Menu Top
Latest Science NCERT Notes and Solutions (Class 6th to 10th)
6th 7th 8th 9th 10th
Latest Science NCERT Notes and Solutions (Class 11th)
Physics Chemistry Biology
Latest Science NCERT Notes and Solutions (Class 12th)
Physics Chemistry Biology

Class 12th (Physics) Chapters
1. Electric Charges And Fields 2. Electrostatic Potential And Capacitance 3. Current Electricity
4. Moving Charges And Magnetism 5. Magnetism And Matter 6. Electromagnetic Induction
7. Alternating Current 8. Electromagnetic Waves 9. Ray Optics And Optical Instruments
10. Wave Optics 11. Dual Nature Of Radiation And Matter 12. Atoms
13. Nuclei 14. Semiconductor Electronics: Materials, Devices And Simple Circuits



12 Atoms



Introduction

By the 19th century, scientific evidence strongly supported the concept of matter being composed of atoms. A key development was J.J. Thomson's experiments on electric discharge through gases in 1897, which led to the discovery of the electron. These experiments showed that atoms of different elements contain identical negatively charged particles (electrons). Since atoms are electrically neutral overall, they must also contain a positive charge to balance the negative charge of the electrons.

The central question then became: How are the positive charge and electrons arranged within an atom? The first attempt to answer this came from J.J. Thomson himself in 1898, with his "plum pudding" model. In this model, the atom was imagined as a sphere of uniformly distributed positive charge, with the negatively charged electrons embedded within it, much like seeds in a watermelon.

However, this initial model did not align with later experimental findings about atomic structure. Furthermore, studies of light emitted by substances provided crucial insights. Condensed matter (solids and liquids) and dense gases emit a continuous spectrum of electromagnetic radiation when heated. In contrast, rarefied gases when heated or electrically excited (like in neon signs) emit radiation only at certain distinct wavelengths, producing a line spectrum (a series of bright lines on a dark background). Each element was found to have its own unique characteristic line spectrum, acting as a "fingerprint".

This relationship between the emitted spectrum and the element's identity suggested a deep connection between the internal structure of an atom and the specific wavelengths of light it can emit. For the simplest element, hydrogen, Johann Jakob Balmer in 1885 found a simple formula that described the wavelengths of a series of lines in its visible spectrum. This empirical formula hinted at an underlying order within the hydrogen atom.

A pivotal experiment to probe the internal structure of the atom was conceived by Ernest Rutherford, a former student of J.J. Thomson. He proposed scattering alpha ($\alpha$)-particles off atoms to investigate their internal arrangement. This experiment, performed by Hans Geiger and Ernest Marsden around 1911, led to significant conclusions about the distribution of mass and charge within the atom, fundamentally altering the view of atomic structure.



Alpha-Particle Scattering And Rutherford’s Nuclear Model Of Atom

To investigate the structure of the atom, Geiger and Marsden, under Rutherford's guidance, conducted experiments involving the scattering of alpha ($\alpha$)-particles by thin metal foils. Alpha particles, known to be positively charged helium nuclei ($^4_2\text{He}^{2+}$), were emitted from a radioactive source (like $^{214}_{83}\text{Bi}$). A beam of these $\alpha$-particles, with kinetic energy around 5.5 MeV, was directed onto a very thin sheet of gold foil (about $2.1 \times 10^{-7}$ m thick) (

Geiger-Marsden scattering experiment setup diagram
Figure 12.2).

The scattered $\alpha$-particles were detected using a rotatable detector comprising a zinc sulfide (ZnS) screen and a microscope. When an $\alpha$-particle struck the ZnS screen, it produced a brief flash of light (scintillation) that could be observed through the microscope. By rotating the detector, they could measure the number of $\alpha$-particles scattered at different angles relative to the original beam direction.

The results of the experiment were surprising and did not fit with the Thomson model:

The large-angle scattering was particularly astonishing. Rutherford famously remarked that it was "quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Rutherford interpreted these results by proposing a new model of the atom, known as the Rutherford nuclear model or the planetary model. According to this model:

The large empty space explained why most $\alpha$-particles passed through undeflected. The large deflections and backscattering of the few $\alpha$-particles that did not pass straight through could be explained by a direct collision or close encounter with the small, massive, positively charged nucleus. The strong electrostatic repulsion between the positive $\alpha$-particle and the positive nucleus would cause a large deflection. The light electrons would not significantly affect the trajectory of the much heavier $\alpha$-particles.


Alpha-Particle Trajectory

The path of an $\alpha$-particle as it interacts with the nucleus can be analyzed using Coulomb's law and Newton's second law. The electrostatic force of repulsion between the $\alpha$-particle (charge $2e$) and the nucleus (charge $Ze$, where Z is the atomic number) at a distance $r$ is given by $F = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r^2}$. Since the gold nucleus is much heavier than the $\alpha$-particle, it is assumed to remain stationary during the interaction.

The trajectory of an $\alpha$-particle is determined by its impact parameter ($b$), which is the perpendicular distance between the initial velocity vector of the $\alpha$-particle and the center of the nucleus (

Impact parameter and scattering angle diagram
Figure 12.4). For a given initial kinetic energy, the scattering angle ($\theta$) depends on $b$:

The small number of $\alpha$-particles scattered at large angles or backscattered implies that head-on or near head-on collisions (small $b$) are rare. This is consistent with the nucleus being a very small target compared to the overall size of the atom. By analyzing the number of scattered particles as a function of angle, Rutherford was able to estimate the size of the nucleus and confirm that it is indeed extremely small.

The distance of closest approach ($d$) in a head-on collision ($b \approx 0$) occurs when the $\alpha$-particle momentarily stops. At this point, its initial kinetic energy is completely converted into electrostatic potential energy. $K_{initial} = U_{final} = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{d}$. This allows for calculating an upper limit for the size of the nucleus.

Example 12.1. In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10–15 m) is analogous to the sun about which the electron move in orbit (radius » 10–10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth’s orbit is about 1.5 ´ 1011 m. The radius of sun is taken as 7 ´ 108 m.

Answer:

In the Rutherford model, the ratio of the electron orbit radius to the nucleus radius is approximately $\frac{10^{-10} \text{ m}}{10^{-15} \text{ m}} = 10^5$. This means the electron orbit is about 100,000 times larger than the nucleus.

Let's apply this same ratio to the solar system, using the sun's radius as the analogue for the nucleus's radius. The given radius of the sun is $R_{sun} = 7 \times 10^8 \text{ m}$. If the Earth's orbit had a radius $R_{earth-orbit}$ that was $10^5$ times the sun's radius, then:

$$R_{earth-orbit\_analogue} = 10^5 \times R_{sun} = 10^5 \times (7 \times 10^8 \text{ m}) = 7 \times 10^{13} \text{ m}$$

The actual radius of the Earth's orbit is given as approximately $1.5 \times 10^{11} \text{ m}$.

Comparing the analogue orbital radius to the actual orbital radius:

$$7 \times 10^{13} \text{ m} \gg 1.5 \times 10^{11} \text{ m}$$

The analogue orbital radius ($7 \times 10^{13}$ m) is more than 100 times larger than the actual orbital radius ($1.5 \times 10^{11}$ m).

Therefore, if the solar system had the same scale proportions as the atom, the Earth would be much farther away from the Sun than it is in reality. This analogy highlights that atoms consist of a much larger fraction of empty space than our solar system does.

Example 12.2. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV a-particle before it comes momentarily to rest and reverses its direction?

Answer:

The situation described is a head-on collision ($b=0$), where the $\alpha$-particle approaches the nucleus directly. At the point of closest approach ($d$), the $\alpha$-particle momentarily stops before being repelled back. At this point, the initial kinetic energy ($K$) of the $\alpha$-particle is completely converted into electrostatic potential energy ($U$) between the $\alpha$-particle and the nucleus. The principle of conservation of mechanical energy applies.

Initial energy $E_i = K$ (potential energy is zero when the $\alpha$-particle is far away).

Final energy $E_f = U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d}$, where $q_1 = 2e$ (charge of $\alpha$-particle) and $q_2 = Ze$ (charge of gold nucleus, $Z=79$).

By conservation of energy, $E_i = E_f$:

$$K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{d}$$

We need to find the distance of closest approach $d$. Rearranging the equation:

$$d = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$$

Given initial kinetic energy $K = 7.7 \text{ MeV}$. We convert this to Joules:

$K = 7.7 \text{ MeV} \times (1.602 \times 10^{-19} \text{ J/eV}) \times (10^6 \text{ eV/MeV}) = 1.23354 \times 10^{-12} \text{ J}$. For simplicity, let's use $1.2 \times 10^{-12}$ J as in the provided original solution calculation.

The Coulomb constant $\frac{1}{4\pi\epsilon_0} \approx 9.0 \times 10^9 \text{ N m}^2/\text{C}^2$.

The elementary charge $e = 1.602 \times 10^{-19} \text{ C}$. Let's use $1.6 \times 10^{-19}$ C.

For gold, the atomic number $Z = 79$.

Substituting these values:

$$d = (9.0 \times 10^9 \text{ N m}^2/\text{C}^2) \times \frac{2 \times 79 \times (1.6 \times 10^{-19} \text{ C})^2}{1.2 \times 10^{-12} \text{ J}}$$

$$d = (9.0 \times 10^9) \times \frac{158 \times (2.56 \times 10^{-38})}{1.2 \times 10^{-12}} \text{ m}$$

$$d = (9.0 \times 10^9) \times \frac{404.48 \times 10^{-38}}{1.2 \times 10^{-12}} \text{ m}$$

$$d = (9.0 \times 10^9) \times (337.067 \times 10^{-26}) \text{ m}$$

$$d = 3033.6 \times 10^{-17} \text{ m} = 3.0336 \times 10^{-14} \text{ m}$$

Using the values from the provided original solution calculation (which might use slightly different constants): $d = 3.84 \times 10^{-16} Z \text{ m}$

For $Z=79$, $d = 3.84 \times 10^{-16} \times 79 \text{ m} = 303.36 \times 10^{-16} \text{ m} = 3.0336 \times 10^{-14} \text{ m}$.

The distance of closest approach is approximately $3.03 \times 10^{-14} \text{ m}$ or 30.3 fm (femtometers), where 1 fm = $10^{-15}$ m.

This value provides an estimate of the maximum possible radius of the gold nucleus (since the $\alpha$-particle did not penetrate it). The actual radius of the gold nucleus is known to be around 6 fm, which is smaller than this calculated closest approach distance, confirming that in these scattering events, the $\alpha$-particle is indeed repelled before "touching" the nucleus.


Electron Orbits

In Rutherford's nuclear model, electrons orbit the positively charged nucleus. For a stable orbit, the centripetal force required to keep the electron in a circular path is provided by the electrostatic force of attraction between the negatively charged electron ($e$) and the positively charged nucleus ($Ze$). For a hydrogen atom ($Z=1$), the nucleus is a proton (charge $e$).

Electrostatic force of attraction $F_e = \frac{1}{4\pi\epsilon_0} \frac{e \cdot e}{r^2} = \frac{e^2}{4\pi\epsilon_0 r^2}$, where $r$ is the orbital radius.

Centripetal force $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the electron and $v$ is its orbital speed.

For a dynamically stable orbit: $F_e = F_c$

$$\frac{e^2}{4\pi\epsilon_0 r^2} = \frac{mv^2}{r}$$

This equation relates the orbital radius and velocity: $mv^2 = \frac{e^2}{4\pi\epsilon_0 r}$.

The kinetic energy of the electron is $K = \frac{1}{2}mv^2 = \frac{1}{2} \left(\frac{e^2}{4\pi\epsilon_0 r}\right) = \frac{e^2}{8\pi\epsilon_0 r}$.

The electrostatic potential energy of the electron-nucleus system is $U = \frac{1}{4\pi\epsilon_0} \frac{(-e)(e)}{r} = -\frac{e^2}{4\pi\epsilon_0 r}$. (Potential energy is negative because the force is attractive).

The total mechanical energy ($E$) of the electron in the orbit is the sum of its kinetic and potential energies:

$$E = K + U = \frac{e^2}{8\pi\epsilon_0 r} - \frac{e^2}{4\pi\epsilon_0 r} = \frac{e^2}{8\pi\epsilon_0 r} - \frac{2e^2}{8\pi\epsilon_0 r} = -\frac{e^2}{8\pi\epsilon_0 r}$$

The total energy $E$ is negative. This indicates that the electron is bound to the nucleus. Positive total energy would mean the electron is not bound and would escape.

However, the Rutherford model faced fundamental problems when viewed from the perspective of classical electromagnetic theory:

1. Atomic Stability: According to classical electromagnetism, an accelerating charged particle, like an electron moving in a circular orbit, must continuously emit electromagnetic radiation and lose energy. As the electron loses energy, its orbit radius would decrease, causing it to spiral inwards and eventually fall into the nucleus. This would lead to the atom's collapse, contradicting the observed stability of atoms.

2. Atomic Spectra: If the electron were to continuously spiral into the nucleus, its orbital frequency would change continuously. According to classical theory, the frequency of the emitted radiation should equal the orbital frequency. Thus, the atom should emit radiation with a continuous range of frequencies (a continuous spectrum). This directly contradicts the observed line spectra, where atoms emit light only at specific, discrete wavelengths/frequencies.

Clearly, classical physics was insufficient to explain the observed properties of atoms.

Example 12.3. It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

Answer:

The energy required to separate a hydrogen atom into a proton and an electron (ionisation energy) is the energy needed to remove the electron from its bound state in the atom to a state where it is free and at rest (energy = 0). Since the total energy of a bound electron is negative, the ionisation energy is the absolute value of the electron's total energy in its ground state.

So, the total energy of the electron in a hydrogen atom is $E = -13.6 \text{ eV}$. We need to convert this energy to Joules:

$E = -13.6 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) = -2.17872 \times 10^{-18} \text{ J}$. Let's use $-2.2 \times 10^{-18}$ J as in the provided solution.

From the Rutherford model (which we are using here to calculate these classical values), the total energy is given by $E = -\frac{e^2}{8\pi\epsilon_0 r}$.

We can find the orbital radius $r$ using this equation:

$$r = -\frac{e^2}{8\pi\epsilon_0 E} = \frac{1}{4\pi\epsilon_0} \frac{-e^2}{2E}$$

Using $\frac{1}{4\pi\epsilon_0} \approx 9.0 \times 10^9 \text{ N m}^2/\text{C}^2$, $e = 1.602 \times 10^{-19} \text{ C}$, and $E = -2.17872 \times 10^{-18} \text{ J}$:

$$r = (9.0 \times 10^9) \times \frac{-(1.602 \times 10^{-19})^2}{2 \times (-2.17872 \times 10^{-18})} \text{ m}$$

$$r = (9.0 \times 10^9) \times \frac{-2.5664 \times 10^{-38}}{-4.35744 \times 10^{-18}} \text{ m}$$

$$r = (9.0 \times 10^9) \times (0.5889 \times 10^{-20}) \text{ m}$$

$$r \approx 5.30 \times 10^{-11} \text{ m}$$

This radius is approximately the Bohr radius ($a_0$), which is the radius of the lowest energy orbit in Bohr's model.

Now, we can find the velocity $v$. From the centripetal force balance, $mv^2 = \frac{e^2}{4\pi\epsilon_0 r}$.

$$v^2 = \frac{e^2}{4\pi\epsilon_0 m r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r}$$

Using $m_e = 9.109 \times 10^{-31} \text{ kg}$, $e = 1.602 \times 10^{-19} \text{ C}$, $r = 5.30 \times 10^{-11} \text{ m}$, and $\frac{1}{4\pi\epsilon_0} \approx 9.0 \times 10^9$:

$$v^2 = (9.0 \times 10^9) \times \frac{(1.602 \times 10^{-19})^2}{(9.109 \times 10^{-31}) \times (5.30 \times 10^{-11})} \text{ (m/s)}^2$$

$$v^2 = (9.0 \times 10^9) \times \frac{2.5664 \times 10^{-38}}{4.8278 \times 10^{-41}} \text{ (m/s)}^2$$

$$v^2 = (9.0 \times 10^9) \times (0.5316 \times 10^3) \text{ (m/s)}^2 = 4.784 \times 10^{12} \text{ (m/s)}^2$$

$$v = \sqrt{4.784 \times 10^{12}} \text{ m/s} \approx 2.187 \times 10^6 \text{ m/s}$$

The orbital radius is approximately $5.3 \times 10^{-11}$ m and the velocity of the electron is approximately $2.19 \times 10^6$ m/s.

Example 12.4. According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.

Answer:

According to classical electromagnetic theory, an electron revolving in an orbit emits radiation at a frequency equal to its frequency of revolution. We can calculate the frequency of revolution ($n_{rev}$) using the electron's velocity ($v$) and orbital radius ($r$). The frequency of revolution is given by $n_{rev} = \frac{\text{speed}}{\text{circumference}} = \frac{v}{2\pi r}$.

From Example 12.3, for the electron in a hydrogen atom (using classical physics approximation for the ground state), the orbital radius is $r \approx 5.3 \times 10^{-11}$ m and the velocity is $v \approx 2.19 \times 10^6$ m/s.

$$n_{rev} = \frac{2.19 \times 10^6 \text{ m/s}}{2\pi \times (5.3 \times 10^{-11} \text{ m})}$$

$$n_{rev} = \frac{2.19 \times 10^6}{33.30 \times 10^{-11}} \text{ Hz}$$

$$n_{rev} \approx 0.0657 \times 10^{17} \text{ Hz} = 6.57 \times 10^{15} \text{ Hz}$$

According to classical electromagnetic theory, the frequency of the emitted light would be equal to this frequency of revolution.

Thus, the initial frequency of light emitted by the classically revolving electron in a hydrogen atom is approximately $6.57 \times 10^{15}$ Hz. As the electron loses energy and spirals inwards, its velocity and radius would change, causing the emitted frequency to change continuously, leading to a continuous spectrum.



Atomic Spectra

As discussed earlier, when a rarefied atomic gas is excited (e.g., by heating or electric discharge), it emits light at specific, discrete wavelengths. Observing this light through a spectrometer produces an emission line spectrum, which appears as a series of bright lines against a dark background. Each element has a unique set of spectral lines.

Conversely, when white light (containing a continuous range of wavelengths) passes through a cold sample of a gas, the gas atoms absorb light at those same specific wavelengths that they would normally emit when excited. Analyzing the transmitted light reveals a continuous spectrum with dark lines at the wavelengths that were absorbed. This is called the absorption line spectrum.

The fact that atoms emit or absorb light only at discrete wavelengths strongly suggests that the energy states within an atom are also discrete or quantized. This provided a crucial clue to understanding the internal structure and behavior of atoms, particularly how they interact with electromagnetic radiation.



Bohr Model Of The Hydrogen Atom

Niels Bohr, working in Rutherford's laboratory, recognized the limitations of the classical Rutherford model, particularly its inability to explain atomic stability and the existence of line spectra. In 1913, Bohr proposed a model for the hydrogen atom by incorporating concepts from the emerging quantum theory. His model is based on three postulates:

(i) Bohr's First Postulate (Postulate of Stable Orbits): An electron in an atom can revolve around the nucleus in certain specific, discrete orbits, called stationary orbits or stable states, without emitting any electromagnetic radiation. Each such stationary state has a definite, fixed total energy. This directly contradicts the classical prediction of continuous energy loss by an orbiting electron.

(ii) Bohr's Second Postulate (Postulate of Quantisation of Angular Momentum): The electron revolves around the nucleus only in those stationary orbits for which its angular momentum ($L$) is an integer multiple of $h/2\pi$.

$$L_n = mvr_n = n \frac{h}{2\pi} \quad \text{where } n = 1, 2, 3, \dots$$

Here, $m$ is the electron mass, $v$ is its speed, $r_n$ is the radius of the $n$-th allowed orbit, $h$ is Planck's constant, and $n$ is an integer called the principal quantum number. This postulate introduces the concept of quantization directly into the atomic model.

(iii) Bohr's Third Postulate (Postulate of Energy Transitions): An electron can transition (jump) from one stationary orbit of higher energy ($E_i$) to another stationary orbit of lower energy ($E_f$). When it does so, the energy difference is emitted as a single photon of frequency $\nu$.

$$h\nu = E_i - E_f \quad \text{where } E_i > E_f$$

Conversely, an atom can absorb a photon of precisely the same energy $h\nu = E_f - E_i$ to cause an electron to jump from a lower energy state ($E_i$) to a higher energy state ($E_f$).

Bohr used these postulates, along with classical mechanics for the forces, to derive expressions for the radii of the allowed orbits and the energies of the stationary states in a hydrogen atom ($Z=1$).

From the centripetal force condition $\frac{mv_n^2}{r_n} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r_n^2}$ and the angular momentum quantization $mv_n r_n = n \frac{h}{2\pi}$, he solved for $r_n$ and $v_n$. The radius of the $n$-th stationary orbit is found to be:

$$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$$

Substituting the fundamental constants, the radius of the first Bohr orbit ($n=1$) is $r_1 = a_0 \approx 5.29 \times 10^{-11}$ m, known as the Bohr radius. The radii of higher orbits are $r_n = n^2 a_0$. Thus, the allowed orbits are discrete and depend on the square of the principal quantum number $n$.

The total energy of the electron in the $n$-th stationary state is given by substituting $r_n$ into the energy expression $E_n = -\frac{e^2}{8\pi\epsilon_0 r_n}$:

$$E_n = -\frac{m e^4}{8 \epsilon_0^2 h^2} \frac{1}{n^2}$$

Substituting the values of the constants, the energy levels for the hydrogen atom are approximately:

$$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \text{ J}$$

Converting to electron volts ($1 \text{ eV} \approx 1.602 \times 10^{-19} \text{ J}$):

$$E_n = -\frac{13.6}{n^2} \text{ eV}$$

The negative sign signifies that the electron is bound to the nucleus. The state with $n=1$ has the lowest (most negative) energy and is called the ground state ($E_1 = -13.6$ eV). States with $n > 1$ are called excited states, having progressively higher (less negative) energies ($E_2 = -3.4$ eV, $E_3 = -1.51$ eV, etc.). The energy levels become closer together as $n$ increases.

The state with $n=\infty$ corresponds to $E_\infty = 0$ eV, which represents a free electron infinitely far from the nucleus. The energy required to remove an electron from the ground state of a hydrogen atom (ionise it) is $0 - (-13.6 \text{ eV}) = 13.6 \text{ eV}$. This theoretical prediction matches the experimental value extremely well.


Energy Levels

The possible discrete energies that an electron in a hydrogen atom can have are called its energy levels. These levels are determined by the principal quantum number $n=1, 2, 3, \dots$. The energy level diagram for hydrogen visualizes these allowed energies as horizontal lines (

Hydrogen atom energy level diagram
Figure 12.7).

An atom in the ground state can be excited to a higher energy state by absorbing energy, typically through collisions with other particles or by absorbing a photon with energy equal to the energy difference between the initial and final states. Excited states are unstable, and the electron quickly transitions back to a lower energy state, emitting a photon in the process.

Example 12.5. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Answer:

The hydrogen atom is initially in the ground state, which corresponds to the principal quantum number $n_i = 1$. It absorbs a photon and is excited to the $n_f = 4$ level.

The energy of the electron in the $n$-th state of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.

Initial energy level ($n_i = 1$): $E_1 = -\frac{13.6}{1^2} \text{ eV} = -13.6 \text{ eV}$.

Final energy level ($n_f = 4$): $E_4 = -\frac{13.6}{4^2} \text{ eV} = -\frac{13.6}{16} \text{ eV} = -0.85 \text{ eV}$.

When an atom absorbs a photon and transitions to a higher energy level, the energy of the absorbed photon ($h\nu$) is equal to the difference between the final and initial energy levels: $h\nu = E_f - E_i$.

Energy of the absorbed photon $h\nu = E_4 - E_1 = (-0.85 \text{ eV}) - (-13.6 \text{ eV}) = -0.85 + 13.6 \text{ eV} = 12.75 \text{ eV}$.

We can find the frequency ($\nu$) using $h\nu = 12.75 \text{ eV}$. First, convert the energy to Joules:

Photon energy $E_{photon} = 12.75 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV}) = 2.04255 \times 10^{-18} \text{ J}$.

Frequency $\nu = \frac{E_{photon}}{h} = \frac{2.04255 \times 10^{-18} \text{ J}}{6.626 \times 10^{-34} \text{ J s}}$.

$$\nu \approx 0.3082 \times 10^{16} \text{ Hz} = 3.082 \times 10^{15} \text{ Hz}$$

The frequency of the absorbed photon is approximately $3.08 \times 10^{15}$ Hz.

We can find the wavelength ($\lambda$) using the relation $c = \nu\lambda$, where $c$ is the speed of light ($3.00 \times 10^8 \text{ m/s}$).

$$\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 \text{ m/s}}{3.082 \times 10^{15} \text{ Hz}}$$

$$\lambda \approx 0.9734 \times 10^{-7} \text{ m}$$

Converting to nanometers ($1 \text{ nm} = 10^{-9} \text{ m}$):

$$\lambda \approx 97.34 \times 10^{-9} \text{ m} = 97.34 \text{ nm}$$

The wavelength of the absorbed photon is approximately 97.3 nm. This falls in the ultraviolet region of the electromagnetic spectrum.



The Line Spectra Of The Hydrogen Atom

The key achievement of Bohr's model was its ability to explain the discrete line spectrum of hydrogen. According to Bohr's third postulate, spectral lines are produced when an electron makes a transition from a higher energy stationary state (with principal quantum number $n_i$) to a lower energy stationary state (with principal quantum number $n_f$, where $n_f < n_i$). The energy difference between these two states is emitted as a single photon with frequency $\nu$ given by:

$$h\nu = E_{n_i} - E_{n_f}$$

Substituting the expression for the energy levels $E_n = -\frac{m e^4}{8 \epsilon_0^2 h^2} \frac{1}{n^2}$, the frequency of the emitted photon is:

$$\nu = \frac{1}{h} \left(E_{n_i} - E_{n_f}\right) = \frac{1}{h} \left(-\frac{m e^4}{8 \epsilon_0^2 h^2 n_i^2} - \left(-\frac{m e^4}{8 \epsilon_0^2 h^2 n_f^2}\right)\right)$$

$$\nu = \frac{m e^4}{8 \epsilon_0^2 h^3} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$

The speed of light $c = \nu\lambda$, so the wavelength $\lambda = c/\nu$.

$$\frac{1}{\lambda} = \frac{\nu}{c} = \frac{m e^4}{8 \epsilon_0^2 h^3 c} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$$

The constant term $\frac{m e^4}{8 \epsilon_0^2 h^3 c}$ is called the Rydberg constant (R) for hydrogen. Its theoretical value is approximately $1.097 \times 10^7 \text{ m}^{-1}$. So, the formula for the wavelength of spectral lines in hydrogen is:

$$\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \quad \text{where } n_f < n_i \text{ are integers.}$$

This formula successfully explains the various series of lines observed in the hydrogen spectrum. Different series correspond to transitions ending in specific lower energy levels ($n_f$):

Each integer value of $n_f$ defines a series, and within each series, different values of $n_i$ produce distinct lines. Since $n_f$ and $n_i$ must be integers, the possible energy differences ($E_{n_i} - E_{n_f}$) are discrete, resulting in the emission of photons with only discrete frequencies (or wavelengths), thus explaining the observed line spectra. Bohr's explanation of the hydrogen spectrum was a monumental achievement in physics.

Absorption spectra occur when an atom in a lower energy state absorbs a photon with energy equal to the difference needed to reach a higher energy state. Only photons with these specific energies (wavelengths) can be absorbed, resulting in dark lines at those wavelengths in a continuous spectrum.



De Broglie’s Explanation Of Bohr’s Second Postulate Of Quantisation

While Bohr's postulates successfully explained the hydrogen spectrum, the second postulate – the quantization of angular momentum ($L = n \frac{h}{2\pi}$) – seemed arbitrary from a classical viewpoint. A physical justification for this quantization rule was provided by Louis de Broglie in 1923, building upon his hypothesis of matter waves.

De Broglie proposed that material particles, like electrons, also possess wave-like properties, with a wavelength $\lambda = h/p$, where $p$ is the particle's momentum. De Broglie suggested that the electron in a Bohr orbit should be thought of as a standing matter wave. For a standing wave to form in a circular orbit, the circumference of the orbit must contain an integer number of wavelengths. If this condition is not met, the wave would destructively interfere with itself over successive orbits, and its amplitude would quickly diminish.

So, for the electron wave to form a stable standing wave in the $n$-th orbit of radius $r_n$, the circumference ($2\pi r_n$) must be equal to an integer multiple of the electron's de Broglie wavelength ($\lambda_n$):

$$2\pi r_n = n \lambda_n \quad \text{where } n = 1, 2, 3, \dots$$

Substituting de Broglie's relation $\lambda_n = h/p_n = h/(mv_n)$ (assuming non-relativistic speed):

$$2\pi r_n = n \frac{h}{mv_n}$$

Rearranging this equation:

$$mv_n r_n = n \frac{h}{2\pi}$$

Since $mv_n r_n$ is the magnitude of the angular momentum ($L_n$) of the electron in the $n$-th orbit, this equation becomes:

$$L_n = n \frac{h}{2\pi}$$

This is precisely Bohr's second postulate. Thus, de Broglie's hypothesis provided a natural and intuitive explanation for why only certain orbits are allowed in the atom: they are the orbits where the electron's matter wave forms a stable standing wave pattern (

Electron standing wave in orbit
Figure 12.8).

De Broglie's explanation linked the wave nature of the electron to the quantization of its angular momentum and energy levels, reinforcing the idea of wave-particle duality and paving the way for modern quantum mechanics.

Despite its success for hydrogen, Bohr's model has limitations:

Although superseded by the more complete theory of quantum mechanics, the Bohr model remains historically important as it successfully introduced quantization into atomic structure and provided a simple, intuitive picture that correctly predicted many features of the hydrogen atom.



Exercises



Question 12.1. Choose the correct alternative from the clues given at the end of the each statement:

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)

(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

Answer:

Question 12.2. Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

Answer:

Question 12.3. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Answer:

Question 12.4. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Answer:

Question 12.5. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Answer:

Question 12.6. (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Answer:

Question 12.7. The radius of the innermost electron orbit of a hydrogen atom is $5.3\times10^{–11}\text{ m}$. What are the radii of the n = 2 and n =3 orbits?

Answer:

Question 12.8. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer:

Question 12.9. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius $1.5 \times 10^{11}\text{ m}$ with orbital speed $3 \times 10^4 \text{ m/s}$. (Mass of earth $= 6.0 \times 10^{24}\text{ kg}$.)

Answer: