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Chapter 2 Structure Of Atom
Discovery Of Sub-Atomic Particles
Ancient philosophers from India and Greece speculated that matter was composed of fundamental, indivisible units called atoms (from the Greek 'a-tomio' meaning 'uncuttable'). These ideas were theoretical until scientific investigations in the late 19th and early 20th centuries revealed that atoms are, in fact, divisible and made up of even smaller particles.
Discovery Of Electron
The first evidence for sub-atomic particles came from experiments involving electrical discharge through gases at very low pressures and high voltages. Michael Faraday's work in the 1830s on electrolysis suggested a particulate nature of electricity.
In the mid-1850s, scientists like Faraday studied electrical discharge in cathode ray discharge tubes. These are sealed glass tubes with electrodes (cathode and anode) connected to a high voltage source. When high voltage is applied at low pressure, a stream of particles flows from the negative electrode (cathode) to the positive electrode (anode). These were named cathode rays.
Using a perforated anode and a fluorescent screen coated with zinc sulfide behind it, scientists observed that cathode rays caused the screen to glow, confirming their path.
Properties of Cathode Rays:
- They originate from the cathode and travel towards the anode.
- They are not visible but cause fluorescent/phosphorescent materials to glow. (Like in old TV screens).
- In the absence of external electric or magnetic fields, they travel in straight lines.
- In the presence of electric or magnetic fields, they are deflected. The direction of deflection corresponds to the behavior of negatively charged particles.
- The characteristics of cathode rays are independent of the electrode material and the type of gas in the tube.
This led to the conclusion that cathode rays consist of fundamental, negatively charged particles, present in all atoms, which were later named electrons.
Charge To Mass Ratio Of Electron
In 1897, J.J. Thomson precisely measured the ratio of the electron's electrical charge ($e$) to its mass ($m_e$), denoted as $e/m_e$. He used a cathode ray tube and applied electric and magnetic fields perpendicular to each other and the electron beam.
By balancing the forces exerted by the electric and magnetic fields, Thomson could determine how much the electron beam deflected. He reasoned that the deflection amount depends on:
- The magnitude of the charge ($e$): Higher charge means greater interaction with fields and more deflection.
- The mass ($m_e$): Lighter particles deflect more easily.
- The strength of the fields: Stronger fields cause greater deflection.
Through accurate measurements, Thomson determined the $e/m_e$ value:
$$ \frac{e}{m_e} = 1.758820 \times 10^{11} \text{ C kg}^{-1} $$Since electrons carry a negative charge, the charge on an electron is actually $-e$.
Charge On The Electron
Between 1906 and 1914, R.A. Millikan conducted his famous oil drop experiment to determine the absolute charge on an electron.
Tiny oil droplets acquired electrical charge (often by colliding with ions produced by X-rays). These charged droplets were introduced into an electric field between two plates. By observing the motion of droplets under the influence of gravity and the electric field, Millikan measured the charges on many droplets. He found that the charge on any droplet ($q$) was always a whole-number multiple of a fundamental unit of charge ($e$), i.e., $q = n \times e$, where $n$ is an integer (1, 2, 3...).
Millikan determined the magnitude of the elementary charge ($e$) to be approximately $1.6 \times 10^{-19}$ C. The currently accepted value is $-1.602176 \times 10^{-19}$ C for a single electron.
Combining Millikan's value for $e$ with Thomson's $e/m_e$ ratio allowed calculation of the electron's mass ($m_e$):
$$ m_e = \frac{e}{(e/m_e)} = \frac{1.602176 \times 10^{-19} \text{ C}}{1.758820 \times 10^{11} \text{ C kg}^{-1}} \approx 9.1094 \times 10^{-31} \text{ kg} $$Discovery Of Protons And Neutrons
Experiments with modified cathode ray tubes containing a perforated cathode showed rays traveling in the opposite direction of cathode rays. These rays passed through the holes in the cathode and were called canal rays. These rays were found to consist of positively charged particles.
Properties of Canal Rays / Positive Particles:
- They originate from the anode side and travel towards the cathode.
- They are positively charged.
- Unlike cathode rays, their mass depends on the nature of the gas in the tube. They are essentially positively charged gaseous ions (formed when electrons are removed from gas atoms).
- Their charge to mass ratio ($e/m$) varies depending on the gas.
The smallest and lightest positively charged particle was obtained when hydrogen gas was used. This particle, carrying a positive charge equal in magnitude to the electron's negative charge, was named the proton. It was characterized in 1919.
Further studies indicated that the mass of an atom was generally more than the combined mass of its protons and electrons. This suggested the existence of another particle in the nucleus. In 1932, James Chadwick discovered this neutral particle by bombarding a thin sheet of beryllium with alpha particles. Electrically neutral particles with a mass slightly greater than protons were emitted. These were called neutrons.
Summary of Properties of Fundamental Particles:
| Name | Symbol | Absolute Charge (C) | Relative Charge | Mass (kg) | Approx. Mass (u) |
|---|---|---|---|---|---|
| Electron | e | $-1.602176 \times 10^{-19}$ | $-1$ | $9.109 \times 10^{-31}$ | $0.00054$ |
| Proton | p | $+1.602176 \times 10^{-19}$ | $+1$ | $1.6726 \times 10^{-27}$ | $1.00727$ |
| Neutron | n | $0$ | $0$ | $1.6749 \times 10^{-27}$ | $1.00867$ |
(Note: 1 u (unified atomic mass unit) $\approx 1.66056 \times 10^{-27}$ kg)
Atomic Models
With the discovery of sub-atomic particles, the view of the atom changed from Dalton's indivisible particle to a composite structure. Scientists then faced the challenge of proposing models to explain how these particles are arranged within an atom, accounting for its stability and properties.
Thomson Model Of Atom
In 1898, J.J. Thomson proposed the first model of the atom after the discovery of the electron. Known as the plum pudding model, it described the atom as:
- A sphere of uniform positive charge (like the pudding).
- With electrons embedded within it (like plums or raisins) in an arrangement that provides electrostatic stability.
Key features of this model:
- The atom has a spherical shape with a radius of about $10^{-10}$ m.
- The positive charge is spread uniformly throughout the sphere.
- The mass of the atom is considered to be uniformly distributed over the atom.
- The atom is overall electrically neutral, as the total positive charge equals the total negative charge from the electrons.
While this model explained the neutrality of the atom, it was later proven inconsistent with experimental results, particularly Rutherford's scattering experiment.
Rutherford’s Nuclear Model Of Atom
Ernest Rutherford, along with his students Hans Geiger and Ernest Marsden, conducted the famous alpha ($\alpha$) particle scattering experiment around 1909. They bombarded a very thin foil of gold with high-energy $\alpha$-particles (which are helium nuclei, carrying a +2 charge and significant mass).
Experimental Setup:
- A source of $\alpha$-particles (e.g., a radioactive element like polonium).
- A collimated beam of $\alpha$-particles directed at a thin gold foil (about 100 nm thick).
- A circular fluorescent screen coated with zinc sulfide (ZnS) placed around the gold foil to detect scattered $\alpha$-particles (they cause flashes of light upon hitting the screen).
Expected Results (based on Thomson's model): Since Thomson's model suggested uniform mass and charge distribution, it was expected that $\alpha$-particles, being energetic, would pass straight through the foil with minimal deflection or slowing down.
Observed Results:
- Most ($\approx 99.9\%$) of the $\alpha$-particles passed straight through the gold foil undeflected.
- A small fraction of the $\alpha$-particles was deflected by small angles.
- A very few $\alpha$-particles (about 1 in 20,000) were deflected by large angles, some even appearing to bounce back (deflection near 180°).
Conclusions based on Observations:
- Since most particles passed through undeflected, most of the space in the atom must be empty.
- The small number of large deflections and the few bounced-back particles indicated that $\alpha$-particles encountered a massive, positively charged region in the atom. The repulsion between the positive $\alpha$-particles and this positive region caused the deflections. This positive charge is not spread uniformly (contradicting Thomson) but concentrated in a very small volume.
- Rutherford calculated that the volume occupied by this central positive region is extremely small compared to the total volume of the atom. (Atomic radius $\approx 10^{-10}$ m, Nuclear radius $\approx 10^{-15}$ m. If an atom were the size of a cricket field, the nucleus would be like a cricket ball at the center).
Based on these conclusions, Rutherford proposed the Nuclear Model of the Atom (also called the Planetary Model):
- The atom consists of a tiny, dense, positively charged center called the nucleus, which contains almost all the mass of the atom.
- The electrons, which are negatively charged, revolve around the nucleus in well-defined circular paths called orbits.
- The electrons and the nucleus are held together by strong electrostatic forces of attraction.
This model is analogous to the solar system, with the nucleus as the sun and electrons as revolving planets.
Atomic Number And Mass Number
Following Rutherford's model, it was understood that the positive charge of the nucleus is due to protons. The number of protons is a fundamental property that defines an element.
- The Atomic Number (Z) of an element is equal to the number of protons in the nucleus of an atom.
- In a neutral atom, the number of electrons is equal to the number of protons (Atomic Number, Z) to maintain electrical neutrality.
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 an atom comes primarily from the protons and neutrons in the nucleus. Protons and neutrons are collectively called nucleons.
- The Mass Number (A) of an atom is the total number of protons and neutrons in its nucleus.
Mass Number (A) = Number of protons (Z) + Number of neutrons ($n$)
An atom can be represented by the symbol $^A_Z X$, where X is the element symbol, A is the mass number (superscript left), and Z is the atomic number (subscript left).
Isobars And Isotopes
Atoms can differ in their number of protons, neutrons, or both.
- Isobars are atoms of different elements that have the same mass number (A) but different atomic numbers (Z). They have a different number of protons and neutrons.
Example: $^{14}_6$C and $^{14}_7$N. Both have A=14, but C has Z=6 (6 protons) and N has Z=7 (7 protons).
- Isotopes are atoms of the same element that have the same atomic number (Z) but different mass numbers (A). They have the same number of protons but a different number of neutrons.
Example: Hydrogen has three common isotopes:
- Protium ($^1_1$H): Z=1, A=1 (1 proton, 0 neutrons). Most common (99.985%).
- Deuterium ($^2_1$H or D): Z=1, A=2 (1 proton, 1 neutron). Less common (0.015%).
- Tritium ($^3_1$H or T): Z=1, A=3 (1 proton, 2 neutrons). Radioactive, trace amounts.
Other examples: Carbon ($^{12}_6$C, $^{13}_6$C, $^{14}_6$C), Chlorine ($^{35}_{17}$Cl, $^{37}_{17}$Cl).
Since the chemical properties of an element are primarily determined by the number of electrons (which equals the number of protons in a neutral atom), isotopes of an element exhibit very similar chemical behavior.
Example 1. Calculate the number of protons, neutrons and electrons in $^{80}_{35}\text{Br}$.
Answer:
The given species is $^{80}_{35}\text{Br}$.
Atomic number, Z = 35 (subscript).
Mass number, A = 80 (superscript).
Since it is a neutral atom, Number of electrons = Number of protons = Z = 35.
Number of neutrons = Mass number (A) - Atomic number (Z) = $80 - 35 = 45$.
So, in $^{80}_{35}\text{Br}$, there are 35 protons, 35 electrons, and 45 neutrons.
Example 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:
Number of protons = 16. This determines the atomic number (Z). Z = 16. The element with atomic number 16 is Sulfur (S).
Number of electrons = 18.
Number of neutrons = 16.
The species is not neutral because the number of electrons (18) is not equal to the number of protons (16). Since there are more electrons than protons, it is a negatively charged ion (anion). The charge is the number of excess electrons = $18 - 16 = 2$. So the charge is 2-.
Mass number (A) = Number of protons + Number of neutrons = $16 + 16 = 32$.
The symbol for the species is $^A_Z \text{X}^{\text{charge}}$.
Symbol = $^{32}_{16}\text{S}^{2-}$
Drawbacks Of Rutherford Model
Despite being a significant improvement, Rutherford's nuclear model had limitations:
- Instability of the Atom: According to classical electromagnetic theory (Maxwell's theory), an electron moving in a circular orbit is an accelerated charged particle. An accelerated charged particle should continuously emit electromagnetic radiation and lose energy. As it loses energy, the electron's orbit should spiral inwards towards the nucleus, eventually collapsing into it. Calculations showed this should happen in about $10^{-8}$ seconds. However, atoms are observed to be stable. Rutherford's model could not explain this stability.
- Electron Distribution: The model did not specify how electrons are distributed around the nucleus or what their energies are. It treated electrons simply as particles orbiting the nucleus, without restrictions on their energy or location.
These fundamental issues indicated that classical physics was insufficient to describe the behavior of electrons within atoms and a new approach was needed.
Developments Leading To The Bohr’s Model Of Atom
The shortcomings of Rutherford's model and the increasing experimental data on the interaction of radiation with matter paved the way for a new atomic model. Two major developments were crucial in the formulation of Bohr's model:
- The dual nature of electromagnetic radiation (possessing both wave and particle properties).
- Experimental results from atomic spectra.
Wave Nature Of Electromagnetic Radiation
Studies of radiation emitted and absorbed by objects (thermal radiation) led to the understanding of electromagnetic waves. James Clerk Maxwell (around 1870) developed the theory that accelerated charged particles produce oscillating electric and magnetic fields that are transmitted as electromagnetic waves or electromagnetic radiation.
Key Properties of Electromagnetic Waves:
- They consist of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
- They do not require a medium to travel and can propagate through vacuum (unlike sound waves).
- There are many types of electromagnetic radiations, differing in wavelength and frequency, forming the electromagnetic spectrum (radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays). Visible light is just a small portion of this spectrum.
- In a vacuum, all types of electromagnetic radiation travel at the same speed, known as the speed of light ($c$), which is approximately $3.0 \times 10^8$ m/s.
Electromagnetic waves are characterized by:
- Wavelength ($\lambda$): The distance between two successive crests or troughs of a wave. SI unit is meter (m), but smaller units like nanometer (nm) or Ångstrom (Å) are often used ($1 \text{ nm} = 10^{-9} \text{ m}$, $1 \text{ Å} = 10^{-10} \text{ m}$).
- Frequency ($\nu$): The number of waves that pass a given point in one second. SI unit is hertz (Hz) or s⁻¹.
The relationship between speed, frequency, and wavelength is:
$$ c = \nu \lambda $$Another related quantity, especially in spectroscopy, is the wavenumber ($\bar{\nu}$), defined as the number of wavelengths per unit length. Its unit is the reciprocal of wavelength, typically cm⁻¹ or m⁻¹.
$$ \bar{\nu} = \frac{1}{\lambda} $$Example 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:
Frequency ($\nu$) = 1368 kHz = $1368 \times 10^3$ Hz = $1.368 \times 10^6$ s⁻¹
Speed of light ($c$) $\approx 3.0 \times 10^8$ m/s
Wavelength ($\lambda$) = $\frac{c}{\nu}$
$\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{1.368 \times 10^6 \text{ s}^{-1}} \approx 219.3 \text{ m}$
Electromagnetic waves with wavelengths of hundreds of meters belong to the radiowave part of the spectrum.
Example 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:
Speed of light ($c$) $\approx 3.0 \times 10^8$ m/s
For Violet light:
Wavelength ($\lambda$) = 400 nm = $400 \times 10^{-9}$ m = $4.00 \times 10^{-7}$ m
Frequency ($\nu$) = $\frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{4.00 \times 10^{-7} \text{ m}} = 0.75 \times 10^{15} \text{ s}^{-1} = 7.50 \times 10^{14}$ Hz
For Red light:
Wavelength ($\lambda$) = 750 nm = $750 \times 10^{-9}$ m = $7.50 \times 10^{-7}$ m
Frequency ($\nu$) = $\frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{7.50 \times 10^{-7} \text{ m}} = 0.40 \times 10^{15} \text{ s}^{-1} = 4.00 \times 10^{14}$ Hz
The visible spectrum ranges from $4.00 \times 10^{14}$ Hz (red) to $7.50 \times 10^{14}$ Hz (violet).
Example 5. Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.
Answer:
Wavelength ($\lambda$) = 5800 Å. We know 1 Å = 10⁻¹⁰ m.
$\lambda = 5800 \times 10^{-10} \text{ m} = 5.800 \times 10^{-7} \text{ m}$.
If converting to cm: $\lambda = 5800 \times 10^{-8} \text{ cm} = 5.800 \times 10^{-6} \text{ cm}$.
(a) Wavenumber ($\bar{\nu}$) = $\frac{1}{\lambda}$
In cm⁻¹: $\bar{\nu} = \frac{1}{5.800 \times 10^{-6} \text{ cm}} \approx 1.724 \times 10^5 \text{ cm}^{-1}$
In m⁻¹: $\bar{\nu} = \frac{1}{5.800 \times 10^{-7} \text{ m}} \approx 1.724 \times 10^6 \text{ m}^{-1}$
(b) Frequency ($\nu$) = $\frac{c}{\lambda}$
Using $\lambda$ in meters and $c = 3.0 \times 10^8$ m/s:
$\nu = \frac{3.0 \times 10^8 \text{ m/s}}{5.800 \times 10^{-7} \text{ m}} \approx 0.517 \times 10^{15} \text{ s}^{-1} = 5.17 \times 10^{14}$ Hz
Particle Nature Of Electromagnetic Radiation: Planck’s Quantum Theory
While wave theory explained phenomena like diffraction and interference, it failed to explain others, including:
- Black-body radiation: The pattern of radiation emitted by an ideal heated object.
- Photoelectric effect: Ejection of electrons from a metal surface when light strikes it.
- Variation of heat capacity of solids with temperature.
- Line spectra of atoms.
These phenomena suggested that energy is absorbed or emitted in discrete amounts, not continuously.
Black-Body Radiation: Heated objects emit radiation. As temperature increases, the distribution of wavelengths shifts towards shorter wavelengths (e.g., from red to white to blue for a heated iron rod). A black body is an idealized object that absorbs and emits all frequencies of radiation uniformly. The intensity of radiation emitted by a black body at a given temperature varies with wavelength, peaking at a specific wavelength which shifts to shorter values as temperature increases.
Classical wave theory couldn't explain the observed shape of these intensity-wavelength curves.
Planck's Quantum Theory (1900): Max Planck proposed that energy is not continuous but is emitted or absorbed in discrete packets or "chunks". He called the smallest such packet a quantum. The energy ($E$) of a quantum of radiation is directly proportional to its frequency ($\nu$).
$$ E = h\nu $$Here, $h$ is Planck's constant, with a value of $6.626 \times 10^{-34}$ J s.
According to Planck, atoms and molecules can only emit or absorb energy in integer multiples of $h\nu$ (i.e., $nh\nu$, where $n = 1, 2, 3...$), not any arbitrary amount. This idea of "quantization" of energy was a revolutionary concept, like being able to stand only on staircase steps, not between them.
Photoelectric Effect: This phenomenon, discovered by H. Hertz, involves the ejection of electrons (photoelectrons) from a metal surface when light of sufficient frequency shines on it.
Key observations of the photoelectric effect that puzzled classical physics:
- Instantaneous ejection: Electrons are ejected immediately when light hits, with no delay.
- Number of electrons vs. intensity: The number of ejected electrons is proportional to the light's intensity (brightness).
- Threshold frequency ($\nu_0$): For each metal, there is a minimum frequency of light below which no electrons are ejected, regardless of intensity.
- Kinetic energy vs. frequency: If the light frequency ($\nu$) is greater than the threshold frequency ($\nu_0$), the ejected electrons have kinetic energy (KE). This KE increases linearly with the frequency of the light, not its intensity. (e.g., dim yellow light ejects electrons from potassium, but bright red light does not, because red light's frequency is below potassium's threshold frequency).
Albert Einstein (1905) explained the photoelectric effect using Planck's quantum theory. He proposed that light consists of particles called photons, each with energy $E = h\nu$.
Explanation based on photons:
- When a photon with energy $h\nu$ strikes a metal electron, it transfers its energy to the electron.
- If the photon's energy ($h\nu$) is greater than the minimum energy required to remove the electron from the metal (called the work function, $W_0 = h\nu_0$, where $\nu_0$ is the threshold frequency), the electron is ejected.
- The excess energy ($h\nu - W_0$) is converted into the kinetic energy of the ejected electron.
This equation explains why KE depends on frequency, not intensity. A more intense beam means more photons, leading to the ejection of more electrons, but the energy of each ejected electron depends only on the energy of the individual photon (and the metal's work function).
Dual Behaviour of Electromagnetic Radiation: The phenomena of interference and diffraction are explained by the wave nature of light, while black-body radiation and the photoelectric effect are explained by its particle nature (photons). This led to the acceptance of the concept that light (and other electromagnetic radiation) exhibits dual behaviour, possessing both wave-like and particle-like properties depending on the experiment.
Example 6. Calculate energy of one mole of photons of radiation whose frequency is $5 \times 10^{14}$ Hz.
Answer:
Frequency ($\nu$) = $5 \times 10^{14}$ Hz = $5 \times 10^{14}$ s⁻¹
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s
Avogadro's number ($N_A$) = $6.022 \times 10^{23}$ mol⁻¹
Energy of one photon ($E$) = $h\nu$
$E = (6.626 \times 10^{-34} \text{ J s}) \times (5 \times 10^{14} \text{ s}^{-1}) = 3.313 \times 10^{-19}$ J
Energy of one mole of photons = $E \times N_A$
Energy per mole = $(3.313 \times 10^{-19} \text{ J/photon}) \times (6.022 \times 10^{23} \text{ photons/mol})$
Energy per mole $\approx 199510 \text{ J/mol} = 199.51$ kJ/mol
Example 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 (This is the total energy emitted per second)
Wavelength ($\lambda$) = 400 nm = $400 \times 10^{-9}$ m = $4 \times 10^{-7}$ m
Speed of light ($c$) = $3.0 \times 10^8$ m/s
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s
First, calculate the energy of a single photon:
$E = h\nu = \frac{hc}{\lambda}$
$E = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{4 \times 10^{-7} \text{ m}} = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}} \text{ J}$
$E \approx 4.9695 \times 10^{-19}$ J/photon
The number of photons emitted per second is the total energy emitted per second divided by the energy of one photon.
Number of photons/second = $\frac{\text{Total Energy/second}}{\text{Energy/photon}} = \frac{100 \text{ J/s}}{4.9695 \times 10^{-19} \text{ J/photon}}$
Number of photons/second $\approx 20.12 \times 10^{19} \text{ s}^{-1} = 2.012 \times 10^{20}$ photons/s
Example 8. When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68 \times 10^5$ 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:
Wavelength of incident radiation ($\lambda$) = 300 nm = $300 \times 10^{-9}$ m = $3 \times 10^{-7}$ m
Kinetic energy of emitted electrons = $1.68 \times 10^5$ J mol⁻¹. This is the KE for one mole of electrons.
To find the KE of a single electron, divide by Avogadro's number:
KE per electron = $\frac{1.68 \times 10^5 \text{ J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} \approx 2.79 \times 10^{-19}$ J
Energy of the incident photon ($E$) = $\frac{hc}{\lambda}$
$h = 6.626 \times 10^{-34}$ J s, $c = 3.0 \times 10^8$ m/s
$E = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{3 \times 10^{-7} \text{ m}} = 6.626 \times 10^{-19}$ J
According to the photoelectric effect equation: $KE = E - W_0$, where $W_0$ is the work function (minimum energy needed to remove an electron).
$W_0 = E - KE$
Minimum energy needed per electron ($W_0$) = $(6.626 \times 10^{-19} \text{ J}) - (2.79 \times 10^{-19} \text{ J}) = 3.836 \times 10^{-19}$ J
To find the minimum energy needed for one mole of electrons (ionization energy), multiply $W_0$ per electron by $N_A$:
$W_0$ per mole = $(3.836 \times 10^{-19} \text{ J/electron}) \times (6.022 \times 10^{23} \text{ electrons/mol}) \approx 2.31 \times 10^5 \text{ J/mol}$
The maximum wavelength that causes emission corresponds to the threshold frequency ($\nu_0$), where the photon energy is exactly equal to the work function ($h\nu_0 = W_0$). The maximum wavelength ($\lambda_0$) is $\lambda_0 = c/\nu_0$.
$W_0 = \frac{hc}{\lambda_0}$
$\lambda_0 = \frac{hc}{W_0} = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{3.836 \times 10^{-19} \text{ J}}$
$\lambda_0 \approx 5.18 \times 10^{-7} \text{ m}$
Converting to nm: $\lambda_0 = 5.18 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 518$ nm.
The maximum wavelength that will cause photoelectron emission is 518 nm.
Example 9. The threshold frequency $\nu_0$ for a metal is $7.0 \times 10^{14}$ s⁻¹. Calculate the kinetic energy of an electron emitted when radiation of frequency $\nu =1.0 \times 10^{15}$ s⁻¹ hits the metal.
Answer:
Threshold frequency ($\nu_0$) = $7.0 \times 10^{14}$ s⁻¹
Frequency of incident radiation ($\nu$) = $1.0 \times 10^{15}$ s⁻¹
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s
Kinetic energy of emitted electron ($KE$) = $h(\nu - \nu_0)$
KE = $(6.626 \times 10^{-34} \text{ J s}) \times (1.0 \times 10^{15} \text{ s}^{-1} - 7.0 \times 10^{14} \text{ s}^{-1})$
Note that $1.0 \times 10^{15} = 10.0 \times 10^{14}$.
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 = $19.878 \times 10^{-20}$ J $= 1.988 \times 10^{-19}$ J
The kinetic energy of the emitted electron is $1.988 \times 10^{-19}$ J.
Evidence For The Quantized Electronic Energy Levels: Atomic Spectra
When white light (like sunlight) passes through a prism, it splits into a continuous spectrum of colors (rainbow, VIBGYOR). This is because white light contains a range of wavelengths, and each wavelength bends differently as it passes through the prism. In a continuous spectrum, all wavelengths within a region are present.
When substances absorb energy (by heating or irradiation), their atoms become "excited". These excited atoms are unstable and emit radiation as they return to lower energy states. If this emitted light is passed through a prism, it produces an emission spectrum.
When white light is passed through a substance, and then the transmitted light is analyzed, it produces an absorption spectrum, where certain wavelengths are missing (absorbed by the substance), appearing as dark lines against a continuous background.
The study of these spectra is called spectroscopy.
Unlike continuous spectra from white light, the emission spectra of atoms in the gas phase consist of only specific, discrete wavelengths. These appear as bright lines on a dark background and are called line spectra or atomic spectra.
Key features of atomic line spectra:
- Each element has a unique line spectrum, like a fingerprint, allowing identification of elements.
- There is a regularity in the lines within an element's spectrum.
Line Spectrum of Hydrogen: The simplest atom, hydrogen, produces a series of lines when an electric discharge is passed through hydrogen gas. These lines fall into different series depending on the region of the electromagnetic spectrum (visible, UV, IR).
J. Balmer found a formula (1885) that described the visible lines (Balmer series) in terms of wavenumber ($\bar{\nu}$):
$$ \bar{\nu} = R_H \left(\frac{1}{2^2} - \frac{1}{n^2}\right), \quad n = 3, 4, 5, ... $$Later, Johannes Rydberg generalized this to include all series in the hydrogen spectrum:
$$ \bar{\nu} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad n_1 = 1, 2, 3, ... \quad \text{and} \quad n_2 = n_1 + 1, n_1 + 2, ... $$Here, $R_H$ is the Rydberg constant for hydrogen ($109,677$ cm⁻¹ or $1.09677 \times 10^7$ m⁻¹). Different $n_1$ values define different series:
| Series | $n_1$ | $n_2$ | Spectral Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, ... | Ultraviolet (UV) |
| Balmer | 2 | 3, 4, 5, ... | Visible |
| Paschen | 3 | 4, 5, 6, ... | Infrared (IR) |
| Brackett | 4 | 5, 6, 7, ... | Infrared (IR) |
| Pfund | 5 | 6, 7, 8, ... | Infrared (IR) |
The fact that atoms only emit or absorb specific frequencies (lines) is strong evidence that the electrons within atoms can only exist in certain discrete energy states. This concept is called quantization of energy levels.
Bohr’s Model For Hydrogen Atom
In 1913, Niels Bohr proposed a model for the hydrogen atom that successfully explained its stability and line spectrum by incorporating Planck's idea of energy quantization. Although not completely accurate by modern quantum mechanics standards, it was a crucial step forward.
Postulates of Bohr's Model for Hydrogen Atom:
- Electron Orbits (Stationary States): Electrons revolve around the nucleus in specific, fixed circular paths called orbits, stationary states, or allowed energy states. These orbits have fixed radii and energies and are arranged concentrically around the nucleus.
- Quantized Energy Levels: Electrons do not radiate energy while in a particular stationary state. The energy of an electron in an orbit remains constant over time. Energy is only absorbed or emitted when an electron moves from one stationary state to another.
- Bohr's Frequency Rule: When an electron transitions from a higher energy state ($E_{higher}$) to a lower energy state ($E_{lower}$), it emits a photon of radiation with energy equal to the difference in energy between the two states. Conversely, energy is absorbed when an electron moves from a lower state to a higher state. The frequency ($\nu$) of the emitted or absorbed radiation is given by: $$ \Delta E = E_{higher} - E_{lower} = h\nu $$
- Quantization of Angular Momentum: The angular momentum of an electron in a given stationary state is quantized. It can only take on values that are integral multiples of $\frac{h}{2\pi}$. $$ \text{Angular Momentum} = m_e v r = n \frac{h}{2\pi} $$ where $m_e$ is the electron mass, $v$ is its velocity, $r$ is the orbit radius, $h$ is Planck's constant, and $n$ is a positive integer (1, 2, 3, ...). This integer $n$ is called the Principal Quantum Number and identifies the orbit/energy level.
Using these postulates, Bohr derived expressions for the radius and energy of the allowed orbits in the hydrogen atom (and hydrogen-like species):
- Radii of Orbits ($r_n$): The radius of the $n$-th orbit is given by: $$ r_n = n^2 a_0 $$ where $a_0$ is the Bohr radius, the radius of the first orbit ($n=1$) in hydrogen, approximately 52.9 pm ($52.9 \times 10^{-12}$ m). As $n$ increases, the radius increases, meaning the electron is further from the nucleus.
- Energies of Orbits ($E_n$): The energy of an electron in the $n$-th orbit is given by: $$ E_n = -R_H \left(\frac{1}{n^2}\right), \quad n = 1, 2, 3, ... $$ where $R_H$ is the Rydberg constant in energy units ($2.18 \times 10^{-18}$ J). The energy is negative, indicating the electron is bound to the nucleus. The lowest energy (most stable, ground state) is for $n=1$. As $n$ increases, $E_n$ becomes less negative (higher energy), approaching zero as $n \rightarrow \infty$ (representing a free electron).
For hydrogen-like species (ions with only one electron, like He⁺, Li²⁺, Be³⁺), the nuclear charge is $+Ze$. Bohr's formulas become:
- Radii: $r_n = a_0 \frac{n^2}{Z} = 52.9 \left(\frac{n^2}{Z}\right)$ pm
- Energies: $E_n = -R_H \frac{Z^2}{n^2} = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right)$ J
Increasing Z makes the energy more negative (more stable) and the radius smaller (electron pulled closer to the nucleus).
Explanation Of Line Spectrum Of Hydrogen
Bohr's model successfully explained the discrete lines in the hydrogen spectrum. According to the model, these lines arise from electrons transitioning between different quantized energy levels (orbits). When an electron moves from a higher orbit ($n_i$) to a lower orbit ($n_f$), it emits a photon with energy $\Delta E = E_{n_i} - E_{n_f}$.
Using the energy expression $E_n = -R_H/n^2$, the energy difference is:
$$ \Delta E = \left(-\frac{R_H}{n_f^2}\right) - \left(-\frac{R_H}{n_i^2}\right) = R_H \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) $$Since $\Delta E = h\nu$, the frequency of the emitted photon is:
$$ \nu = \frac{\Delta E}{h} = \frac{R_H}{h} \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right) $$And the wavenumber is $\bar{\nu} = \frac{\nu}{c} = \frac{R_H}{hc} \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$.
The term $\frac{R_H}{hc}$ is the Rydberg constant in wavenumber units ($109,677$ cm⁻¹ or $1.09677 \times 10^7$ m⁻¹). Thus:
$$ \bar{\nu} = R_H (\text{wavenumber}) \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) $$This formula exactly matches the empirical Rydberg formula, with $n_f$ corresponding to $n_1$ and $n_i$ corresponding to $n_2$ (where $n_i > n_f$ for emission). Each line in the spectrum corresponds to a specific transition between energy levels. For absorption, the electron moves from a lower orbit ($n_i$) to a higher orbit ($n_f$), so $n_f > n_i$, and $\Delta E$ is positive.
Example 12. 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:
Initial orbit ($n_i$) = 5
Final orbit ($n_f$) = 2
This transition ends at $n=2$, so it belongs to the Balmer series (visible region).
Energy difference ($\Delta E$) = $E_{n_i} - E_{n_f} = E_5 - E_2$
Using $E_n = -2.18 \times 10^{-18} / n^2$ J:
$E_5 = -2.18 \times 10^{-18} / 5^2 = -2.18 \times 10^{-18} / 25 = -0.0872 \times 10^{-18}$ J
$E_2 = -2.18 \times 10^{-18} / 2^2 = -2.18 \times 10^{-18} / 4 = -0.545 \times 10^{-18}$ J
$\Delta E = (-0.0872 \times 10^{-18} \text{ J}) - (-0.545 \times 10^{-18} \text{ J}) = (0.545 - 0.0872) \times 10^{-18} \text{ J} = 0.4578 \times 10^{-18}$ J
For emission, the energy is released, so $\Delta E$ is positive in magnitude of the photon.
Frequency ($\nu$) = $\Delta E / h$
$\nu = \frac{0.4578 \times 10^{-18} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \approx 0.06908 \times 10^{16} \text{ s}^{-1} = 6.908 \times 10^{14}$ Hz
Wavelength ($\lambda$) = $c / \nu$
$\lambda = \frac{3.0 \times 10^8 \text{ m/s}}{6.908 \times 10^{14} \text{ s}^{-1}} \approx 4.343 \times 10^{-7} \text{ m}$
Converting to nm: $\lambda = 4.343 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} \approx 434.3$ nm
This wavelength (434.3 nm) is in the visible region (Balmer series).
Example 13. Calculate the energy associated with the first orbit of He⁺. What is the radius of this orbit?
Answer:
He⁺ is a hydrogen-like species (one electron). Atomic number of He is Z = 2.
We need to calculate energy and radius for the first orbit, so Principal Quantum Number (n) = 1.
Energy of n-th orbit for hydrogen-like species: $E_n = -R_H \frac{Z^2}{n^2} = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right)$ J
For the first orbit of He⁺ (n=1, Z=2):
$E_1 = -2.18 \times 10^{-18} \left(\frac{2^2}{1^2}\right) \text{ J} = -2.18 \times 10^{-18} \times 4 \text{ J} = -8.72 \times 10^{-18}$ J
Radius of n-th orbit for hydrogen-like species: $r_n = a_0 \frac{n^2}{Z} = 52.9 \left(\frac{n^2}{Z}\right)$ pm
For the first orbit of He⁺ (n=1, Z=2):
$r_1 = 52.9 \left(\frac{1^2}{2}\right) \text{ pm} = 52.9 \times \frac{1}{2} \text{ pm} = 26.45$ pm
The energy of the first orbit of He⁺ is $-8.72 \times 10^{-18}$ J, and its radius is 26.45 pm.
Limitations Of Bohr’s Model
While Bohr's model successfully explained the hydrogen spectrum and the stability of the hydrogen atom, it had significant limitations:
- It could not explain the spectra of multi-electron atoms (atoms with more than one electron), such as Helium.
- It failed to account for the fine structure of the hydrogen spectrum (the observation that spectral lines, when viewed with high-resolution spectroscopes, are actually composed of two or more closely spaced lines).
- It could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
- It did not explain the ability of atoms to form chemical bonds and create molecules.
These limitations indicated that Bohr's model was incomplete and a more sophisticated theory was needed to describe atomic structure accurately, especially for atoms beyond hydrogen.
Towards Quantum Mechanical Model Of The Atom
To overcome the limitations of the Bohr model, new developments in physics were incorporated into atomic theory. The two most important were the discovery of the dual behavior of matter and the formulation of the uncertainty principle.
Dual Behaviour Of Matter
Inspired by the wave-particle duality of light, French physicist Louis de Broglie proposed in 1924 that matter, like radiation, also exhibits dual behaviour. This means that particles such as electrons, protons, atoms, and molecules should also possess wave-like properties in addition to their particle-like properties.
De Broglie related the wavelength ($\lambda$) of a material particle to its momentum ($p = mv$, where $m$ is mass and $v$ is velocity) using a relationship analogous to that for photons:
$$ \lambda = \frac{h}{p} = \frac{h}{mv} $$where $h$ is Planck's constant.
De Broglie's hypothesis was experimentally confirmed later when electron beams were shown to undergo diffraction, a phenomenon characteristic of waves. This wave nature of electrons is utilized in electron microscopes, which achieve much higher magnifications than optical microscopes.
The de Broglie wavelength is significant for microscopic particles because their mass ($m$) is very small, resulting in a measurable wavelength. For macroscopic objects (like a ball or a car), the mass is large, making the de Broglie wavelength extremely small, practically undetectable. This is why classical mechanics, which treats objects solely as particles, works well for macroscopic objects.
Example 14. What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s⁻¹ ?
Answer:
Mass ($m$) = 0.1 kg
Velocity ($v$) = 10 m/s
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s = $6.626 \times 10^{-34}$ kg m² s⁻¹ (since J = kg m² s⁻²)
Using de Broglie equation: $\lambda = \frac{h}{mv}$
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(0.1 \text{ kg}) \times (10 \text{ m s}^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{1 \text{ kg m s}^{-1}} = 6.626 \times 10^{-34}$ m
The wavelength is extremely small ($6.626 \times 10^{-34}$ m), far too small to be observed or measured. This confirms that the wave nature of macroscopic objects is negligible.
Example 15. 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:
Mass of electron ($m$) = $9.1 \times 10^{-31}$ kg
Kinetic Energy (KE) = $3.0 \times 10^{-25}$ J
We need the velocity ($v$) to use the de Broglie equation ($\lambda = h/mv$). KE is related to velocity by $KE = \frac{1}{2}mv^2$.
$v^2 = \frac{2 \times KE}{m}$
$v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25} \text{ J}}{9.1 \times 10^{-31} \text{ kg}}}$
Using J = kg m² s⁻²:
$v = \sqrt{\frac{6.0 \times 10^{-25} \text{ kg m}^2 \text{ s}^{-2}}{9.1 \times 10^{-31} \text{ kg}}} = \sqrt{\frac{6.0}{9.1} \times 10^{(-25 - (-31))} \text{ m}^2 \text{ s}^{-2}}$
$v = \sqrt{0.6593 \times 10^6 \text{ m}^2 \text{ s}^{-2}} \approx \sqrt{65.93 \times 10^4 \text{ m}^2 \text{ s}^{-2}}$
$v \approx 8.12 \times 10^2$ m/s = 812 m/s
Now calculate the wavelength using the de Broglie equation:
$\lambda = \frac{h}{mv}$
$h = 6.626 \times 10^{-34}$ J s = $6.626 \times 10^{-34}$ kg m² s⁻¹
$\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(9.1 \times 10^{-31} \text{ kg}) \times (812 \text{ m s}^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 812 \times 10^{-31}} \text{ m} = \frac{6.626 \times 10^{-34}}{7390.2 \times 10^{-31}} \text{ m} \approx 0.000896 \times 10^{-3} \text{ m}$
$\lambda \approx 8.96 \times 10^{-7}$ m
Converting to nm: $\lambda = 8.96 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 896$ nm
This wavelength is in the range of visible light and is experimentally detectable, highlighting the significant wave nature of microscopic particles like electrons.
Example 16. Calculate the mass of a photon with wavelength 3.6 Å.
Answer:
Wavelength ($\lambda$) = 3.6 Å = $3.6 \times 10^{-10}$ m
Photons travel at the speed of light ($c$) = $3.0 \times 10^8$ m/s.
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s = $6.626 \times 10^{-34}$ kg m² s⁻¹
While photons are often considered massless particles in the traditional sense, their momentum and energy are related by $E=pc$ and $E=h\nu$. Using de Broglie's relation ($\lambda = h/p$), we can find the momentum of the photon: $p = h/\lambda$.
Momentum ($p$) = Mass ($m$) $\times$ velocity ($v$). For a photon, $v=c$, so $p = mc$.
Equating the two expressions for momentum: $mc = \frac{h}{\lambda}$
So, the "effective mass" of a photon can be calculated as $m = \frac{h}{\lambda c}$.
$m = \frac{6.626 \times 10^{-34} \text{ J s}}{(3.6 \times 10^{-10} \text{ m}) \times (3.0 \times 10^8 \text{ m/s})}$
$m = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{10.8 \times 10^{-2} \text{ m}^2 \text{ s}^{-1}} = \frac{6.626 \times 10^{-34}}{0.108} \text{ kg}$
$m \approx 61.35 \times 10^{-34}$ kg $= 6.135 \times 10^{-33}$ kg
This is the equivalent mass associated with the energy of the photon.
Heisenberg’s Uncertainty Principle
The dual nature of matter implies a fundamental limitation in simultaneously determining certain properties of a particle. In 1927, German physicist Werner Heisenberg formulated the Uncertainty Principle.
Statement: It is impossible to determine simultaneously the exact position and the exact momentum (or velocity) of a microscopic particle like an electron.
Mathematically, this is expressed as:
$$ \Delta x \cdot \Delta p_x \ge \frac{h}{4\pi} $$or, since $\Delta p_x = m \Delta v_x$ (assuming mass is constant):
$$ \Delta x \cdot m \Delta v_x \ge \frac{h}{4\pi} $$where:
- $\Delta x$ is the uncertainty in the position of the particle.
- $\Delta p_x$ is the uncertainty in the momentum in the x-direction.
- $\Delta v_x$ is the uncertainty in the velocity in the x-direction.
- $h$ is Planck's constant.
The principle states that the product of the uncertainty in position and the uncertainty in momentum is always greater than or equal to a constant value ($h/4\pi$). This means:
- If the position of an electron is known very accurately (small $\Delta x$), its momentum (and velocity) is highly uncertain (large $\Delta v_x$).
- If the momentum (or velocity) of an electron is known very accurately (small $\Delta v_x$), its position is highly uncertain (large $\Delta x$).
It's impossible to know both with perfect precision at the same time.
Significance of the Uncertainty Principle:
- The Uncertainty Principle directly challenges the classical concept of a particle having a definite trajectory or path. A trajectory requires knowing both position and momentum simultaneously at every point, which is forbidden for microscopic particles like electrons. Therefore, we cannot speak of definite orbits for electrons in atoms.
- The effects of the Uncertainty Principle are significant only for microscopic objects (like electrons). For macroscopic objects, the mass ($m$) is so large that even a very small uncertainty in velocity leads to an undetectable uncertainty in position (or vice versa), as the product $\Delta x \cdot \Delta v_x$ remains very small compared to the object's scale.
This principle, along with the dual nature of matter, forms the basis of quantum mechanics, which describes the behavior of microscopic particles in terms of probabilities rather than certainties.
Example 17. 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:
Uncertainty in position ($\Delta x$) = 0.1 Å = $0.1 \times 10^{-10}$ m = $1 \times 10^{-11}$ m
Mass of electron ($m$) = $9.109 \times 10^{-31}$ kg (using a more precise value)
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s
Using Heisenberg's Uncertainty Principle: $\Delta x \cdot m \Delta v_x \ge \frac{h}{4\pi}$
We want to find the minimum uncertainty in velocity ($\Delta v_x$), so we use the equality:
$\Delta x \cdot m \Delta v_x = \frac{h}{4\pi}$
$\Delta v_x = \frac{h}{4\pi m \Delta x}$
$\Delta v_x = \frac{6.626 \times 10^{-34} \text{ J s}}{4 \times 3.1416 \times (9.109 \times 10^{-31} \text{ kg}) \times (1 \times 10^{-11} \text{ m})}$
Using J = kg m² s⁻²:
$\Delta v_x = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{114.45 \times 10^{(-31-11)} \text{ kg m}} = \frac{6.626 \times 10^{-34}}{114.45 \times 10^{-42}} \text{ m s}^{-1}$
$\Delta v_x \approx 0.05789 \times 10^{(-34 - (-42))} \text{ m s}^{-1} = 0.05789 \times 10^8 \text{ m s}^{-1} = 5.789 \times 10^6$ m/s
The uncertainty in the velocity of the electron is approximately $5.79 \times 10^6$ m/s. This is a very large uncertainty, demonstrating that if the position is known quite precisely, the velocity is highly uncertain.
Example 18. 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:
Mass ($m$) = 40 g = 0.040 kg
Speed ($v$) = 45 m/s
Accuracy in speed measurement = 2%
Uncertainty in speed ($\Delta v$) = 2% of 45 m/s = $0.02 \times 45$ m/s = 0.9 m/s
Planck's constant ($h$) = $6.626 \times 10^{-34}$ J s
Using Heisenberg's Uncertainty Principle: $\Delta x \cdot m \Delta v \ge \frac{h}{4\pi}$
We calculate the minimum uncertainty in position ($\Delta x$):
$\Delta x = \frac{h}{4\pi m \Delta v}$
$\Delta x = \frac{6.626 \times 10^{-34} \text{ J s}}{4 \times 3.1416 \times (0.040 \text{ kg}) \times (0.9 \text{ m/s})}$
Using J = kg m² s⁻²:
$\Delta x = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{1.131 \text{ kg m s}^{-1}}$ (since $4 \times 3.1416 \times 0.040 \times 0.9 \approx 1.131$)
$\Delta x \approx 5.859 \times 10^{-34}$ m
The uncertainty in the position of the golf ball is approximately $5.86 \times 10^{-34}$ m. This value is exceedingly small, far below any practical measurement limit, and is insignificant compared to the size of the golf ball. This shows why quantum mechanical effects are not observable for macroscopic objects.
Reasons for the Failure of the Bohr Model:
Bohr's model treated the electron as a particle orbiting the nucleus in a well-defined path. However, this concept of a definite trajectory is inconsistent with:
- The wave character of the electron (de Broglie's hypothesis).
- Heisenberg's Uncertainty Principle, which states that a particle cannot simultaneously have a precisely known position and momentum needed to define a trajectory.
Therefore, Bohr's model needed to be replaced by a theory that could account for the wave-particle duality of matter and the Uncertainty Principle.
Quantum Mechanical Model Of Atom
Classical mechanics, based on Newton's laws, works well for macroscopic objects but fails for microscopic ones due to their significant wave nature and the limitations imposed by the Uncertainty Principle. A new branch of science, quantum mechanics, was developed to describe the motion and properties of microscopic objects that exhibit both wave-like and particle-like behavior.
Quantum mechanics is a theoretical science that describes the behavior of microscopic particles in terms of probabilities. It was developed independently by Werner Heisenberg and Erwin Schrödinger in 1926.
Schrödinger formulated a fundamental equation (the Schrödinger equation) that incorporates the wave-particle duality of matter. Solving this complex mathematical equation for an atom provides crucial information about the electron.
For a system whose energy doesn't change with time (like an electron in a stable atom), the Schrödinger equation can be written as $\hat{H}\psi = E\psi$, where $\hat{H}$ is the Hamiltonian operator (representing the total energy of the system), $\psi$ is the wave function, and $E$ is the energy.
Hydrogen Atom and the Schrödinger Equation: When the Schrödinger equation is solved for the hydrogen atom, the solutions give:
- The possible quantized energy levels that the electron can occupy.
- The corresponding wave functions ($\psi$) associated with each energy level.
These quantized energy states and wave functions arise naturally from the mathematical solution and are characterized by a set of three integer values called quantum numbers: the principal quantum number ($n$), the azimuthal quantum number ($l$), and the magnetic quantum number ($m_l$). These wave functions ($\psi$) for one-electron systems are called atomic orbitals.
The wave function $\psi$ itself does not have a direct physical meaning. However, the square of the wave function, $|\psi|^2$, represents the probability density of finding the electron at a particular point in space within the atom. A higher value of $|\psi|^2$ means a higher probability of finding the electron at that point.
The quantum mechanical model successfully predicts all aspects of the hydrogen atom's spectrum and structure, including those phenomena that the Bohr model couldn't explain.
Applying the Schrödinger equation to multi-electron atoms is more complex and often requires approximate methods. However, these calculations show that the concept of atomic orbitals is still valid, although their energies are affected by electron-electron repulsions and vary not just with $n$ but also with $l$.
Important Features of the Quantum Mechanical Model of Atom:
- The energy of electrons in atoms is quantized, meaning electrons can only exist in specific, discrete energy levels.
- The existence of quantized energy levels is a direct consequence of the electron's wave nature and the solutions to the Schrödinger equation.
- The exact position and velocity of an electron cannot be known simultaneously (Heisenberg Uncertainty Principle). Therefore, the concept of a definite electron path or orbit is invalid.
- An atomic orbital is a mathematical wave function ($\psi$) that describes the state of an electron in an atom. It characterizes a region of space around the nucleus where the probability of finding the electron is high. An atom contains many such orbitals. Each orbital has a definite energy.
- An atomic orbital can hold a maximum of two electrons, provided they have opposite spins (Pauli exclusion principle).
- In multi-electron atoms, electrons fill the available atomic orbitals in order of increasing energy.
- The probability of finding an electron at a point within an atom is proportional to the square of the wave function ($|\psi|^2$) at that point. $|\psi|^2$ is called the probability density. Plotting $|\psi|^2$ helps visualize the shape and distribution of electron density in an orbital.
Orbitals And Quantum Numbers
Atomic orbitals are distinguished by their size, shape, and orientation in space. These properties are defined by a set of three quantum numbers obtained from the solution of the Schrödinger equation for the hydrogen atom.
Each orbital is uniquely identified by a set of values for the following quantum numbers:
- Principal Quantum Number ($n$):
- Takes positive integer values: $n = 1, 2, 3, ...$
- Determines the size and largely the energy of the orbital. Higher $n$ means a larger orbital, further from the nucleus, and higher energy.
- Identifies the main energy shell or level. Shells are often denoted by letters: K (for $n=1$), L (for $n=2$), M (for $n=3$), N (for $n=4$), etc.
- The total number of orbitals in a given principal shell ($n$) is equal to $n^2$.
- Azimuthal Quantum Number ($l$) or Orbital Angular Momentum Quantum Number:
- Takes integer values from 0 to $n-1$ for a given value of $n$.
- Defines the three-dimensional shape of the orbital.
- Identifies subshells within a shell. The number of subshells in a shell is equal to $n$.
- Subshells are denoted by letters corresponding to $l$ values:
- $l=0$: s subshell (shape is spherical)
- $l=1$: p subshell (shape is dumbbell)
- $l=2$: d subshell (shapes are more complex)
- $l=3$: f subshell (shapes are even more complex)
- and so on (g, h, ...)
- Also influences the energy of the orbital in multi-electron atoms.
- Magnetic Orbital Quantum Number ($m_l$):
- Takes integer values from $-l$ to $+l$, including 0, for a given value of $l$.
- Describes the spatial orientation of the orbital in a magnetic field (or in the absence of a field, simply the different possible orientations in space).
- The number of possible $m_l$ values for a given $l$ is $2l+1$. This indicates the number of orbitals within a subshell:
- $l=0$ (s subshell): $m_l = 0$ (1 orbital)
- $l=1$ (p subshell): $m_l = -1, 0, +1$ (3 orbitals)
- $l=2$ (d subshell): $m_l = -2, -1, 0, +1, +2$ (5 orbitals)
- $l=3$ (f subshell): $m_l = -3, -2, -1, 0, +1, +2, +3$ (7 orbitals)
Each specific combination of $n$, $l$, and $m_l$ defines a unique atomic orbital (e.g., $n=2, l=1, m_l=0$ describes one of the 2p orbitals). The total number of orbitals in a shell with principal quantum number $n$ is the sum of orbitals in all its subshells: $\sum_{l=0}^{n-1} (2l+1) = n^2$.
Electron Spin Quantum Number ($m_s$): While the first three quantum numbers describe the orbital, a fourth quantum number, $m_s$, is needed to fully describe an electron within an orbital. Introduced by Uhlenbeck and Goudsmit (1925), it accounts for the intrinsic angular momentum of an electron, often visualized as the electron spinning on its own axis.
- Takes values of $+\frac{1}{2}$ or $-\frac{1}{2}$.
- Represents the two possible spin orientations of an electron.
- Two electrons in the same orbital must have opposite spins (one $+\frac{1}{2}$, one $-\frac{1}{2}$). This is a key aspect of the Pauli Exclusion Principle.
Summary of Information from Quantum Numbers:
- $n$: Determines the main energy level (shell), size, and overall energy.
- $l$: Determines the subshell and the shape of the orbital, also affects energy in multi-electron atoms.
- $m_l$: Determines the specific orbital within a subshell and its spatial orientation.
- $m_s$: Determines the spin orientation of the electron (either "spin up" or "spin down").
Orbit vs. Orbital: It's important to distinguish between Bohr's "orbit" and the quantum mechanical "orbital." Bohr's orbit was a definite, fixed circular path, a concept invalidated by the Uncertainty Principle. An atomic orbital, on the other hand, is a probability distribution describing where an electron is *likely* to be found in space around the nucleus. It is not a trajectory.
Example 19. What is the total number of orbitals associated with the principal quantum number n = 3 ?
Answer:
Given principal quantum number $n = 3$.
For $n=3$, the possible values for the azimuthal quantum number ($l$) are $0, 1, 2$ (since $l$ ranges from 0 to $n-1$).
- If $l=0$ (s subshell), there is $2l+1 = 2(0)+1 = 1$ orbital ($m_l=0$). This is the 3s orbital.
- If $l=1$ (p subshell), there are $2l+1 = 2(1)+1 = 3$ orbitals ($m_l=-1, 0, +1$). These are the 3p orbitals.
- If $l=2$ (d subshell), there are $2l+1 = 2(2)+1 = 5$ orbitals ($m_l=-2, -1, 0, +1, +2$). These are the 3d orbitals.
Total number of orbitals for $n=3$ = (number of 3s orbitals) + (number of 3p orbitals) + (number of 3d orbitals) = $1 + 3 + 5 = 9$ orbitals.
This matches the general rule that the total number of orbitals in a shell with principal quantum number $n$ is $n^2$, so for $n=3$, the total orbitals are $3^2 = 9$.
Example 20. 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 $n$[subshell symbol based on $l$].
(a) $n = 2, l = 1$: $l=1$ corresponds to the 'p' subshell. So this is a 2p orbital.
(b) $n = 4, l = 0$: $l=0$ corresponds to the 's' subshell. So this is a 4s orbital.
(c) $n = 5, l = 3$: $l=3$ corresponds to the 'f' subshell. So this is a 5f orbital.
(d) $n = 3, l = 2$: $l=2$ corresponds to the 'd' subshell. So this is a 3d orbital.
Shapes Of Atomic Orbitals
The shape of an atomic orbital is represented by the region in space where the probability of finding the electron ($|\psi|^2$) is high. While $|\psi|^2$ never drops strictly to zero at any finite distance from the nucleus, we typically visualize orbital shapes using a boundary surface diagram that encloses a region where the probability of finding the electron is significant (e.g., 90%).
s Orbitals ($l=0$):
- There is 1 s orbital in each shell ($m_l=0$).
- s orbitals are spherically symmetric. The probability of finding the electron is the same in all directions at a given distance from the nucleus.
- The size of s orbitals increases with increasing principal quantum number ($n$) (e.g., 4s > 3s > 2s > 1s).
- For $n>1$, s orbitals have radial nodes, where the probability density drops to zero. The number of radial nodes in an ns orbital is $n-1$.
p Orbitals ($l=1$):
- There are 3 p orbitals in each shell for $n \ge 2$ ($m_l = -1, 0, +1$).
- p orbitals have a dumbbell shape, consisting of two lobes on opposite sides of the nucleus.
- The three p orbitals in a given subshell are oriented along the x, y, and z axes and are denoted as $p_x$, $p_y$, and $p_z$.
- They are degenerate (have the same energy) in the absence of external fields.
- There is a nodal plane passing through the nucleus between the two lobes where the probability density is zero. For $p_z$, the xy-plane is a nodal plane. The number of angular nodes in a p orbital is $l=1$.
- The size and energy of p orbitals increase with increasing principal quantum number (e.g., 4p > 3p > 2p).
- Number of radial nodes in an np orbital is $n-l-1 = n-1-1 = n-2$.
d Orbitals ($l=2$):
- There are 5 d orbitals in each shell for $n \ge 3$ ($m_l = -2, -1, 0, +1, +2$).
- d orbitals have more complex shapes, typically with four lobes (except for $d_{z^2}$).
- The five d orbitals are designated as $d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, and $d_{z^2}$.
- In a given subshell, all five d orbitals are degenerate.
- Most d orbitals have two nodal planes passing through the nucleus. The number of angular nodes in a d orbital is $l=2$.
- The size and energy of d orbitals increase with increasing principal quantum number (e.g., 4d > 3d).
- Number of radial nodes in an nd orbital is $n-l-1 = n-2-1 = n-3$.
The total number of nodes (radial + angular) for any orbital is $n-1$.
Energies Of Orbitals
In Hydrogen Atom (and hydrogen-like species):
- The energy of an orbital depends only on the principal quantum number ($n$).
- Orbitals within the same shell (same $n$) but different subshells (different $l$) have the same energy. These are called degenerate orbitals.
- The order of energy is $1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < ...$
- The lowest energy state ($n=1$) is the ground state; higher energy states ($n>1$) are excited states.
In Multi-electron Atoms:
- The energy of an orbital depends on both the principal quantum number ($n$) and the azimuthal quantum number ($l$).
- Orbitals within the same shell (same $n$) but different subshells (different $l$) have different energies.
- Within a given shell, the energy of subshells increases in the order: $s < p < d < f$.
This difference in energy for subshells within the same shell in multi-electron atoms is due to:
- Electron-electron repulsion: Electrons in multi-electron atoms repel each other, which affects their energy.
- Shielding (Screening): Inner shell electrons 'shield' or screen the outer shell electrons from the full positive charge of the nucleus. The outer electrons experience an effective nuclear charge ($Z_{eff}e$) which is less than the actual nuclear charge ($Ze$).
- Penetration: The extent to which an electron in a particular orbital penetrates the region close to the nucleus affects the Z_eff it experiences. s orbitals penetrate the nucleus most effectively, followed by p, then d, then f. This means an electron in an s orbital experiences a higher effective nuclear charge and has lower energy than an electron in a p orbital (of the same shell), and so on ($E_s < E_p < E_d < E_f$ within the same shell).
The relative order of orbital energies in multi-electron atoms generally follows the (n + l) rule:
- Orbitals with a lower value of (n + l) have lower energy.
- If two orbitals have the same (n + l) value, the orbital with the lower $n$ value has lower energy.
| Orbital | $n$ | $l$ | $n+l$ | Energy Order |
|---|---|---|---|---|
| 1s | 1 | 0 | 1 | Lowest |
| 2s | 2 | 0 | 2 | |
| 2p | 2 | 1 | 3 | |
| 3s | 3 | 0 | 3 | 2s < 2p (lower n for same n+l) |
| 3p | 3 | 1 | 4 | 3s < 3p |
| 4s | 4 | 0 | 4 | 3p < 4s (lower n for same n+l) |
| 3d | 3 | 2 | 5 | 4s < 3d |
| 4p | 4 | 1 | 5 | 3d < 4p (lower n for same n+l) |
| 5s | 5 | 0 | 5 | 4p < 5s (lower n for same n+l) |
| 4d | 4 | 2 | 6 | 5s < 4d |
| 5p | 5 | 1 | 6 | 4d < 5p (lower n for same n+l) |
This rule helps determine the approximate order of filling orbitals.
Filling Of Orbitals In Atom
The distribution of electrons into the atomic orbitals of an atom follows specific rules, collectively used in the Aufbau principle ("building up" principle).
The filling order is based on:
- Aufbau Principle: In the ground state of an atom, electrons occupy the orbitals in order of increasing energy. Electrons fill the lowest energy orbitals available first before entering higher energy ones. The approximate energy order is typically: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s, ... (This order can be visualized using a diagram).
- Pauli Exclusion Principle: No two electrons in the same atom can have the identical set of all four quantum numbers ($n, l, m_l, m_s$). An equivalent statement is: An atomic orbital can hold a maximum of two electrons, and these two electrons must have opposite spins ($m_s = +1/2$ and $m_s = -1/2$). This principle limits the number of electrons per orbital and per subshell.
- Hund's Rule of Maximum Multiplicity: For degenerate orbitals (orbitals within the same subshell, having the same energy), electron pairing does not occur until each orbital in the subshell is singly occupied with electrons of parallel spins. This rule maximizes the total spin angular momentum, leading to higher multiplicity (number of unpaired electrons). This preference for unpaired electrons in separate orbitals within a subshell minimizes electron-electron repulsion.
Electronic Configuration Of Atoms
The electronic configuration of an atom describes how its electrons are distributed among the atomic orbitals. It reflects the lowest energy arrangement of electrons in the ground state.
Electronic configurations can be represented in two ways:
- $ns^a np^b ...$ notation: The principal quantum number ($n$) is followed by the subshell symbol (s, p, d, f), and a superscript indicates the number of electrons in that subshell (e.g., 1s², 2p³).
- Orbital diagram: Each orbital is represented by a box or circle, and electrons are shown as arrows (↑ for one spin, ↓ for the opposite spin). This representation explicitly shows the number of electrons and their spins in each orbital.
Filling examples:
- Hydrogen (Z=1): 1 electron. Configuration: 1s¹. Orbital diagram: [↑] (1s)
- Helium (Z=2): 2 electrons. Configuration: 1s². Orbital diagram: [↑↓] (1s) (Pauli principle: two electrons in 1s with opposite spins)
- Lithium (Z=3): 3 electrons. 1s is full. Next is 2s. Configuration: 1s² 2s¹. Orbital diagram: [↑↓] (1s) [↑] (2s)
- Beryllium (Z=4): 4 electrons. Configuration: 1s² 2s². Orbital diagram: [↑↓] (1s) [↑↓] (2s)
- Boron (Z=5): 5 electrons. 2s is full. Next are 2p orbitals. Configuration: 1s² 2s² 2p¹. Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑][ ][ ] (2p) (Electron goes into one of the 2p orbitals)
- Carbon (Z=6): 6 electrons. Configuration: 1s² 2s² 2p². Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑][↑][ ] (2p) (Hund's rule: second electron goes into a different 2p orbital with parallel spin)
- Nitrogen (Z=7): 7 electrons. Configuration: 1s² 2s² 2p³. Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑][↑][↑] (2p) (Hund's rule: all three 2p orbitals singly occupied with parallel spins - half-filled 2p)
- Oxygen (Z=8): 8 electrons. Configuration: 1s² 2s² 2p⁴. Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑↓][↑][↑] (2p) (Hund's rule: pairing starts in 2p)
- Fluorine (Z=9): 9 electrons. Configuration: 1s² 2s² 2p⁵. Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑↓][↑↓][↑] (2p)
- Neon (Z=10): 10 electrons. Configuration: 1s² 2s² 2p⁶. Orbital diagram: [↑↓] (1s) [↑↓] (2s) [↑↓][↑↓][↑↓] (2p) (Fully filled 2p)
For heavier elements, the electronic configuration can be simplified by using the configuration of the preceding noble gas in square brackets, representing the core electrons (electrons in filled shells). The remaining electrons are valence electrons (in the outermost shell or incompletely filled subshells).
Example: Sodium (Z=11): [Ne] 3s¹ (Neon's configuration is 1s² 2s² 2p⁶)
Potassium (Z=19): [Ar] 4s¹ (Argon's configuration is 1s² 2s² 2p⁶ 3s² 3p⁶). Note that 4s is filled before 3d according to the Aufbau order.
There are some exceptions to the strict Aufbau order, particularly involving d and f subshells, due to the extra stability associated with completely filled or half-filled subshells.
Stability Of Completely Filled And Half Filled Subshells
In certain cases, like Chromium (Cr, Z=24) and Copper (Cu, Z=29), the observed electronic configuration deviates from the strict Aufbau filling order (which would predict Cr: [Ar] 3d⁴ 4s² and Cu: [Ar] 3d⁹ 4s²).
- Actual configuration of Cr: [Ar] 3d⁵ 4s¹
- Actual configuration of Cu: [Ar] 3d¹⁰ 4s¹
This occurs because completely filled and exactly half-filled subshells have extra stability. This enhanced stability is attributed to two main factors:
- Symmetrical Distribution of Electrons: Half-filled and completely filled subshells have electrons symmetrically distributed among the degenerate orbitals. Symmetry leads to stability. Electrons in the same subshell screen each other less effectively, allowing them to be more strongly attracted by the nucleus.
- Exchange Energy: Electrons with the same spin in degenerate orbitals can exchange their positions. Each such exchange releases energy, called exchange energy, which has a stabilizing effect. The number of possible exchanges is maximized when the subshell is exactly half-filled or completely filled. Thus, the exchange energy is highest, leading to maximum stability for these configurations.
For example, in a d⁵ configuration (like in Cr³⁺ or half-filled 3d in Cr), there are 10 possible exchanges among the five electrons with parallel spin. In a d¹⁰ configuration (like in Cu⁺ or filled 3d in Cu), there are exchanges possible among the five spin-up electrons and 10 exchanges among the five spin-down electrons, leading to even greater stability.
The chemical behavior and properties of elements are largely determined by their electronic configurations, particularly the arrangement of valence electrons.
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?
$Na^+, K^+, Mg^{2+}, Ca^{2+}, S^{2-}, Ar$.
Answer:
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] $4s^2 3d^1$.
Answer:
Question 2.24. What is the lowest value of n that allows g orbitals to exist?
Answer:
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.
Answer:
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.
Answer:
Question 2.27. Give the number of electrons in the species $H^+_2$, $H_2$ and $O^+_2$
Answer:
Question 2.28.
(i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Answer:
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.
Answer:
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, $m_s$ = +1/2
(b) n = 1, l = 0, $m_l$ = 0, $m_s$ = –1/2
(c) n = 1, l = 1, $m_l$ = 0, $m_s$ = +1/2
(d) n = 2, l = 1, $m_l$ = 0, $m_s$ = –1/2
(e) n = 3, l = 3, $m_l$ = –3, $m_s$ = +1/2
(f) n = 3, l = 1, $m_l$ = 0, $m_s$ = +1/2
Answer:
Question 2.31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, $m_s$ = –1/2
(b) n = 3, l = 0
Answer:
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.
Answer:
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 ?
Answer:
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$^{-1}$
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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 ?
Answer:
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.
Answer:
Question 2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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 |
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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
Answer:
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 ?
Answer:
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.
Answer:
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 ?
Answer:
Question 2.66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer:
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 ?
Answer: