NCERT Exemplar Solutions Class 12 Physics Atoms — Very Short, Short and Long Answer Questions Q14–29

NCERT Exemplar Problems, Class 12 Physics — Atoms, Very Short Answer Questions 14 to 18, Short Answer Questions 19 to 23 and Long Answer Questions 24 to 29. All 16 questions solved. The book numbers them 12.14 to 12.29.

Three ideas carry this page. The Bohr model is a Coulomb-force problem with quantised angular momentum, so $r_n = n^2a_0/Z$, $E_n = -13.6\,Z^2/n^2$ eV and $v_n \propto Z/n$ follow from $mv^2/r = $ force and $mvr = n\hbar$; whenever the force law or the masses change, redo those two equations rather than the formulas. Reduced mass replaces the electron mass when the nucleus moves, which is why positronium, deuterium and hydrogen differ slightly. Line energies are differences of level energies, so every spectral question reduces to $13.6Z^2(1/n^2 – 1/m^2)$ eV.

Key insight. Write the centripetal equation and the Bohr quantisation condition for the force law in the question, solve for $r$, and only then compute $v$, kinetic energy and potential energy. Questions 26 and 29 are the same method with a different force inside a small radius.

Very Short Answer Questions

Question 14

The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?

Solution. The hydrogen atom is a bound system. Its total energy is $-13.6$ eV, lower than that of a free proton and electron at rest by the binding energy $B = 13.6$ eV. By mass–energy equivalence this energy deficit shows up as a mass deficit, $$m_H = m_p + m_e – \frac{B}{c^2},\qquad \frac{B}{c^2} = \frac{13.6\times1.6\times10^{-19}}{(3\times10^{8})^2}\approx 2.4\times10^{-35}\ \text{kg}.$$

Question 15

Imagine removing one electron from He$^4$ and He$^3$. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

Solution. After one electron is removed, each is a one-electron ion with $Z = 2$. The Bohr energy is $E_n = -\dfrac{13.6\,Z^2}{n^2}\dfrac{\mu}{m_e}$ eV, where $\mu$ is the reduced mass of the electron and nucleus. Both nuclei are thousands of times heavier than the electron, so $\mu = \dfrac{m_eM}{m_e+M}\approx m_e$ in both cases, with a difference of only about one part in $10^4$. The charge is the same, so the levels are almost identical.

Question 16

When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy?

Solution. The electron is bound to the nucleus only by the electromagnetic force and it has no other interaction through which it could hand energy over. It carries no thermal or mechanical coupling inside an isolated atom either. Energy can therefore leave only as an electromagnetic quantum, a photon.

Question 17

Would the Bohr formula for the H-atom remain unchanged if proton had a charge $(+4/3)e$ and electron a charge $(-3/4)e$, where $e = 1.6\times10^{-19}$ C. Give reasons for your answer.

Solution. In the Bohr model the charges enter only through the product in the Coulomb force, $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_pq_e}{r^2}$. Here $$q_pq_e = \tfrac43e\times\tfrac34e = e^2,$$ the same product as for ordinary charges. The force, the radii and the energies are therefore unchanged.

Question 18

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

Solution. In the Bohr model $E_n = -13.6/n^2$ eV and $L = nh/2\pi$. Different energies mean different values of $n$, and different $n$ mean different angular momenta. So the two cannot be separated.

Short Answer Questions

Question 19

Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Solution. The two particles have equal mass $m_e$ and orbit their common centre of mass, so the reduced mass is $$\mu = \frac{m_e\cdot m_e}{m_e+m_e} = \frac{m_e}{2}.$$ The Bohr energy is proportional to the mass, $E_n = -\dfrac{\mu e^4}{8\varepsilon_0^2h^2n^2}$, so with $\mu = m_e/2$ $$E_1 = \frac12\times(-13.6\ \text{eV}) = -6.8\ \text{eV}.$$

Question 20

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

Solution. With no electron–electron repulsion each electron moves independently in the field of the nucleus of charge $+2e$, as in the ion He$^+$. Its ground state energy is $$E_1 = -13.6\,Z^2 = -13.6\times4 = -54.4\ \text{eV}.$$ Both electrons occupy this ground state (with opposite spins), so the atom’s energy is the sum of the two, $$E = 2\times(-54.4) = -108.8\ \text{eV}.$$

Printed key error. The key writes the total as $-(4\times13.6)$ eV $= -54.4$ eV. That is the energy of one electron only; the question asks for the atom, which has two.

Question 21

Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Solution. In the ground state the electron has speed $v = 2.18\times10^6$ m/s on an orbit of radius $a_0 = 0.53\times10^{-10}$ m. Its period is $T = 2\pi a_0/v$, and the charge $e$ passes any point of the orbit once per period, so the current is charge per unit time, $$I = \frac{e}{T} = \frac{ev}{2\pi a_0} = \frac{1.6\times10^{-19}\times2.18\times10^{6}}{2\pi\times0.53\times10^{-10}}\approx1.05\times10^{-3}\ \text{A}.$$

Printed key error. The key writes the number of revolutions per unit time as $2\pi a_0/v$ and the current as $(2\pi a_0/v)\,e$. But $2\pi a_0/v$ is the period, not the frequency, so the correct frequency is $v/(2\pi a_0)$. The key also stops without a number.

Question 22

Show that the first few frequencies of light that is emitted when electrons fall to the $n^{\text{th}}$ level from levels higher than $n$, are approximate harmonics (i.e. in the ratio $1:2:3\ldots$) when $n\gg1$.

Solution. A transition from level $m = n+p$ to level $n$ has frequency $$\nu_{mn} = cRZ^2\left[\frac1{n^2}-\frac1{(n+p)^2}\right] = cRZ^2\,\frac{(n+p)^2-n^2}{n^2(n+p)^2} = cRZ^2\,\frac{2np+p^2}{n^2(n+p)^2}.$$ For the first few lines $p\ll n$ when $n\gg1$, so $2np+p^2\approx2np$ and $(n+p)^2\approx n^2$: $$\nu_{mn}\approx\frac{2cRZ^2\,p}{n^3}.$$ This is proportional to $p = 1, 2, 3,\dots$, so the frequencies are in the ratio $1:2:3:\dots$, like the harmonics of a fixed fundamental $2cRZ^2/n^3$.

Question 23

What is the minimum energy that must be given to a H atom in ground state so that it can emit an $H_\gamma$ line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such $H_\gamma$ photon?

Solution. $H_\gamma$ is the transition $n = 5\to n = 2$, so the atom must first be raised to $n = 5$. The minimum energy is that of the $n = 1\to5$ excitation, $$\Delta E = 13.6\left(1-\frac1{25}\right) = 13.6\times0.96 = 13.06\ \text{eV}.$$ In the Bohr model $L_n = n\hbar$. If the angular momentum of the system is conserved, the photon carries away the difference, $$L = L_5 – L_2 = (5-2)\hbar = 3\hbar = \frac{3\times6.63\times10^{-34}}{2\pi}\approx3.2\times10^{-34}\ \text{kg m}^2/\text{s}.$$ (This follows the Bohr picture the question asks for. A real photon carries $\hbar$ of spin angular momentum, which is why the full theory needs extra rules.)

Long Answer Questions

Question 24

The first four spectral lines in the Lyman series of a H-atom are $\lambda = 1218$ Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Solution. When the nucleus moves, the electron’s mass is replaced by the reduced mass $\mu = \dfrac{m_eM}{m_e+M}\approx m_e\left(1-\dfrac{m_e}{M}\right)$. The Rydberg constant, and so every transition energy, is proportional to $\mu$, so $\lambda\propto1/\mu$. Deuterium has $M_D\approx2M_H$. Then $$\frac{\lambda_D}{\lambda_H} = \frac{\mu_H}{\mu_D}\approx1-\frac{m_e}{M_H}+\frac{m_e}{M_D} = 1-\frac{m_e}{2M_H}.$$ With $\dfrac{m_e}{M_H} = \dfrac{9.1\times10^{-31}}{1.67\times10^{-27}} = 5.45\times10^{-4}$, we get $\lambda_D = 0.99973\,\lambda_H$. So each Deuterium line is shorter by $2.7\times10^{-4}$ of its wavelength:

Question 25

Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in $^1$H and $^2$H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass $\mu$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mu = m_eM/(m_e+M)$ where $M$ is the nuclear mass and $m_e$ is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in $^1$H and $^2$H. (Mass of $^1$H nucleus is $1.6725\times10^{-27}$ kg, Mass of $^2$H nucleus is $3.3374\times10^{-27}$ kg, Mass of electron $= 9.109\times10^{-31}$ kg.)

Solution. Since $\lambda\propto1/\mu$ and $\mu = \dfrac{m_e}{1+m_e/M}\approx m_e\left(1-\dfrac{m_e}{M}\right)$, $$\lambda\propto1+\frac{m_e}{M},\qquad\frac{\Delta\lambda}{\lambda}\approx m_e\left(\frac1{M_H}-\frac1{M_D}\right).$$ Numerically, $$\frac1{M_H}-\frac1{M_D} = \frac1{1.6725\times10^{-27}}-\frac1{3.3374\times10^{-27}} = (5.979-2.996)\times10^{26} = 2.983\times10^{26}\ \text{kg}^{-1},$$ $$\frac{\Delta\lambda}{\lambda} = 9.109\times10^{-31}\times2.983\times10^{26} = 2.72\times10^{-4}.$$ As a percentage this is about $2.7\times10^{-2}\,\%$, with the hydrogen line the longer one.

Question 26

If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) $R = 0.1$ Å, and (ii) $R = 10$ Å.

Solution. Let $k = \dfrac{e^2}{4\pi\varepsilon_0} = 14.4$ eV Å and let $r_B = \dfrac{\hbar^2}{mke^2} = 0.53$ Å be the ordinary Bohr radius.

(i) $R = 0.1$ Å. This is smaller than $r_B = 0.53$ Å, so the electron orbits outside the charged sphere, where a uniform sphere acts as a point charge. Nothing changes: $E = -13.6$ eV.

(ii) $R = 10$ Å. This is much larger than $r_B$, so the electron is inside the sphere. At radius $r$ the enclosed charge is $e\,r^3/R^3$, so the attractive force is $$F = \frac{ke^2\,r}{R^3}\quad(\text{here } ke^2 = 14.4\ \text{eV Å}).$$ Centripetal condition: $mv^2 = \dfrac{ke^2r^2}{R^3}$. Bohr quantisation: $m^2v^2r^2 = \hbar^2$. Multiplying the first by $m r^2$: $$\hbar^2 = m\cdot\frac{ke^2r^2}{R^3}\cdot r^2\ \Rightarrow\ r’^4 = \frac{\hbar^2R^3}{mke^2} = r_BR^3,$$ $$r’ = (0.53\times1000)^{1/4}\ \text{Å}\approx4.8\ \text{Å}\ (<R,\ \text{consistent}).$$ Kinetic energy: $$KE = \tfrac12mv^2 = \frac{ke^2r’^2}{2R^3} = \frac{14.4\times23.0}{2\times1000}\approx0.17\ \text{eV}.$$ Potential energy of an electron at $r<R$ in a uniformly charged sphere (the potential is $\dfrac{k(3R^2-r^2)}{2R^3}$, times $-e$): $$PE = -\frac{ke^2\,(3R^2-r’^2)}{2R^3} = -\frac{14.4\times(300-23)}{2\times1000}\approx-2.0\ \text{eV}.$$ Total energy: $$E = KE + PE\approx0.17-1.99\approx-1.8\ \text{eV}.$$

Printed key error. In part (ii) the key writes the potential energy correctly with $2R^3$ in the denominator but then drops the 2 in the numbers, getting $-3.83$ eV, which is twice the true $-1.99$ eV. It also reports this as the ground state energy without adding the kinetic energy.

Question 27

In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an $n = 4$ Auger electron emitted by Chromium by absorbing the energy from a $n = 2$ to $n = 1$ transition.

Solution. Chromium has $Z = 24$. The $n=2\to1$ transition releases $$\Delta E = 13.6\,Z^2\left(\frac1{1^2}-\frac1{2^2}\right) = 13.6\times576\times\frac34 = 5875.2\ \text{eV}.$$ Treating the $n=4$ electron as hydrogen-like with the same $Z$ (as the question intends), its binding energy is $$13.6\,Z^2\times\frac1{4^2} = \frac{13.6\times576}{16} = 489.6\ \text{eV}.$$ The kinetic energy of the ejected electron is the energy received minus the binding energy, $$KE = 5875.2-489.6 = 5385.6\ \text{eV}.$$ This is the same as $13.6\,Z^2\left(\frac34-\frac1{16}\right)$.

Question 28

The inverse square law in electrostatics is $|\mathbf F| = \dfrac{e^2}{(4\pi\varepsilon_0)r^2}$ for the force between an electron and a proton. The $\left(\dfrac1r\right)$ dependence of $|\mathbf F|$ can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass $m_p$, force would be modified to $|\mathbf F| = \dfrac{e^2}{(4\pi\varepsilon_0)r^2}\left[\dfrac1{r^2}+\dfrac\lambda r\right]\times\exp(-\lambda r)$ where $\lambda = m_pc/\hbar$ and $\hbar = \dfrac h{2\pi}$. Estimate the change in the ground state energy of a H-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

Solution. The printed force law has the form $F = \dfrac{e^2}{4\pi\varepsilon_0}\left[\dfrac1{r^2}+\dfrac\lambda r\right]e^{-\lambda r}$ (the extra $r^2$ in the denominator in the printed expression is a misprint; the bracket carries the dependence on $r$). Notice that $$\frac{d}{dr}\left(-\frac{e^{-\lambda r}}{r}\right) = \left[\frac1{r^2}+\frac\lambda r\right]e^{-\lambda r},$$ so the force comes from the potential energy $$V(r) = -\frac{e^2}{4\pi\varepsilon_0}\,\frac{e^{-\lambda r}}{r}.$$ Here $\lambda$ is tiny: $\lambda^{-1} = \dfrac{\hbar}{m_pc} = \dfrac{\hbar}{10^{-6}m_ec}\approx3.9\times10^{-7}$ m, which is thousands of times larger than the Bohr radius $r_B = 5.3\times10^{-11}$ m. So $\lambda r\ll1$ over the whole atom and $e^{-\lambda r}\approx1-\lambda r$: $$V(r)\approx-\frac{e^2}{4\pi\varepsilon_0 r}+\frac{e^2\lambda}{4\pi\varepsilon_0}.$$ The extra term is a constant. It does not change the force to first order, so the orbit and kinetic energy are unaltered, and the energy simply rises by $$\Delta E = \frac{e^2\lambda}{4\pi\varepsilon_0} = \frac{e^2}{4\pi\varepsilon_0r_B}\,\lambda r_B = 27.2\,\lambda r_B\ \text{eV}.$$ Numerically $\lambda r_B = \dfrac{5.3\times10^{-11}}{3.9\times10^{-7}} = 1.4\times10^{-4}$, so $$\Delta E\approx27.2\times1.4\times10^{-4}\approx3.7\times10^{-3}\ \text{eV}.$$

Question 29

The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge $+q_1$, $-q_2$ is modified to

$$|\mathbf F| = \frac{q_1q_2}{4\pi\varepsilon_0}\frac1{r^2},\quad r\ge R_0$$

$$|\mathbf F| = \frac{q_1q_2}{4\pi\varepsilon_0}\frac1{R_0^2}\left(\frac{R_0}r\right)^{\varepsilon},\quad r\le R_0$$

Calculate in such a case, the ground state energy of a H-atom, if $\varepsilon = 0.1$, $R_0 = 1$ Å.

Solution. First guess which region the orbit lies in. The ordinary Bohr radius is $0.53$ Å, less than $R_0 = 1$ Å, so try $r<R_0$, where $$F = \frac{ke^2}{R_0^{2}}\left(\frac{R_0}{r}\right)^{\varepsilon},\qquad k e^2 = 14.4\ \text{eV Å}.$$ Centripetal condition: $mv^2 = rF = ke^2R_0^{\varepsilon-2}\,r^{1-\varepsilon}$. Bohr condition: $m^2v^2r^2 = \hbar^2$. Combining, $$m\cdot ke^2R_0^{\varepsilon-2}r^{1-\varepsilon}\cdot r^2 = \hbar^2\ \Rightarrow\ r^{3-\varepsilon} = a_0R_0^{2-\varepsilon},$$ $$r = \left(0.529\times1^{1.9}\right)^{1/2.9}\ \text{Å}\approx0.80\ \text{Å} = 0.08\ \text{nm}.$$ This is less than $R_0$, so the assumption holds. Then $$KE = \tfrac12mv^2 = \tfrac12ke^2R_0^{\varepsilon-2}r^{1-\varepsilon} = \tfrac12\times14.4\times0.80^{0.9}\approx5.9\ \text{eV},\qquad v = \sqrt{\frac{2\,KE}m}\approx1.44\times10^{6}\ \text{m/s}.$$ For the potential energy, take zero at infinity. From infinity to $R_0$ the force is the usual Coulomb force, giving $-ke^2/R_0$. From $R_0$ in to $r$ one integrates the modified force: $$PE = -ke^2\left[\frac1{R_0}+\frac{R_0^{\varepsilon-2}\left(R_0^{1-\varepsilon}-r^{1-\varepsilon}\right)}{1-\varepsilon}\right] = -14.4\left[1+\frac{1-0.80^{0.9}}{0.9}\right]\approx-17.3\ \text{eV}.$$ Total energy: $$E = KE+PE\approx5.9-17.3\approx-11.4\ \text{eV}.$$