NCERT Exemplar Solutions Class 12 Physics Dual Nature of Radiation and Matter — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Dual Nature of Radiation and Matter, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 11.1 to 11.13; here they are Questions 1 to 13. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• de Broglie wavelength. $\lambda = h/p = h/\sqrt{2mE}$, so at equal energy a heavier particle has a shorter wavelength, and at equal momentum all particles share one wavelength.
• Photon energy. $E = h\nu = hc/\lambda$, with $hc \approx 1240\ \text{eV nm}$, which turns any wavelength into an energy at once.
• Forces and wavelength. A force changes the momentum, so it changes $\lambda$; a force that does no work (a magnetic one) leaves the speed, and so $\lambda$, unchanged.
• Relativity. Relativistic corrections matter once the momentum is comparable to $m_ec$, that is for wavelengths near or below $h/m_ec \approx 2.4\times10^{-3}$ nm.

Key insight. Nearly every question is the single relation $\lambda = h/p$ in disguise. Decide what happens to the momentum, and the wavelength follows. Questions 3 and 13 are the ones where you must reason about the physics rather than substitute.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

A particle is dropped from a height $H$. The de Broglie wavelength of the particle as a function of height is proportional to

Solution. The book prints the label of the fourth option as “(b)” a second time; it is option (d), and that is how it is called here. After falling through a height $H$ the speed is $v = \sqrt{2gH}$, so the momentum is $p = m\sqrt{2gH}$ and

$$\lambda = \frac{h}{p} = \frac{h}{m\sqrt{2g}}\,H^{-1/2}.$$

The wavelength falls as the particle gains speed, and varies as $H^{-1/2}$.

Question 2

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with $1$ MeV energy is nearly

Solution. The photon must carry at least the binding energy, $E = 1\ \text{MeV} = 10^6$ eV. Then

$$\lambda = \frac{hc}{E} = \frac{1240\ \text{eV nm}}{10^{6}\ \text{eV}} = 1.24\times10^{-3}\ \text{nm}.$$

This is a gamma-ray wavelength, as one expects for nuclear energies.

Question 3

Consider a beam of electrons (each electron with energy $E_0$) incident on a metal surface kept in an evacuated chamber. Then

Solution. Option (a) is false: nothing restricts the ejection of electrons to photons, and an electron striking a metal can certainly knock electrons out. The electrons that leave the surface are of two kinds. Some are incident electrons scattered back; an elastic reflection returns an electron with its full energy $E_0$, and a partly inelastic one returns it with less. Others are secondary electrons knocked out after the incident electron has shared its energy with the metal in a series of collisions. Because the energy is divided in so many ways, the emerging electrons have a continuous range of energies, from nearly zero up to the greatest possible value $E_0$. That rules out (b), which asks for a single energy.

Option (c) wrongly borrows the photon rule $K_{max} = h\nu – \phi$. A photon is absorbed completely, so the work function has to be paid out of its energy. An elastically reflected electron is not absorbed, and need not pay $\phi$ at all; it can come back with $E_0$ intact.

Question 4

Consider Fig. 11.7 in the NCERT textbook of Physics for Class XII. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of $\theta$ that

Solution. The figure is the textbook’s Davisson–Germer set-up: electrons accelerated through a potential difference $V$ by the electrode A strike a nickel crystal, and the scattered intensity is measured as the detector angle $\theta$ is varied. The accelerated electrons have $\lambda = h/\sqrt{2meV} \propto V^{-1/2}$. A higher voltage therefore gives a shorter wavelength. For a given crystal spacing, the diffraction maximum condition $d\sin\theta = n\lambda$ ties the angle to the wavelength, so a shorter $\lambda$ puts the maximum at a smaller angle. The angle depends on the target only through the fixed spacing, and the question asks how it changes when $V$ rises, so (d) is not the answer.

Question 5

A proton, a neutron, an electron and an $\alpha$-particle have the same energy. Then their de Broglie wavelengths compare as

Solution. For a given kinetic energy $E$, $p = \sqrt{2mE}$, so

$$\lambda = \frac{h}{\sqrt{2mE}} \propto \frac{1}{\sqrt{m}}.$$

The heavier the particle, the shorter its wavelength. The proton and neutron have almost the same mass, so $\lambda_p = \lambda_n$ (to the accuracy of the question); the $\alpha$-particle is about four times heavier and has the shortest wavelength; the electron is lightest and has the longest. Hence $\lambda_\alpha < \lambda_p = \lambda_n < \lambda_e$.

Question 6

An electron is moving with an initial velocity $\mathbf{v} = v_0\hat{\mathbf i}$ and is in a magnetic field $\mathbf{B} = B_0\hat{\mathbf j}$. Then its de Broglie wavelength

Solution. The magnetic force $-e\,\mathbf{v}\times\mathbf{B}$ is always perpendicular to the velocity, so it does no work and cannot change the speed. The electron simply turns along a helix or circle. Its direction changes but the magnitude of the momentum, $mv_0$, stays the same, and since $\lambda = h/(mv_0)$ depends only on that magnitude, the wavelength is constant.

Question 7

An electron (mass $m$) with an initial velocity $\mathbf{v} = v_0\hat{\mathbf i}$ ($v_0 > 0$) is in an electric field $\mathbf{E} = -E_0\hat{\mathbf i}$ ($E_0 = $ constant $> 0$). Its de Broglie wavelength at time $t$ is given by

Solution. The electron has charge $-e$, so the force is $-e\mathbf{E} = +eE_0\hat{\mathbf i}$, along its motion. It speeds up uniformly:

$$v = v_0 + \frac{eE_0}{m}t.$$

With $\lambda_0 = h/(mv_0)$,

$$\lambda = \frac{h}{mv} = \frac{h}{mv_0\left(1 + \dfrac{eE_0t}{mv_0}\right)} = \frac{\lambda_0}{1 + \dfrac{eE_0t}{mv_0}}.$$

The wavelength shrinks as the electron is accelerated, which excludes (b), (c) and (d).

Question 8

An electron (mass $m$) with an initial velocity $\mathbf{v} = v_0\hat{\mathbf i}$ is in an electric field $\mathbf{E} = E_0\hat{\mathbf j}$. If $\lambda_0 = h/mv_0$, its de Broglie wavelength at time $t$ is given by

Solution. Now the force $-eE_0\hat{\mathbf j}$ is perpendicular to the initial velocity. The $x$-component of velocity stays $v_0$ while a $y$-component grows as $v_y = eE_0t/m$ in magnitude. The speed is

$$v = \sqrt{v_0^2 + \frac{e^2E_0^2t^2}{m^2}} = v_0\sqrt{1 + \frac{e^2E_0^2t^2}{m^2v_0^2}},$$

so

$$\lambda = \frac{h}{mv} = \frac{\lambda_0}{\sqrt{1 + \dfrac{e^2E_0^2t^2}{m^2v_0^2}}}.$$

The wavelength decreases, but more slowly than in Question 7, because only the square root of the extra term enters.

Question 9

Relativistic corrections become necessary when the expression for the kinetic energy $\tfrac12 mv^2$ becomes comparable to $mc^2$, where $m$ is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron?

Solution. Relativity matters when the momentum is of the order of $m_ec$, which is when $\tfrac12 mv^2 \sim mc^2$. The corresponding wavelength is

$$\lambda \lesssim \frac{h}{m_ec} = \frac{6.63\times10^{-34}}{(9.11\times10^{-31})(3\times10^{8})}\ \text{m} = 2.4\times10^{-12}\ \text{m} = 2.4\times10^{-3}\ \text{nm}.$$

Wavelengths of $10$ nm and $0.1$ nm are far longer, so such electrons are slow and non-relativistic. The wavelengths $10^{-4}$ nm and $10^{-6}$ nm are below $2.4\times10^{-3}$ nm, so (c) and (d) are right.

Question 10

Two particles $A_1$ and $A_2$ of masses $m_1, m_2$ $(m_1 > m_2)$ have the same de Broglie wavelength. Then

Solution. Since $\lambda = h/p$, equal wavelengths mean equal momenta: (a) is true. The kinetic energy is $E = p^2/2m$, so for the same $p$ the heavier particle has the smaller energy. As $m_1 > m_2$, $E_1 < E_2$: (c) is true, while (b) and (d) are false.

Question 11

The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is $v_e = \dfrac{c}{100}$. Then

Solution. Here $E_p$ and $p_p$ stand for the photon’s energy and momentum. From $\lambda_p = 2\lambda_e$ and $\lambda = h/p$, the photon has half the electron’s momentum: $p_e = 2p_p$.

The electron is slow, so $p_e = m_ev_e$ and

$$\frac{p_e}{m_ec} = \frac{v_e}{c} = 10^{-2},$$

which is (c). For the energies, $E_e = \tfrac12 m_ev_e^2$ and $E_p = p_pc = \tfrac12 p_ec = \tfrac12 m_ev_ec$. Then

$$\frac{E_e}{E_p} = \frac{\tfrac12 m_ev_e^2}{\tfrac12 m_ev_ec} = \frac{v_e}{c} = 10^{-2},$$

which is (b). Options (a) and (d) would need a ratio of $10^{-4}$, the square of the true value, and are false.

Question 12

Photons absorbed in matter are converted to heat. A source emitting $n$ photons/sec of frequency $\nu$ is used to convert $1$ kg of ice at $0^\circ$C to water at $0^\circ$C. Then the time $T$ taken for the conversion

Solution. Melting $1$ kg of ice needs a fixed amount of heat, $Q = mL$. The source delivers energy at the rate $nh\nu$ per second, so

$$T = \frac{Q}{nh\nu}.$$

$T$ falls when $n$ rises with $\nu$ fixed (a), and when $\nu$ rises with $n$ fixed (b). It depends on $n$ and $\nu$ only through the product $n\nu$, so it is unchanged when $n\nu$ is constant (c). A larger $n\nu$ means a faster supply of energy and a shorter time, so (d) is false.

Question 13

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values $\lambda_1, \lambda_2$ with $\lambda_1 > \lambda_2$. Which of the following statements are true?

Solution. A varying wavelength means a varying speed. In a circular orbit about the origin the central force does no work and the speed is constant, so $\lambda$ would be constant: (a) is false. An elliptic orbit with the origin at a focus (as for a planet about the Sun, with an attractive central force) has a speed that changes cyclically, so (b) is possible.

In such an orbit the particle moves fastest when nearest the focus, as in Kepler’s second law. A larger speed means a larger momentum and a smaller wavelength. The smaller wavelength, $\lambda_2$, therefore occurs when the particle is nearer the origin, which is (d); (c) states the reverse and is false.

Common mistakes

• Mixing up the options in Question 1. The fourth option is printed “(b)” in the book; it is (d), $H^{-1/2}$. The wavelength falls with height fallen.
• Applying $K_{max} = h\nu – \phi$ to electrons in Question 3. A photon is absorbed whole and pays $\phi$; an incident electron can bounce back with all of $E_0$.
• Reading the wavelength of equal-energy particles as equal in Question 5. Equal energy means unequal momentum; the lighter the particle, the longer the wavelength.
• Thinking a magnetic field changes $\lambda$ in Question 6. It changes direction only, never the magnitude of the momentum.
• Treating the sideways acceleration like the forward one in Questions 7 and 8. A force along the motion changes $v$ linearly; a perpendicular force adds in quadrature, giving the square root.
• Choosing the long wavelengths in Question 9. Relativistic electrons have very short wavelengths, below about $2.4\times10^{-3}$ nm.
• Equating energies when momenta are equal in Question 10. Equal $p$ leaves $E = p^2/2m$ different for different masses.
• Taking a circular orbit in Question 13. A circle has a constant speed and hence a constant wavelength.

Practise next

• Exemplar Dual Nature of Radiation and Matter: Very Short, Short and Long Answer Questions — where the same ideas are applied to the photoelectric effect, stopping potentials and photon counts, with printed key errors corrected.
• NCERT textbook Chapter 11 — Dual Nature of Radiation and Matter for the theory of the photoelectric effect, Einstein’s equation and the de Broglie hypothesis.
• Exemplar Wave Optics — the previous unit, whose interference and diffraction ideas the electron-diffraction questions here lean on.
• Exemplar Atoms — the next unit, where de Broglie waves are used to explain the allowed Bohr orbits.