The existence of discrete lines in atomic spectra is strong evidence that electrons within atoms can only occupy specific energy levels, implying that their energy is quantised.
Bohr’s Model For Hydrogen Atom
In 1913, Niels Bohr combined Rutherford's nuclear model with Planck's quantum theory to propose a model for the hydrogen atom that could explain its stability and line spectrum quantitatively.
Bohr's model is based on the following postulates:
- Quantised Orbits: Electrons revolve around the nucleus in specific circular paths called orbits, stationary states, or allowed energy states. These orbits have fixed radii and fixed energies. They are arranged concentrically around the nucleus.
- Stationary States: Electrons in these specific orbits do not radiate energy, so their energy remains constant. Energy is only absorbed or emitted when an electron moves from one stationary state to another.
- Energy Transitions: When an electron moves from a lower energy orbit ($E_1$) to a higher energy orbit ($E_2$), energy is absorbed ($\Delta E = E_2 - E_1 > 0$). When it moves from a higher energy orbit to a lower one, energy is emitted ($\Delta E = E_1 - E_2 < 0$). The frequency ($\nu$) of the emitted or absorbed radiation is given by Bohr's frequency rule:
$\nu = \frac{|\Delta E|}{h} = \frac{|E_2 - E_1|}{h}$
This implies that energy transitions are not continuous but occur only between discrete energy levels. - Quantisation of Angular Momentum: The angular momentum of an electron in a stationary orbit is quantised. It must be an integral multiple of $\frac{h}{2\pi}$:
$m_e v r = n \frac{h}{2\pi}$, where $n = 1, 2, 3, \dots$
$m_e$ is the electron mass, $v$ is its velocity, $r$ is the orbit radius, and $n$ is the principal quantum number (an integer). This condition selects only certain allowed orbits out of many classically possible ones.
According to Bohr's theory for the hydrogen atom (derivation complex, based on balancing electrostatic attraction and centripetal force, plus the angular momentum quantisation):
- The stationary states are numbered by the principal quantum number ($n$), where $n = 1, 2, 3, \dots$. $n=1$ is the lowest energy state (ground state).
- The radii of the stationary states are given by $r_n = n^2 a_0$, where $a_0$ is the Bohr radius ($52.9$ pm for $n=1$). The radius increases as $n$ increases.
- The energy of the stationary state is given by $E_n = -\frac{R_H}{n^2}$, where $R_H = 2.18 \times 10^{-18}$ J. The energy levels are negative, indicating the electron is bound to the nucleus. The lowest energy ($n=1$) is the most negative and thus most stable. As $n \rightarrow \infty$, $E_n \rightarrow 0$, which corresponds to a free electron (ionised atom).
Bohr's theory also applies to hydrogen-like species (ions with only one electron, like $\text{He}^+, \text{Li}^{2+}$). For these, the energy and radius depend on the atomic number ($Z$):
$E_n = -R_H \frac{Z^2}{n^2}$ J/atom
$r_n = a_0 \frac{n^2}{Z}$
Increasing $Z$ makes the energy more negative (tighter binding) and the radius smaller.
Explanation Of Line Spectrum Of Hydrogen
Bohr's model explained the observed line spectrum of hydrogen. Transitions between energy levels lead to the absorption or emission of photons with specific frequencies.
The energy difference between an initial state ($n_i$) and a final state ($n_f$) is:
$\Delta E = E_{n_f} - E_{n_i} = (-\frac{R_H}{n_f^2}) - (-\frac{R_H}{n_i^2}) = R_H (\frac{1}{n_i^2} - \frac{1}{n_f^2})$
= $2.18 \times 10^{-18} \text{ J } (\frac{1}{n_i^2} - \frac{1}{n_f^2})$
For emission ($n_i > n_f$), $\Delta E$ is negative, indicating energy is released as a photon. For absorption ($n_f > n_i$), $\Delta E$ is positive, meaning energy is absorbed.
The frequency ($\nu$) of the photon is $\nu = \frac{|\Delta E|}{h}$.
The wavenumber ($\bar{\nu}$) is $\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c} = \frac{|\Delta E|}{hc} = \frac{R_H}{hc} (\frac{1}{n_i^2} - \frac{1}{n_f^2})$ (for $n_i > n_f$ in emission) or $\frac{R_H}{hc} (\frac{1}{n_f^2} - \frac{1}{n_i^2})$ (for $n_f > n_i$ in absorption).
Substituting the values of $R_H$, $h$, and $c$, the factor $\frac{R_H}{hc}$ matches the Rydberg constant for hydrogen ($\approx 1.09677 \times 10^7 \text{ m}^{-1}$ or $109677 \text{ cm}^{-1}$). This quantitatively explains the Rydberg formula and the existence of spectral series (Lyman, Balmer, etc.), where each line corresponds to a specific transition between two discrete energy levels.
The intensity of spectral lines relates to the number of photons emitted or absorbed during those specific transitions.
Problem 2.10. What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Answer:
Given transition from $n_i = 5$ to $n_f = 2$. This transition ends at $n_f = 2$, which corresponds to the Balmer series, located in the visible region of the spectrum.
The energy difference ($\Delta E$) is given by:
$\Delta E = E_{n_f} - E_{n_i} = R_H (\frac{1}{n_i^2} - \frac{1}{n_f^2})$
Using $R_H = 2.18 \times 10^{-18}$ J, $n_i = 5$, and $n_f = 2$:
$\Delta E = 2.18 \times 10^{-18} \text{ J } (\frac{1}{5^2} - \frac{1}{2^2}) = 2.18 \times 10^{-18} \text{ J } (\frac{1}{25} - \frac{1}{4})$
$\Delta E = 2.18 \times 10^{-18} \text{ J } (0.04 - 0.25) = 2.18 \times 10^{-18} \text{ J } (-0.21)$
$\Delta E = -4.578 \times 10^{-19} \text{ J}$. The negative sign indicates energy is emitted.
The energy of the emitted photon is $|\Delta E| = 4.578 \times 10^{-19}$ J.
The frequency ($\nu$) of the emitted photon is $\nu = \frac{|\Delta E|}{h}$.
Using $h = 6.626 \times 10^{-34}$ J s:
$\nu = \frac{4.578 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} = \frac{4.578}{6.626} \times 10^{(-19 - (-34))} \text{ s}^{-1} = 0.6909 \times 10^{15} \text{ Hz}$
$\nu \approx 6.909 \times 10^{14}$ Hz.
The wavelength ($\lambda$) of the emitted photon is $\lambda = \frac{c}{\nu}$.
Using $c = 3.0 \times 10^8$ m s$^{-1}$:
$\lambda = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{6.909 \times 10^{14} \text{ s}^{-1}} = \frac{3.0}{6.909} \times 10^{(8 - 14)} \text{ m} = 0.4342 \times 10^{-6} \text{ m}$.
Converting to nanometers: $0.4342 \times 10^{-6} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 434.2 \text{ nm}$.
So, the frequency is approximately $6.909 \times 10^{14}$ Hz and the wavelength is approximately 434.2 nm. This is a line in the Balmer series, in the violet part of the visible spectrum.
Problem 2.11. Calculate the energy associated with the first orbit of He+. What is the radius of this orbit?
Answer:
The species is He$^+$. This is a hydrogen-like ion (one electron). The atomic number ($Z$) for Helium is 2.
We need to calculate the energy and radius of the first orbit ($n=1$).
Using the formula for energy of hydrogen-like species: $E_n = -R_H \frac{Z^2}{n^2}$ J/atom.
Given $R_H = 2.18 \times 10^{-18}$ J, $Z=2$, and $n=1$:
$E_1 = -(2.18 \times 10^{-18} \text{ J}) \times \frac{2^2}{1^2} = -(2.18 \times 10^{-18} \text{ J}) \times 4$
$E_1 = -8.72 \times 10^{-18}$ J/atom.
The energy associated with the first orbit of He$^+$ is $-8.72 \times 10^{-18}$ J.
Using the formula for the radius of hydrogen-like species: $r_n = a_0 \frac{n^2}{Z}$.
Using $a_0 = 52.9$ pm, $n=1$, and $Z=2$:
$r_1 = 52.9 \text{ pm} \times \frac{1^2}{2} = 52.9 \text{ pm} \times \frac{1}{2} = 26.45 \text{ pm}$.
The radius of the first orbit of He$^+$ is 26.45 pm.
Limitations Of Bohr’s Model
While Bohr's model successfully explained the hydrogen atom and hydrogen-like ions, it had limitations:
- It could not explain the spectra of multi-electron atoms (atoms with more than one electron).
- It failed to account for the finer details (like the splitting into closely spaced lines, or doublets) observed in the hydrogen spectrum using high-resolution spectroscopes.
- It could not explain the Zeeman effect (splitting of spectral lines in a magnetic field) or the Stark effect (splitting in an electric field).
- It did not explain how atoms form chemical bonds to create molecules.
These limitations showed that Bohr's model, despite its success for hydrogen, was not a complete description of atomic structure and behavior, highlighting the need for a more advanced theory.
Towards Quantum Mechanical Model Of The Atom
The development of a more comprehensive atomic model was influenced by two significant ideas that emerged after Bohr's model:
- The confirmation of the dual behaviour of matter (that particles also exhibit wave-like properties).
- The formulation of the Heisenberg uncertainty principle.
Dual Behaviour Of Matter
Inspired by the wave-particle duality of light, French physicist Louis de Broglie proposed in 1924 that all matter, including particles like electrons, should also exhibit dual nature – possessing both particle-like and wave-like properties.
He suggested a relationship between the wavelength ($\lambda$) associated with a moving particle and its momentum ($p$). This is the de Broglie relation:
$\lambda = \frac{h}{p} = \frac{h}{mv}$
where $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is its velocity.
De Broglie's hypothesis was experimentally confirmed a few years later when electron beams were shown to undergo diffraction, a phenomenon characteristic of waves. This wave nature of electrons is used in devices like the electron microscope, which achieves much higher magnifications than light microscopes.
According to de Broglie, every object in motion has a wavelength. However, for macroscopic objects (like a baseball or a car), the mass ($m$) is very large, making the de Broglie wavelength ($\lambda = h/mv$) extremely small – too small to be detected or have any significant effect on their behavior. For microscopic particles like electrons, protons, or neutrons, the mass is very small, resulting in a detectable wavelength that influences their behavior significantly.
Problem 2.12. What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s$^{-1}$ ?
Answer:
Given mass ($m$) = 0.1 kg.
Given velocity ($v$) = 10 m s$^{-1}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
Using the de Broglie equation: $\lambda = \frac{h}{mv}$.
The unit for J is kg m$^2$ s$^{-2}$. So, the unit for $h$ (J s) is kg m$^2$ s$^{-1}$.
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(0.1 \text{ kg}) \times (10 \text{ m s}^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{1.0 \text{ kg m s}^{-1}}$
$\lambda = 6.626 \times 10^{-34} \text{ m}$.
The de Broglie wavelength of the ball is $6.626 \times 10^{-34}$ m. This wavelength is exceedingly small, far below the limits of any measurement technique, which is why the wave nature of macroscopic objects is not observable.
Problem 2.13. The mass of an electron is 9.1×10$^{-31}$ kg. If its K.E. is 3.0×10$^{-25}$ J, calculate its wavelength.
Answer:
Given mass ($m$) = $9.1 \times 10^{-31}$ kg.
Given kinetic energy (KE) = $3.0 \times 10^{-25}$ J.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
To use the de Broglie equation ($\lambda = h/mv$), we first need to find the velocity ($v$) from the kinetic energy.
KE = $\frac{1}{2} mv^2$
$v^2 = \frac{2 \times \text{KE}}{m} = \frac{2 \times (3.0 \times 10^{-25} \text{ J})}{9.1 \times 10^{-31} \text{ kg}}$.
Using units J = kg m$^2$ s$^{-2}$:
$v^2 = \frac{6.0 \times 10^{-25} \text{ kg m}^2 \text{ s}^{-2}}{9.1 \times 10^{-31} \text{ kg}} = \frac{6.0}{9.1} \times 10^{(-25 - (-31))} \text{ m}^2 \text{ s}^{-2}$
$v^2 \approx 0.6593 \times 10^6 \text{ m}^2 \text{ s}^{-2}$.
$v = \sqrt{0.6593 \times 10^6} \text{ m s}^{-1} = \sqrt{0.6593} \times \sqrt{10^6} \text{ m s}^{-1} \approx 0.812 \times 10^3 \text{ m s}^{-1} = 812 \text{ m s}^{-1}$.
Now, calculate the de Broglie wavelength:
$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \text{ J s}}{(9.1 \times 10^{-31} \text{ kg}) \times (812 \text{ m s}^{-1})}$.
Using units J s = kg m$^2$ s$^{-1}$:
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(9.1 \times 812) \times 10^{(-31)} \text{ kg m s}^{-1}} = \frac{6.626}{7389.2} \times 10^{(-34 - (-31))} \text{ m}$
$\lambda \approx 0.0008967 \times 10^{-3} \text{ m} = 8.967 \times 10^{-7} \text{ m}$.
Converting to nanometers: $8.967 \times 10^{-7} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}} = 896.7 \text{ nm}$.
The wavelength associated with the electron is approximately 896.7 nm, which is in the infrared region of the electromagnetic spectrum. This wavelength is large enough to be significant at the atomic scale.
Problem 2.14. Calculate the mass of a photon with wavelength 3.6 Å.
Answer:
Given wavelength ($\lambda$) = 3.6 Å.
Conversion: $1 \text{ Å} = 10^{-10} \text{ m}$. So, $\lambda = 3.6 \times 10^{-10} \text{ m}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
A photon is a particle of light and travels at the speed of light ($c = 3.0 \times 10^8$ m s$^{-1}$).
We can use the de Broglie relation, $\lambda = \frac{h}{p}$, where $p$ is the momentum of the photon.
For a photon, momentum $p = mc$, where $m$ is the relativistic mass of the photon (photons have zero rest mass). However, it's more direct to use the energy-momentum relation for photons, $E = pc$, and $E = h\nu = hc/\lambda$.
So, $pc = hc/\lambda$, which simplifies to $p = h/\lambda$.
Using $p = mc$ (relativistic mass), we have $mc = h/\lambda$.
We can calculate the mass ($m$) of the photon from this relationship:
$m = \frac{h}{c\lambda}$.
$m = \frac{6.626 \times 10^{-34} \text{ J s}}{(3.0 \times 10^8 \text{ m s}^{-1}) \times (3.6 \times 10^{-10} \text{ m})}$.
Using units J = kg m$^2$ s$^{-2}$, so J s = kg m$^2$ s$^{-1}$.
$m = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(3.0 \times 3.6) \times 10^{(8 - 10)} \text{ m}^2 \text{ s}^{-1}}$
$m = \frac{6.626 \times 10^{-34}}{10.8 \times 10^{-2}} \text{ kg} = \frac{6.626}{10.8} \times 10^{(-34 - (-2))} \text{ kg}$
$m \approx 0.6135 \times 10^{-32} \text{ kg} = 6.135 \times 10^{-33} \text{ kg}$.
The relativistic mass of a photon with wavelength 3.6 Å is approximately $6.135 \times 10^{-33}$ kg.
Heisenberg’s Uncertainty Principle
The wave-particle duality of matter has a profound consequence, described by Werner Heisenberg's Uncertainty Principle (1927):
It is impossible to determine simultaneously and precisely both the exact position and the exact momentum (or velocity) of a sub-atomic particle like an electron.
Mathematically, this is expressed as:
$\Delta x \cdot \Delta p_x \ge \frac{h}{4\pi}$
where:
- $\Delta x$ is the uncertainty in the position of the particle along the x-axis.
- $\Delta p_x$ is the uncertainty in the momentum along the x-axis ($\Delta p_x = m \Delta v_x$).
- $h$ is Planck's constant.
This principle means that the more accurately we know the position of an electron ($\Delta x$ is small), the less accurately we know its momentum or velocity ($\Delta p_x$ or $\Delta v_x$ is large), and vice versa. Our measurements are inherently limited; we get a "fuzzy" picture of the electron's state.
To illustrate, imagining trying to find an electron's position using light: The light used must have a wavelength smaller than the electron's "size" to locate it precisely. Photons of such short wavelength have high energy and momentum ($p=h/\lambda$). When a photon collides with the electron to determine its position, it transfers some of its momentum, inevitably changing the electron's velocity. So, while we might locate the electron, we lose certainty about its velocity after the observation.
Significance of Uncertainty Principle
The uncertainty principle is not noticeable for macroscopic objects. For a large object, even if there's a small uncertainty in momentum, because the mass is large, the uncertainty in velocity ($\Delta v_x = \Delta p_x / m$) becomes extremely small, and the uncertainty in position ($\Delta x$) remains effectively negligible for everyday purposes. Classical mechanics and the concept of a precise trajectory work perfectly for these objects.
However, for microscopic particles like electrons, the mass is tiny. A small uncertainty in momentum translates to a large uncertainty in velocity. Consequently, it's impossible to define a precise path or trajectory for an electron in an atom. The idea of electrons moving in fixed, well-defined orbits (like in Bohr's model) is incompatible with the uncertainty principle and the wave nature of matter.
Instead of trajectories, quantum mechanics describes the probability of finding an electron in a certain region of space around the nucleus.
Problem 2.15. A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?
Answer:
Given uncertainty in position ($\Delta x$) = 0.1 Å.
Conversion: $1 \text{ Å} = 10^{-10} \text{ m}$. So, $\Delta x = 0.1 \times 10^{-10} \text{ m} = 1 \times 10^{-11} \text{ m}$.
Mass of an electron ($m_e$) = $9.11 \times 10^{-31}$ kg.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
Heisenberg uncertainty principle: $\Delta x \cdot \Delta p_x \ge \frac{h}{4\pi}$.
Since $\Delta p_x = m_e \Delta v_x$, we have $\Delta x \cdot m_e \Delta v_x \ge \frac{h}{4\pi}$.
We want to find the minimum uncertainty in velocity, so we use the equality:
$\Delta x \cdot m_e \Delta v_x = \frac{h}{4\pi}$.
Rearranging to solve for $\Delta v_x$:
$\Delta v_x = \frac{h}{4\pi \cdot m_e \cdot \Delta x}$.
Substitute the given values:
$\Delta v_x = \frac{6.626 \times 10^{-34} \text{ J s}}{4 \times 3.14159 \times (9.11 \times 10^{-31} \text{ kg}) \times (1 \times 10^{-11} \text{ m})}$.
Using units J s = kg m$^2$ s$^{-1}$:
$\Delta v_x = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(4 \times 3.14159 \times 9.11 \times 1) \times 10^{(-31 - 11)} \text{ kg m}}}$
$\Delta v_x = \frac{6.626 \times 10^{-34}}{114.45 \times 10^{-42}} \text{ m s}^{-1}$
$\Delta v_x = \frac{6.626}{114.45} \times 10^{(-34 - (-42))} \text{ m s}^{-1} = 0.0579 \times 10^8 \text{ m s}^{-1}$
$\Delta v_x \approx 5.79 \times 10^6 \text{ m s}^{-1}$.
The uncertainty in the velocity measurement is very large, approximately $5.79 \times 10^6$ m s$^{-1}$. This demonstrates that precisely locating an electron significantly increases the uncertainty in its velocity, making a defined trajectory impossible.
Problem 2.16. A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
Answer:
Given mass ($m$) = 40 g.
Conversion: $40 \text{ g} = 40 \times 10^{-3} \text{ kg} = 0.040 \text{ kg}$.
Given speed ($v$) = 45 m/s.
Accuracy of speed measurement = 2%.
This means the uncertainty in speed ($\Delta v_x$) is 2% of the speed.
$\Delta v_x = 2\% \text{ of } 45 \text{ m/s} = \frac{2}{100} \times 45 \text{ m/s} = 0.02 \times 45 \text{ m/s} = 0.9 \text{ m/s}$.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s.
Using the Heisenberg uncertainty principle: $\Delta x \cdot \Delta p_x \ge \frac{h}{4\pi}$.
Using $\Delta p_x = m \Delta v_x$ and the equality for minimum uncertainty:
$\Delta x \cdot m \Delta v_x = \frac{h}{4\pi}$.
Rearranging to solve for $\Delta x$:
$\Delta x = \frac{h}{4\pi \cdot m \cdot \Delta v_x}$.
Substitute the given values:
$\Delta x = \frac{6.626 \times 10^{-34} \text{ J s}}{4 \times 3.14159 \times (0.040 \text{ kg}) \times (0.9 \text{ m s}^{-1})}$.
Using units J s = kg m$^2$ s$^{-1}$:
$\Delta x = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(4 \times 3.14159 \times 0.040 \times 0.9) \text{ kg m s}^{-1}}$
$\Delta x = \frac{6.626 \times 10^{-34}}{0.4524} \text{ m}$
$\Delta x \approx 14.65 \times 10^{-34} \text{ m} = 1.465 \times 10^{-33} \text{ m}$.
The uncertainty in the position of the golf ball is approximately $1.465 \times 10^{-33}$ m. This value is exceedingly small, far beyond the scale of observation even for individual atoms, demonstrating why the uncertainty principle is negligible for macroscopic objects.
The limitations of the Bohr model (ignoring wave nature and contradicting uncertainty principle) highlighted the need for a new theory that could account for the dual behaviour of matter and be consistent with the uncertainty principle. This led to the development of quantum mechanics.
Quantum Mechanical Model Of Atom
Classical mechanics, based on Newton's laws, successfully describes the motion of macroscopic objects. However, it fails when applied to microscopic particles because it doesn't account for their wave-like properties or the uncertainty principle. Quantum mechanics is the scientific framework that describes the motion and behavior of these microscopic objects, incorporating their dual nature.
Quantum mechanics was developed independently by Werner Heisenberg and Erwin Schrödinger in 1926. Schrödinger's approach, based on wave motion, is particularly relevant to understanding atomic structure. He developed the fundamental equation of quantum mechanics, the Schrödinger equation, which incorporates the wave-particle duality proposed by de Broglie.
For a system whose energy doesn't change over time (like an atom in a stable state), the time-independent Schrödinger equation is written as:
$\hat{H}\psi = E\psi$
where:
- $\hat{H}$ is the Hamiltonian operator, representing the total energy (kinetic + potential) of the system's particles.
- $\psi$ (psi) is the wave function, a mathematical function that describes the state of the system.
- $E$ is the total energy of the system.
Solving the Schrödinger equation for an atom provides the possible energy levels that electrons can occupy and the corresponding wave functions ($\psi$) for each energy level.
Hydrogen Atom and the Schrödinger Equation
Solving the Schrödinger equation for the hydrogen atom yields specific, quantised energy states and corresponding wave functions. These solutions are characterised by a set of three numbers called quantum numbers (principal $n$, azimuthal $l$, and magnetic $m_l$). The quantisation of energy levels arises naturally from the mathematics of the wave equation.
The wave function ($\psi$) itself doesn't have a direct physical meaning. However, the square of the wave function, $|\psi|^2$, at a particular point in space gives the probability density of finding the electron at that point. $|\psi|^2$ is always a positive value.
For one-electron systems like hydrogen or He$^+$, these wave functions are called atomic orbitals. Each atomic orbital corresponds to a specific energy level and describes a region in space where the probability of finding the electron is high.
For multi-electron atoms, the Schrödinger equation is more complex and cannot be solved exactly. Approximate methods are used, which show that the orbitals in multi-electron atoms are similar to hydrogen orbitals but are affected by the increased nuclear charge and electron-electron repulsions. Notably, in multi-electron atoms, the energy of an orbital depends not just on the principal quantum number ($n$) but also on the azimuthal quantum number ($l$).
Important Features of the Quantum Mechanical Model of Atom
The quantum mechanical model provides a more accurate description of atomic structure:
- Electron energy in atoms is quantised (can only have specific discrete values).
- The existence of these quantised energy levels is a direct result of the electron's wave-like properties, as described by the Schrödinger equation.
- It's impossible to determine the exact position and velocity of an electron simultaneously (Heisenberg Uncertainty Principle). Therefore, the concept of a definite electron path or trajectory is replaced by the concept of probability.
- An atomic orbital is the wave function ($\psi$) for an electron in an atom. It represents a region of space where the electron is likely to be found, not a fixed path. An atom can have many atomic orbitals, each corresponding to a different possible state for an electron. Each orbital can hold a maximum of two electrons with opposite spins. In multi-electron atoms, electrons occupy these orbitals starting from the lowest energy ones.
- $|\psi|^2$ represents the probability density of finding the electron at a point. A boundary surface enclosing a region where the probability of finding the electron is high (e.g., 90%) is often used to depict the shape of an orbital.
Orbitals And Quantum Numbers
Atomic orbitals can be distinguished by their size, shape, and orientation in space. These characteristics are defined by a set of three quantum numbers ($n, l, m_l$) obtained from the solution of the Schrödinger equation for the hydrogen atom.
- Principal Quantum Number ($n$):
- Represents the main energy level or shell the electron occupies.
- Can be any positive integer: $n = 1, 2, 3, \dots$.
- Determines the size and primarily the energy of the orbital. Higher $n$ means a larger orbital, further from the nucleus, and higher energy.
- Shells are also designated by letters: K ($n=1$), L ($n=2$), M ($n=3$), N ($n=4$), etc.
- The total number of orbitals within a shell $n$ is $n^2$.
- Azimuthal Quantum Number ($l$):
- Also called the orbital angular momentum quantum number or subsidiary quantum number.
- Defines the shape of the orbital (and thus the subshell).
- For a given $n$, $l$ can have integer values from 0 to $n-1$. So, there are $n$ possible $l$ values for a given $n$.
- Different $l$ values correspond to different subshells:
- $l=0$: s subshell (spherical shape)
- $l=1$: p subshell (dumbbell shape)
- $l=2$: d subshell (more complex shapes)
- $l=3$: f subshell (even more complex shapes)
- And so on (g, h, etc. for $l=4, 5, \dots$)
- Magnetic Orbital Quantum Number ($m_l$):
- Describes the spatial orientation of the orbital in a magnetic field.
- For a given $l$, $m_l$ can have integer values from $-l$ through 0 to $+l$.
- The number of possible $m_l$ values for a given $l$ is $2l+1$. This is the number of orbitals within a subshell:
- $l=0$ (s subshell): $m_l = 0$ (1 s orbital)
- $l=1$ (p subshell): $m_l = -1, 0, +1$ (3 p orbitals)
- $l=2$ (d subshell): $m_l = -2, -1, 0, +1, +2$ (5 d orbitals)
- $l=3$ (f subshell): $m_l = -3, -2, -1, 0, +1, +2, +3$ (7 f orbitals)
- Electron Spin Quantum Number ($m_s$):
- An intrinsic property of the electron, like mass and charge.
- Represents the electron's inherent angular momentum, often visualised as spinning on its axis.
- Can only have two possible values: $+\frac{1}{2}$ or $-\frac{1}{2}$, representing the two possible spin orientations (often denoted by arrows $\uparrow$ and $\downarrow$).
These four quantum numbers ($n, l, m_l, m_s$) completely describe the state of an electron in an atom. The Pauli Exclusion Principle states that no two electrons in the same atom can have the exact same set of all four quantum numbers. This limits the capacity of orbitals and shells.
Quantum Number | Symbol | Allowed Values | Describes |
Principal | $n$ | 1, 2, 3, ... | Shell, Size, Main energy level |
Azimuthal (Orbital angular momentum) | $l$ | 0, 1, ..., $n-1$ | Subshell, Shape |
Magnetic | $m_l$ | $-l, ..., 0, ..., +l$ | Orientation |
Spin | $m_s$ | $+1/2, -1/2$ | Electron spin orientation |
Maximum number of electrons:
- In an orbital: 2 (with opposite spins).
- In a subshell $l$: $2 \times (2l+1)$.
- In a shell $n$: $2 \times n^2$.
For example, a shell with $n=2$ has $l=0$ and $l=1$. $l=0$ has $2(0)+1=1$ orbital (2s), holding 2 electrons. $l=1$ has $2(1)+1=3$ orbitals (2p), holding $3 \times 2 = 6$ electrons. Total electrons in $n=2$ shell = $2+6=8$, which is $2 \times 2^2 = 8$.
Orbit vs. Orbital: It's crucial to distinguish between Bohr's concept of a circular orbit (a definite path) and the quantum mechanical concept of an orbital (a region of probability). Due to the uncertainty principle, a definite trajectory for an electron cannot exist. An atomic orbital is a mathematical description ($\psi$) related to the probability of finding an electron in a certain volume of space around the nucleus.
Problem 2.17. What is the total number of orbitals associated with the principal quantum number n = 3 ?
Answer:
Given principal quantum number $n=3$.
For a given $n$, the possible values of the azimuthal quantum number $l$ are $0, 1, \dots, n-1$.
For $n=3$, the possible values of $l$ are 0, 1, and 2.
Now, find the number of orbitals for each $l$ value using $2l+1$:
- When $l=0$ (s subshell): Number of orbitals = $2(0)+1 = 1$. This is the 3s orbital.
- When $l=1$ (p subshell): Number of orbitals = $2(1)+1 = 3$. These are the three 3p orbitals.
- When $l=2$ (d subshell): Number of orbitals = $2(2)+1 = 5$. These are the five 3d orbitals.
The total number of orbitals associated with $n=3$ is the sum of orbitals in each subshell:
Total orbitals = (Number of 3s orbitals) + (Number of 3p orbitals) + (Number of 3d orbitals) = 1 + 3 + 5 = 9.
Alternatively, the total number of orbitals in a shell with principal quantum number $n$ is given by $n^2$.
For $n=3$, total orbitals = $3^2 = 9$.
So, there are 9 orbitals associated with the principal quantum number $n=3$.
Problem 2.18. Using s, p, d, f notations, describe the orbital with the following quantum numbers (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5, l = 3, (d) n = 3, l = 2.
Answer:
The notation for an orbital is generally written as (principal quantum number)(subshell notation). The subshell notation (s, p, d, f, ...) corresponds to the azimuthal quantum number ($l$).
- $l=0$ corresponds to s orbitals.
- $l=1$ corresponds to p orbitals.
- $l=2$ corresponds to d orbitals.
- $l=3$ corresponds to f orbitals.
Now, let's describe the orbitals based on the given quantum numbers:
(a) $n=2, l=1$: $n=2$ indicates the second shell. $l=1$ corresponds to a p orbital.
The orbital is a 2p orbital.
(b) $n=4, l=0$: $n=4$ indicates the fourth shell. $l=0$ corresponds to an s orbital.
The orbital is a 4s orbital.
(c) $n=5, l=3$: $n=5$ indicates the fifth shell. $l=3$ corresponds to an f orbital.
The orbital is a 5f orbital.
(d) $n=3, l=2$: $n=3$ indicates the third shell. $l=2$ corresponds to a d orbital.
The orbital is a 3d orbital.
Shapes Of Atomic Orbitals
While the wave function $\psi$ itself has no physical meaning, the probability density $|\psi|^2$ is important. Plots of $|\psi|^2$ show the regions where an electron is most likely to be found.
For 1s and 2s orbitals, the probability density varies with the distance from the nucleus. For 1s, it's highest at the nucleus and decreases outwards. For 2s, it decreases to zero, increases to a small peak, and then decreases again. The region where the probability density is zero is called a node or nodal surface. For ns orbitals, there are $(n-1)$ radial nodes.
To visualize orbital shapes, boundary surface diagrams are used. These surfaces enclose a region where the probability of finding the electron is high (e.g., 90%).
s Orbitals (l=0):
- s orbitals are spherically symmetric. The probability of finding the electron depends only on the distance from the nucleus, not the direction.
- The boundary surface diagram of an s orbital is a sphere centered on the nucleus.
- The size of s orbitals increases with the principal quantum number $n$ (e.g., 4s > 3s > 2s > 1s).
p Orbitals (l=1):
- p orbitals are not spherically symmetric. They have a dumbbell shape consisting of two lobes on either side of a plane passing through the nucleus.
- The plane where the two lobes meet and the probability density is zero is a nodal plane.
- For $l=1$, there are $2l+1 = 3$ p orbitals, oriented along the x, y, and z axes, designated as $p_x$, $p_y$, and $p_z$. They are identical in shape, size, and energy but differ in orientation.
- The size and energy of p orbitals increase with increasing $n$ (e.g., 4p > 3p > 2p).
- Number of radial nodes for np orbitals = $n-2$. Number of angular nodes = $l = 1$. Total nodes = $(n-1)$.
d Orbitals (l=2):
- d orbitals exist starting from $n=3$ (since $l \le n-1$).
- For $l=2$, there are $2l+1 = 5$ d orbitals.
- Four of the d orbitals ($d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}$) have similar shapes, typically depicted as four lobes. The fifth orbital ($d_{z^2}$) has a distinct shape (two lobes along the z-axis and a torus around the center).
- In an isolated atom, the five d orbitals within the same subshell are equivalent in energy (degenerate).
- The size and energy of d orbitals increase with increasing $n$.
- Number of radial nodes for nd orbitals = $n-3$. Number of angular nodes = $l = 2$. Total nodes = $(n-1)$.
In general, the total number of nodes in an orbital is $(n-1)$, which is the sum of radial nodes $(n-l-1)$ and angular nodes ($l$).
Energies Of Orbitals
The energy of an electron in an orbital depends on the attraction between the electron and the nucleus and the repulsion between the electron and other electrons.
- Hydrogen Atom (One-electron species): The energy of an orbital depends ONLY on the principal quantum number ($n$). For a given $n$, all orbitals ($s, p, d, f, \dots$) have the same energy; they are degenerate. Energy increases as $1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < \dots$.
- Multi-electron Atoms: The energy of an orbital depends on BOTH $n$ and $l$. This is due to electron-electron repulsions and the shielding effect. Inner electrons shield the outer electrons from the full positive charge of the nucleus. The net positive charge experienced by an outer electron is the effective nuclear charge ($Z_{eff}$), which is less than the actual nuclear charge ($Z$).
Different subshells within the same shell ($n$) experience different amounts of shielding. Electrons in s orbitals penetrate closer to the nucleus than p, d, or f electrons in the same shell. This means s electrons experience a higher $Z_{eff}$ and are more strongly attracted, having lower energy than p electrons, which have lower energy than d electrons, and so on.
Within a given shell, the order of increasing energy is typically:
$s < p < d < f$
For higher principal quantum numbers, energy levels of different shells can overlap or stagger (e.g., the 4s orbital has lower energy than the 3d orbitals in many atoms).
A useful rule to estimate the relative energies of orbitals in multi-electron atoms is the $(n+l)$ rule:
The orbital with the lower value of $(n+l)$ has lower energy. If two orbitals have the same $(n+l)$ value, the one with the lower principal quantum number ($n$) has lower energy.
Orbital | $n$ | $l$ | $n+l$ | Order of Increasing Energy |
---|---|---|---|---|
1s | 1 | 0 | 1 | 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d ... |
2s | 2 | 0 | 2 | |
2p | 2 | 1 | 3 | |
3s | 3 | 0 | 3 | |
3p | 3 | 1 | 4 | |
4s | 4 | 0 | 4 | |
3d | 3 | 2 | 5 | |
4p | 4 | 1 | 5 | |
5s | 5 | 0 | 5 | |
4d | 4 | 2 | 6 | |
5p | 5 | 1 | 6 |
(Note: When $n+l$ is the same, e.g., 2p ($n=2, l=1, n+l=3$) and 3s ($n=3, l=0, n+l=3$), the orbital with lower $n$ (2p) has lower energy than the one with higher $n$ (3s)).
Energy of orbitals within the same subshell decreases with increasing atomic number ($Z$), as the nuclear attraction increases (though $Z_{eff}$ is affected by shielding).
Filling Of Orbitals In Atom
The distribution of electrons among the orbitals of an atom is called its electronic configuration. This filling process follows three fundamental rules:
- Aufbau Principle: Electrons fill orbitals in the order of increasing energy. Lower energy orbitals are filled before higher energy ones. The approximate order of increasing energy (and thus filling order) is: $1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, \dots$ This order can be remembered using a diagram.
- Pauli Exclusion Principle: No two electrons in the same atom can have the identical set of all four quantum numbers ($n, l, m_l, m_s$). This implies that an atomic orbital can hold a maximum of two electrons, and these two electrons must have opposite spins (one with $m_s = +1/2$, the other with $m_s = -1/2$).
- Hund’s Rule of Maximum Multiplicity: For orbitals within the same subshell (which are degenerate, meaning they have the same energy, e.g., the three 2p orbitals), electron pairing does not occur until each orbital in the subshell is singly occupied. When orbitals are singly occupied, the electrons in them have parallel spins (all $+\frac{1}{2}$ or all $-\frac{1}{2}$) as far as possible. This arrangement minimises electron-electron repulsion and results in greater stability.
Applying these rules allows us to determine the electronic configuration of any atom.
Electronic Configuration Of Atoms
Electronic configuration can be written in two ways:
- $nl^x$ notation: The principal quantum number ($n$) and subshell letter ($l$) are followed by a superscript ($x$) indicating the number of electrons in that subshell (e.g., $1s^2$).
- Orbital diagram: Each orbital is represented by a box or a line. Electrons are represented by arrows ($\uparrow$ for spin $+1/2$, $\downarrow$ for spin $-1/2$). Paired electrons in an orbital are shown with opposite arrows ($\uparrow\downarrow$). Electrons in different orbitals of the same subshell are shown with parallel arrows ($\uparrow \uparrow \uparrow$) according to Hund's rule before pairing occurs ($\uparrow\downarrow \uparrow \uparrow$).
Examples:
- Hydrogen (Z=1): 1 electron. Lowest energy orbital is 1s. Configuration: $1s^1$. Orbital diagram: $\boxed{\uparrow}_{1s}$
- Helium (Z=2): 2 electrons. Both can go in 1s (Pauli principle allows opposite spins). Configuration: $1s^2$. Orbital diagram: $\boxed{\uparrow\downarrow}_{1s}$
- Lithium (Z=3): 3 electrons. $1s$ is full. Next lowest is 2s. Configuration: $1s^2 2s^1$. Orbital diagram: $\boxed{\uparrow\downarrow}_{1s} \boxed{\uparrow}_{2s}$
- Carbon (Z=6): 6 electrons. $1s^2 2s^2$. Remaining 2 electrons go in 2p. 2p has 3 orbitals. According to Hund's rule, they go into separate orbitals with parallel spins. Configuration: $1s^2 2s^2 2p^2$. Orbital diagram: $\boxed{\uparrow\downarrow}_{1s} \boxed{\uparrow\downarrow}_{2s} \boxed{\uparrow}_{2p_x} \boxed{\uparrow}_{2p_y} \boxed{\phantom{\uparrow}}_{2p_z}$
Electrons in the outermost shell (highest $n$) are called valence electrons, as they are involved in chemical bonding. Electrons in inner, filled shells are called core electrons. Electronic configurations can be abbreviated by using the noble gas configuration for the core electrons (e.g., [He] represents $1s^2$, [Ne] represents $1s^2 2s^2 2p^6$).
Filling continues following the energy order: $1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow \dots$. The 4s orbital is filled before 3d because it has lower energy according to the $(n+l)$ rule ($4+0=4$ for 4s, $3+2=5$ for 3d). This explains why Potassium ([Ar]$4s^1$) and Calcium ([Ar]$4s^2$) fill the 4s orbital.
Then, the 3d orbitals begin filling for the transition metals (Sc to Zn). Exceptions occur where small energy differences and the stability of half-filled or fully-filled subshells lead to slightly different configurations (e.g., Chromium: [Ar]$3d^5 4s^1$ instead of [Ar]$3d^4 4s^2$; Copper: [Ar]$3d^{10} 4s^1$ instead of [Ar]$3d^9 4s^2$).
Filling proceeds through the p orbitals (Ga to Kr), then the 5s, 4d, 5p, 6s, 4f (Lanthanides), 5d, 6p, 7s, 5f (Actinides), 6d, and 7p.
Understanding electronic configuration is fundamental in chemistry because it explains the chemical behavior and properties of elements, such as why some elements are reactive while others are inert, and how atoms form bonds.
See Table 2.6 for electronic configurations of elements (as provided in the original text).
Stability Of Completely Filled And Half Filled Subshells
Electronic configurations with completely filled or exactly half-filled subshells ($p^3, p^6, d^5, d^{10}, f^7, f^{14}$) tend to have extra stability compared to configurations with one electron more or less than half-filled/fully-filled. This extra stability is primarily due to two factors:
- Symmetrical Distribution of Electrons: Configurations like $s^2, p^3, p^6, d^5, d^{10}, f^7, f^{14}$ have electrons symmetrically distributed within the degenerate orbitals of the subshell. This symmetry contributes to stability. Electrons in the same subshell have similar energies and spatial distributions. Their shielding of each other is relatively effective, but compared to unevenly filled subshells, the electron-electron repulsion is somewhat minimized, leading to slightly stronger attraction to the nucleus on average.
- Exchange Energy: This stabilizing effect arises when electrons with the same spin can exchange positions within a set of degenerate orbitals. The energy released during these exchanges is called exchange energy. The more possible exchanges, the greater the exchange energy and the higher the stability. The number of possible exchanges is maximised when the subshell is exactly half-filled or completely filled.
For a $d^5$ configuration (5 electrons in 5 degenerate d orbitals, each with parallel spin), each electron can exchange positions with any of the other 4 electrons with the same spin. The number of such exchanges for electrons with the same spin is given by $\frac{n_e (n_e - 1)}{2}$, where $n_e$ is the number of electrons with parallel spins in the degenerate set. For $d^5$, if all 5 have parallel spin, exchanges = $\frac{5(5-1)}{2} = 10$. For a $d^{10}$ configuration, there are 5 spin-up and 5 spin-down electrons. Exchanges for spin-up = $\frac{5(5-1)}{2} = 10$. Exchanges for spin-down = $\frac{5(5-1)}{2} = 10$. Total exchanges = 20. For a $d^9$ configuration (say, 5 up, 4 down), exchanges = $\frac{5(4)}{2} + \frac{4(3)}{2} = 10 + 6 = 16$, less than $d^{10}$.
Hund's rule itself is a consequence of this exchange energy; placing electrons in separate degenerate orbitals with parallel spins maximises the number of same-spin electrons, thus maximising stabilizing exchange energy.
In summary, the extra stability of half-filled and completely filled subshells is attributed to their symmetrical electron distribution, relatively lower electron-electron repulsion, and maximum exchange energy.
Exercises
Question 2.1 (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
Question 2.2 (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $^{14}C$. (Assume that mass of a neutron = $1.675 \times 10^{–27}$ kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of $NH_3$ at STP.
Will the answer change if the temperature and pressure are changed ?
Answer:
Question 2.3 How many neutrons and protons are there in the following nuclei ?
$^{13}_6C, \ ^{16}_8O, \ ^{24}_{12}Mg, \ ^{56}_{26}Fe, \ ^{88}_{38}Sr$
Answer:
Question 2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.
Answer:
Question 2.5 Yellow light emitted from a sodium lamp has a wavelength ($\lambda$) of 580 nm. Calculate the frequency ($\nu$) and wavenumber ($\bar{\nu}$) of the yellow light.
Answer:
Question 2.6 Find energy of each of the photons which
(i) correspond to light of frequency $3 \times 10^{15}$ Hz.
(ii) have wavelength of 0.50 Å.
Answer:
Question 2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0 \times 10^{–10}$ s.
Answer:
Question 2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
Answer:
Question 2.9 A photon of wavelength $4 \times 10^{–7}$ m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron ($1 \ eV= 1.6020 \times 10^{–19} J$).
Answer:
Question 2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ $mol^{–1}$.
Answer:
Question 2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57 \mu m$. Calculate the rate of emission of quanta per second.
Answer:
Question 2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ($\nu_0$) and work function ($W_0$) of the metal.
Answer:
Question 2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
Answer:
Question 2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom ( energy required to remove the electron from n =1 orbit).
Answer:
Question 2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer:
Question 2.16 (i) The energy associated with the first orbit in the hydrogen atom is $–2.18 \times 10^{–18} \ J \ atom^{–1}$. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
Question 2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
Question 2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $–2.18 \times 10^{–11}$ ergs.
Answer:
Question 2.19 The electron energy in hydrogen atom is given by $E_n = (–2.18 \times 10^{–18} )/n^2 \ J$. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:
Question 2.20 Calculate the wavelength of an electron moving with a velocity of $2.05 \times 10^7 \ m \ s^{–1}$.
Answer:
Question 2.21 The mass of an electron is $9.1 \times 10^{–31}$ kg. If its K.E. is $3.0 \times 10^{–25}$ J, calculate its wavelength.
Answer:
Question 2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?
$Na^+, K^+, Mg^{2+}, Ca^{2+}, S^{2–}, Ar$.
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Question 2.23 (i) Write the electronic configurations of the following ions: (a) $H^–$ (b) $Na^+$ (c) $O^{2–}$ (d) $F^–$
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) $3s^1$ (b) $2p^3$ and (c) $3p^5$ ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He] $2s^1$ (b) [Ne] $3s^2 3p^3$ (c) [Ar] $4s^2 3d^1$.
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Question 2.24 What is the lowest value of n that allows g orbitals to exist?
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Question 2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and $m_l$ for this electron.
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Question 2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
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Question 2.27 Give the number of electrons in the species $H_2^+, H_2$ and $O_2^+$.
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Question 2.28 (i) An atomic orbital has n = 3. What are the possible values of l and $m_l$ ?
(ii) List the quantum numbers ($m_l$ and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
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Question 2.29 Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.
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Question 2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, $m_l$ = 0, $m_s$ = + ½
(b) n = 1, l = 0, $m_l$ = 0, $m_s$ = – ½
(c) n = 1, l = 1, $m_l$ = 0, $m_s$ = + ½
(d) n = 2, l = 1, $m_l$ = 0, $m_s$ = – ½
(e) n = 3, l = 3, $m_l$ = –3, $m_s$ = + ½
(f) n = 3, l = 1, $m_l$ = 0, $m_s$ = + ½
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Question 2.31 How many electrons in an atom may have the following quantum numbers?
(a) n = 4, $m_s$ = – ½ (b) n = 3, l = 0
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Question 2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
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Question 2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of $He^+$ spectrum ?
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Question 2.34 Calculate the energy required for the process
$He^+ (g) \rightarrow He^{2+} (g) + e^–$
The ionization energy for the H atom in the ground state is $2.18 \times 10^{–18} \ J \ atom^{–1}$
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Question 2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
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Question 2.36 $2 \times 10^8$ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
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Question 2.37 The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
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Question 2.38 A certain particle carries $2.5 \times 10^{–16}C$ of static electric charge. Calculate the number of electrons present in it.
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Question 2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $–1.282 \times 10^{–18}C$, calculate the number of electrons present on it.
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Question 2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the $\alpha$-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
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Question 2.41 Symbols $^{79}_{35}Br$ and $^{79}Br$ can be written, whereas symbols $^{35}_{79}Br$ and $^{35}Br$ are not acceptable. Answer briefly.
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Question 2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
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Question 2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion conatins 11.1% more neutrons than the electrons, find the symbol of the ion.
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Question 2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
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Question 2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
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Question 2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is $5.6 \times 10^{24}$, calculate the power of this laser.
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Question 2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
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Question 2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{–18}$ J from the radiations of 600 nm, calculate the number of photons received by the detector.
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Question 2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is $2.5 \times 10^{15}$, calculate the energy of the source.
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Question 2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.
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Question 2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
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Question 2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
$\lambda$ (nm) | $v \times 10^{–5} (cm \ s^{–1})$ |
---|---|
500 | 2.55 |
450 | 4.35 |
400 | 5.35 |
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Question 2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
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Question 2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of $1.5 \times 10^7 \ m \ s^{–1}$, calculate the energy with which it is bound to the nucleus.
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Question 2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as $v = 3.29 \times 10^{15} (Hz) [ 1/3^2 – 1/n^2]$
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
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Question 2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
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Question 2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is $1.6 \times 10^6 \ ms^{–1}$, calculate de Broglie wavelength associated with this electron.
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Question 2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
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Question 2.59 If the velocity of the electron in Bohr’s first orbit is $2.19 \times 10^6 \ ms^{–1}$, calculate the de Broglie wavelength associated with it.
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Question 2.60 The velocity associated with a proton moving in a potential difference of 1000 V is $4.37 \times 10^5 \ ms^{–1}$. If the hockey ball of mass 0.1 kg is moving with this velocity, calcualte the wavelength associated with this velocity.
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Question 2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $h/4\pi m \times 0.05$ nm, is there any problem in defining this value.
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Question 2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, $m_l$ = –2 , $m_s$ = –1/2
2. n = 3, l = 2, $m_l$ = 1 , $m_s$ = +1/2
3. n = 4, l = 1, $m_l$ = 0 , $m_s$ = +1/2
4. n = 3, l = 2, $m_l$ = –2 , $m_s$ = –1/2
5. n = 3, l = 1, $m_l$ = –1 , $m_s$ = +1/2
6. n = 4, l = 1, $m_l$ = 0 , $m_s$ = +1/2
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Question 2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ?
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Question 2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
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Question 2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
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Question 2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
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Question 2.67 (a) How many subshells are associated with n = 4 ? (b) How many electrons will be present in the subshells having $m_s$ value of –1/2 for n = 4 ?
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