NCERT Class 12 Physics — Atoms, Chapter 12 Exercises. All 9 questions solved.

The exercises at the end of Chapter 12 fall into two groups. Questions 12.1 and 12.2 compare Thomson’s and Rutherford’s models of the atom and ask what alpha-particle scattering can reveal. Questions 12.3 to 12.9 use Bohr’s model of the hydrogen atom: its energy levels, the photons emitted or absorbed between them, and the radii, speeds and periods of its orbits. The arithmetic runs on three results from the chapter:

• Bohr’s quantisation condition — the angular momentum of the electron is $L = mvr = \dfrac{nh}{2\pi}$, with $n = 1, 2, 3, \ldots$ (Eq. 12.5). Combined with the Coulomb force supplying the centripetal force, it gives orbits of radius $r_n = n^2r_1$, with $r_1 = 5.3\times10^{-11}\ \text{m}$ (Eq. 12.7).
• The energy levels of hydrogen — $E_n = -\dfrac{13.6}{n^2}\ \text{eV}$ (Eq. 12.10).
• Bohr’s frequency condition — a jump between levels emits or absorbs a single photon of energy $h\nu = E_i – E_f$ (Eq. 12.6), with $h = 6.63\times10^{-34}\ \text{J s}$ and $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

Key insight. Nearly everything in Bohr’s model follows from two equations: the Coulomb attraction supplying the centripetal force, and angular momentum coming in steps of $h/2\pi$. Together they fix how every quantity scales with $n$ — the radius as $n^2$, the speed as $1/n$, the period as $n^3$ and the energy as $-1/n^2$. Each numerical question here (12.4 to 12.9) is either one of those scalings or an energy difference turned into a photon.

Question 12.1

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

Question 12.1 (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)

Solution. Both models have to describe atoms of the size actually measured, about $10^{-10}\ \text{m}$ across, so the overall size is the same in both. What differs is the arrangement inside. Thomson spreads the positive charge through a sphere as large as the whole atom, with the electrons embedded in it. Rutherford packs the positive charge into a nucleus about $10^{-15}\ \text{m}$ across, with the electrons moving round it at atomic distances. The nucleus is tiny; the atom is not.

Question 12.1 (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)

Solution. In Thomson’s model each electron sits at rest inside the sphere of positive charge, at a point where the forces on it balance. If it is displaced, the positive charge pulls it back, so the equilibrium is stable and the net force is zero. In Rutherford’s model an electron at rest would simply fall into the nucleus, so the electrons must orbit it. Motion in a circle needs a centripetal force, and the Coulomb attraction of the nucleus supplies it — so there is always a net force on every electron.

Question 12.1 (c)

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

Solution. An electron moving in a circle is accelerating, and classical electromagnetic theory says that an accelerating charge radiates electromagnetic waves. The orbiting electron of Rutherford’s atom would therefore lose energy continuously and spiral into the nucleus. This is exactly the difficulty Bohr’s postulates were introduced to remove. Thomson’s electrons are at rest in equilibrium, so classically they do not radiate.

Question 12.1 (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)

Solution. In Thomson’s model the positive charge, and with it the mass, fills the whole atom evenly, so the mass is spread almost uniformly. In Rutherford’s model nearly all the mass is in a nucleus about $10^{-5}$ of the atom’s radius, and the rest of the atom is almost empty space — which is why most alpha particles went straight through the gold foil.

Question 12.1 (e)

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

Solution. Electrons are very light — about $1/1836$ of the mass of a proton — so in any model almost all the mass of the atom must go with the positive charge. In Thomson’s model that is the uniform positive sphere; in Rutherford’s it is the nucleus. The models differ in how that mass is spread out, not in which part carries it.

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 $14\ \text{K}$.) What results do you expect?

Solution. In Rutherford’s experiment, the large-angle scattering — including the few alpha particles that bounced straight back — came from close encounters with a gold nucleus, which is about fifty times as heavy as an alpha particle and carries a charge of $79e$. In solid hydrogen the nucleus is a single proton, and that changes the outcome.

The proton’s mass is $1.67\times10^{-27}\ \text{kg}$, only about a quarter of the alpha particle’s $6.64\times10^{-27}\ \text{kg}$. When a heavy particle strikes a lighter one at rest, conservation of momentum and energy do not allow it to reverse: even in a head-on collision it carries on forwards, pushing the light particle ahead of it, as a football does when it hits a tennis ball at rest. For an elastic collision the largest angle through which the heavier particle can be turned is given by

$$\sin\theta_{\max} = \frac{m_p}{m_\alpha} = \frac{1.67\times10^{-27}}{6.64\times10^{-27}} \approx 0.25,\qquad \theta_{\max}\approx15^\circ$$

The proton’s charge is also only $+e$, so its Coulomb push on a passing alpha particle is far weaker than a gold nucleus’s. The alpha particles would pass through the sheet with small deflections, and the experiment would show none of the large-angle scattering that revealed the nucleus.

Question 12.3

A difference of $2.3\ \text{eV}$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Solution. By Bohr’s third postulate the atom emits a single photon carrying away exactly the energy difference, $h\nu = E_i – E_f$. The energy has to be in joules before dividing by $h$: $2.3\ \text{eV} = 2.3\times1.6\times10^{-19} = 3.68\times10^{-19}\ \text{J}$.

$$\nu = \frac{E_i – E_f}{h} = \frac{3.68\times10^{-19}}{6.63\times10^{-34}} = 5.55\times10^{14}\ \text{Hz}$$

This is visible light, a wavelength of about $540\ \text{nm}$.

Question 12.4

The ground state energy of hydrogen atom is $-13.6\ \text{eV}$. What are the kinetic and potential energies of the electron in this state?

Solution. The chapter gives the two energies of an electron in an orbit of radius $r$ (just before Eq. 12.4):

$$K = \frac{e^2}{8\pi\varepsilon_0 r},\qquad U = -\frac{e^2}{4\pi\varepsilon_0 r}$$

So $U = -2K$, and the total energy is $E = K + U = -K$. That fixes both energies from the total alone, whatever the radius:

$$K = -E = 13.6\ \text{eV},\qquad U = 2E = -27.2\ \text{eV}$$

The kinetic energy is positive, as it must be, and the potential energy is twice as large and negative; their sum is the $-13.6\ \text{eV}$ we started with.

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.

Solution. A photon is absorbed only if its energy equals the gap between the two levels. From $E_n = -13.6/n^2\ \text{eV}$, $E_1 = -13.6\ \text{eV}$ and $E_4 = -13.6/16 = -0.85\ \text{eV}$, so

$$h\nu = E_4 – E_1 = -0.85 – (-13.6) = 12.75\ \text{eV} = 12.75\times1.6\times10^{-19} = 2.04\times10^{-18}\ \text{J}$$

$$\nu = \frac{2.04\times10^{-18}}{6.63\times10^{-34}} = 3.08\times10^{15}\ \text{Hz},\qquad \lambda = \frac{c}{\nu} = \frac{3\times10^8}{3.08\times10^{15}} = 9.75\times10^{-8}\ \text{m}$$

The wavelength is about $97\ \text{nm}$, in the ultraviolet. The answer key’s $9.7\times10^{-8}\ \text{m}$ is the same result; with more precise values of $h$, $c$ and $e$ it comes to $9.72\times10^{-8}\ \text{m}$.

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.

Question 12.6 (a)

Solution. The Coulomb attraction supplies the centripetal force (Eq. 12.3):

$$\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}\quad\Longrightarrow\quad (mvr)\,v = \frac{e^2}{4\pi\varepsilon_0}$$

Writing it this way puts the angular momentum $mvr$ in view, and Bohr’s condition fixes it at $nh/2\pi$. Substituting,

$$\frac{nh}{2\pi}\,v_n = \frac{e^2}{4\pi\varepsilon_0}\quad\Longrightarrow\quad v_n = \frac{e^2}{2\varepsilon_0 nh}$$

With $\varepsilon_0 = 8.854\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$:

$$v_1 = \frac{(1.6\times10^{-19})^2}{2(8.854\times10^{-12})(6.63\times10^{-34})} = \frac{2.56\times10^{-38}}{1.174\times10^{-44}} = 2.18\times10^6\ \text{m/s}$$

The speed falls as $1/n$, so $v_2 = v_1/2 = 1.09\times10^6\ \text{m/s}$ and $v_3 = v_1/3 = 7.27\times10^5\ \text{m/s}$.

Question 12.6 (b)

Solution. One orbit is a distance $2\pi r_n$ covered at speed $v_n$, so $T_n = 2\pi r_n/v_n$. From Eq. 12.7 the radius grows as $n^2$: $r_1 = 5.3\times10^{-11}\ \text{m}$, $r_2 = 2.12\times10^{-10}\ \text{m}$, $r_3 = 4.77\times10^{-10}\ \text{m}$.

$$T_1 = \frac{2\pi(5.3\times10^{-11})}{2.18\times10^6} = 1.53\times10^{-16}\ \text{s}$$

$$T_2 = \frac{2\pi(2.12\times10^{-10})}{1.09\times10^6} = 1.22\times10^{-15}\ \text{s},\qquad T_3 = \frac{2\pi(4.77\times10^{-10})}{7.27\times10^5} = 4.12\times10^{-15}\ \text{s}$$

Because $r_n \propto n^2$ and $v_n \propto 1/n$, the period grows as $n^3$: $T_2 = 8T_1$ and $T_3 = 27T_1$, which the figures bear out. The answer key’s $1.52\times10^{-16}$ and $4.11\times10^{-15}\ \text{s}$ differ from these only in the last digit, through rounding.

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?

Solution. In Eq. 12.7, $r_n = \left(\dfrac{n^2}{m}\right)\left(\dfrac{h}{2\pi}\right)^2\dfrac{4\pi\varepsilon_0}{e^2}$, everything except $n^2$ is a constant, so $r_n = n^2r_1$:

$$r_2 = 4\times5.3\times10^{-11} = 2.12\times10^{-10}\ \text{m},\qquad r_3 = 9\times5.3\times10^{-11} = 4.77\times10^{-10}\ \text{m}$$

Question 12.8

A $12.5\ \text{eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution. At room temperature practically every hydrogen atom is in its ground state, $n = 1$, with $E_1 = -13.6\ \text{eV}$. In a collision an electron of $12.5\ \text{eV}$ can hand over at most $12.5\ \text{eV}$, so it can raise an atom no higher than $-13.6 + 12.5 = -1.1\ \text{eV}$. The levels above the ground state are

$$E_2 = -3.40\ \text{eV},\qquad E_3 = -1.51\ \text{eV},\qquad E_4 = -0.85\ \text{eV}$$

Reaching $n = 3$ takes $13.6 – 1.51 = 12.09\ \text{eV}$, which the electron has; reaching $n = 4$ takes $12.75\ \text{eV}$, which it does not. So atoms are excited to $n = 2$ or $n = 3$. Unlike a photon, which is absorbed only if its energy matches a gap exactly, an electron can give up just part of its kinetic energy and keep the rest, which is why $12.5\ \text{eV}$ can excite both levels even though it matches neither gap.

An atom in $n = 3$ can fall straight to $n = 1$, or go to $n = 2$ first and then to $n = 1$. That gives three lines. The wavelength of each is $\lambda = hc/\Delta E$, with $hc = 1240\ \text{eV nm}$:

$$3\to1:\ \lambda = \frac{1240}{12.09} = 102.6\ \text{nm},\qquad 2\to1:\ \lambda = \frac{1240}{10.20} = 121.6\ \text{nm},\qquad 3\to2:\ \lambda = \frac{1240}{1.89} = 656\ \text{nm}$$

The two lines ending on $n = 1$ belong to the Lyman series, in the ultraviolet; the line ending on $n = 2$ is the red first line of the Balmer series.

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 $1.5\times10^{11}\ \text{m}$ with orbital speed $3\times10^4\ \text{m/s}$. (Mass of earth $= 6.0\times10^{24}\ \text{kg}$.)

Solution. Bohr’s condition says any orbit’s angular momentum is a whole number of units of $h/2\pi$, so the quantum number is the angular momentum measured in those units: $mvr = nh/2\pi$ gives

$$n = \frac{2\pi mvr}{h} = \frac{2\pi(6.0\times10^{24})(3\times10^4)(1.5\times10^{11})}{6.63\times10^{-34}} = \frac{1.70\times10^{41}}{6.63\times10^{-34}} = 2.6\times10^{74}$$

A number this large means the allowed orbits are packed so closely that neighbouring ones are indistinguishable, and the quantisation of the Earth’s orbit can never be noticed. At the scale of planets, classical mechanics is all that is needed.

Common mistakes

• Question 12.1(a): answering “much greater than”. Students picture Rutherford’s atom as tiny because its nucleus is tiny. But the nucleus is not the atom: the electrons orbit at atomic distances, so both models describe atoms about $10^{-10}\ \text{m}$ across.
• Question 12.4: giving the kinetic energy as $-13.6\ \text{eV}$. Kinetic energy is never negative. The total energy is negative because the potential energy, $-27.2\ \text{eV}$, outweighs the kinetic energy, $+13.6\ \text{eV}$.
• Question 12.5: using the energy of the $n = 4$ level as the photon’s energy. The photon supplies the difference between the levels, $12.75\ \text{eV}$, not $0.85\ \text{eV}$; using $0.85\ \text{eV}$ gives a wavelength fifteen times too long.
• Question 12.6(b): taking the period to grow as $n^2$. The radius grows as $n^2$ but the speed also falls as $1/n$, so the electron has farther to go and goes more slowly: the period grows as $n^3$.
• Question 12.7: scaling the radius by $n$. The radius goes as $n^2$, so the $n = 2$ orbit is four times, not twice, the innermost one.
• Question 12.8: including $n = 4$, or leaving out the $3\to2$ line. Reaching $n = 4$ needs $12.75\ \text{eV}$, more than the beam has. And an atom in $n = 3$ can come down in two steps as well as one, so there are three lines, not one.
• Question 12.9: dividing by $h$ instead of $h/2\pi$. The unit of angular momentum in Bohr’s condition is $h/2\pi$; forgetting the $2\pi$ gives an answer about six times too small.

Practise next

• Chapter 13, Nuclei — from the electrons of the atom to its nucleus, where the same energy bookkeeping runs in MeV instead of eV.
• Chapter 11, Dual Nature of Radiation and Matter — the photon energy $E = h\nu$ used throughout this exercise, and the de Broglie waves that explain Bohr’s quantisation condition.