NCERT Exemplar Solutions Class 12 Physics Dual Nature of Radiation and Matter — Very Short, Short and Long Answer Questions Q14–29

NCERT Exemplar Problems, Class 12 Physics — Dual Nature of Radiation and Matter, Very Short Answer Questions 14 to 18, Short Answer Questions 19 to 24 and Long Answer Questions 25 to 29. All 16 questions solved. The book numbers them 11.14 to 11.29.

Three ideas carry this page. Light comes in quanta: a photon of frequency $\nu$ carries energy $h\nu$ and momentum $h/\lambda$, and a single photon gives all its energy to a single electron. Matter has a wavelength: a particle of momentum $p$ has $\lambda = h/p$, and for a charge $q$ accelerated through $V$ this is $\lambda = h/\sqrt{2mqV}$. Conservation laws still apply: momentum in collisions, energy in the photoelectric equation $K_{\max} = h\nu – \phi$. Throughout, $hc = 1240$ eV nm is used.

Key insight. Convert every wavelength to a photon energy $E = 1240/\lambda$ (eV, with $\lambda$ in nm) or a momentum $h/\lambda$ first. Questions 18, 20, 22, 25 and 29 then reduce to counting photons or comparing energies, and Questions 14, 23, 24 and 28 to the single relation $\lambda = h/p$.

Very Short Answer Questions

Question 14

A proton and an $\alpha$-particle are accelerated, using the same potential difference. How are the de Broglie wavelengths $\lambda_p$ and $\lambda_\alpha$ related to each other?

Solution. A charge $q$ accelerated through $V$ gains kinetic energy $qV$, so its momentum is $p = \sqrt{2mqV}$ and $$\lambda = \frac{h}{\sqrt{2mqV}}.$$ For the same $V$, $\lambda \propto 1/\sqrt{mq}$. The $\alpha$-particle has four times the proton’s mass and twice its charge, so $$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{4\times2} = \sqrt8 = 2\sqrt2.$$

Question 15

(i) In the explanation of photo electric effect, we assume one photon of frequency $\nu$ collides with an electron and transfers its energy. This leads to the equation for the maximum energy $E_{\max}$ of the emitted electron as $E_{\max} = h\nu – \phi_0$, where $\phi_0$ is the work function of the metal. If an electron absorbs 2 photons (each of frequency $\nu$) what will be the maximum energy for the emitted electron? (ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Solution. (i) Two photons bring in $2h\nu$, and the electron still has to pay the work function once to leave the metal, so $$E_{\max} = 2h\nu – \phi_0.$$

(ii) The chance that one particular electron absorbs a second photon before it gives up the energy of the first is very small. At the intensities used in the ordinary photoelectric experiment, photons arrive so sparsely at any one electron that single-photon absorption is overwhelmingly the usual event, and the stopping potential is therefore set by $h\nu – \phi_0$. (Very intense laser light is different, and two-photon emission is then observed.)

Question 16

There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.

Solution. No. The first kind of material absorbs a photon of high energy and emits one of lower energy, and the balance goes into heat, so energy is conserved. The second kind would take in a low-energy photon and send out a high-energy one. The extra energy would have to come out of the substance itself, and a stable substance has no store of energy to give. (It can happen only if the substance is already excited, or through multi-photon processes at very high intensity.)

Question 17

Do all the electrons that absorb a photon come out as photoelectrons?

Solution. No. An electron deep inside the metal that absorbs a photon loses its energy in collisions with other electrons and the lattice before it reaches the surface, so it stays in the metal as heat. Only the electrons absorbing photons very near the surface escape, and then only if the energy left on arrival exceeds the work function. That is why the photoelectric yield is only a small fraction of the photons absorbed.

Question 18

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.

Solution. In time $t$ a source of power $P$ emits energy $Pt$. Each photon has energy $hc/\lambda$, so the number of photons is $N = Pt\lambda/hc$. With equal power the number emitted is proportional to $\lambda$: $$\frac{N_x}{N_v} = \frac{\lambda_x}{\lambda_v} = \frac{1\ \text{nm}}{500\ \text{nm}} = \frac{1}{500}.$$ Each X-ray photon carries 500 times the energy, so far fewer are needed for the same power.

Short Answer Questions

Question 19

Consider Fig. 11.1 for photoemission. How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Solution. The figure shows light striking a metal plate from below at a slant, and an electron leaving in a different direction. The photon’s momentum is not shared by the electron alone. The electron is bound in the metal, so the metal as a whole (its ion cores and lattice) takes up the difference in momentum. Because the metal is so massive, this recoil carries a negligible share of the energy, which is why the energy equation $K_{\max} = h\nu – \phi$ works without it. Momentum is conserved for the system of photon, electron and metal.

Question 20

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

(The book prints “400 snm”, a misprint for 400 nm.)

Solution. The photon energies are $E_1 = \dfrac{1240}{600} = 2.067$ eV and $E_2 = \dfrac{1240}{400} = 3.10$ eV. The maximum kinetic energy is $K = E – \phi$, and the second is double the first: $$E_2 – \phi = 2(E_1 – \phi) \;\Rightarrow\; \phi = 2E_1 – E_2 = 4.133 – 3.10 = 1.03\ \text{eV}.$$

The key prints $1.02$ eV because it uses $hc = 1230$ eV nm. That is a difference of rounding, not an error.

Question 21

Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (Ref Eq 11.12 of NCERT Textbook). You can assume the uncertainty in position $\Delta x$ as 1 nm. Assuming $p = \Delta p$, find the energy of the electron in electron volts.

Solution. The uncertainty relation used in the textbook is $\Delta x\,\Delta p \approx \hbar$, so $$\Delta p \approx \frac{\hbar}{\Delta x} = \frac{1.055\times10^{-34}}{10^{-9}} \approx 1.05\times10^{-25}\ \text{kg m/s}.$$ Taking $p = \Delta p$, the kinetic energy is $$E = \frac{p^2}{2m} = \frac{(1.05\times10^{-25})^2}{2\times9.11\times10^{-31}} = 6.1\times10^{-21}\ \text{J} = \frac{6.1\times10^{-21}}{1.6\times10^{-19}}\ \text{eV} \approx 3.8\times10^{-2}\ \text{eV}.$$ Confining an electron to atomic-size boxes therefore costs only a few hundredths of an electron volt per nanometre of box; smaller boxes cost far more, because $E \propto 1/(\Delta x)^2$.

Question 22

Two monochromatic beams A and B of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?

Solution. The intensity is the energy arriving per unit area per second, which is the number of photons per second times the energy of each: $$I = n_A\,h\nu_A = n_B\,h\nu_B.$$ Given $n_A = 2n_B$ and equal $I$, $$2n_B\,h\nu_A = n_B\,h\nu_B \;\Rightarrow\; \nu_B = 2\nu_A.$$ Beam A has more photons, each of lower energy, so beam B has twice the frequency.

Question 23

Two particles A and B of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Solution. The momenta are $p_1 = h/\lambda_1$ and $p_2 = h/\lambda_2$, and in one dimension they can point the same way or in opposite ways.

Same direction. $p_C = p_1 + p_2$, so $$\frac{h}{\lambda_C} = \frac{h}{\lambda_1} + \frac{h}{\lambda_2} \;\Rightarrow\; \lambda_C = \frac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}.$$

Opposite directions. $p_C = p_1 – p_2$ in magnitude, so $$\lambda_C = \frac{\lambda_1\lambda_2}{|\lambda_1 – \lambda_2|}.$$ (If $\lambda_1 = \lambda_2$ the momenta cancel and C is at rest, so its wavelength is infinite.)

Question 24

A neutron beam of energy $E$ scatters from atoms on a surface with a spacing $d = 0.1$ nm. The first maximum of intensity in the reflected beam occurs at $\theta = 30^\circ$. What is the kinetic energy $E$ of the beam in eV?

Solution. This is Bragg reflection: $2d\sin\theta = n\lambda$, and for the first maximum $n = 1$: $$\lambda = 2\times0.1\ \text{nm}\times\sin30^\circ = 0.1\ \text{nm}.$$ The neutron’s momentum is $$p = \frac{h}{\lambda} = \frac{6.6\times10^{-34}}{10^{-10}} = 6.6\times10^{-24}\ \text{kg m/s},$$ and its energy, with $m_n = 1.7\times10^{-27}$ kg, is $$E = \frac{p^2}{2m_n} = \frac{(6.6\times10^{-24})^2}{2\times1.7\times10^{-27}} = 1.28\times10^{-20}\ \text{J} = \frac{1.28\times10^{-20}}{1.6\times10^{-19}}\ \text{eV} \approx 0.08\ \text{eV}.$$ This is a thermal-neutron energy, which is why neutron beams from a reactor diffract from crystals.

Printed key error. The key gives $0.21$ eV. Its own numbers do not produce that: $6.6^2/(2\times1.7\times1.6)\times10^{-2} = 0.080$ eV, so the printed $0.21$ eV is wrong.

Long Answer Questions

Question 25

Consider a thin target ($10^{-2}$ m square, $10^{-3}$ m thickness) of sodium, which produces a photocurrent of $100\ \mu$A when a light of intensity $100\ \text{W/m}^2$ ($\lambda = 660$ nm) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na $= 0.97\ \text{kg/m}^3$].

Solution. (The density of sodium is $0.97\times10^{3}\ \text{kg/m}^3$; the book’s $0.97\ \text{kg/m}^3$ is the same value in $\text{g/cm}^3$ units. The number of atoms in the target below, $2.5\times10^{18}$, is what the printed key also uses.)

Atoms in the target. The volume is $10^{-2}\times10^{-2}\times10^{-3} = 10^{-7}\ \text{m}^3$, so the mass is $0.97\times10^{3}\times10^{-7} = 9.7\times10^{-5}$ kg. With molar mass $23\times10^{-3}$ kg, $$N_{\text{atoms}} = \frac{9.7\times10^{-5}}{23\times10^{-3}}\times6.02\times10^{23} \approx 2.5\times10^{18}.$$

Photons arriving per second. The area is $10^{-4}\ \text{m}^2$, so the power on the target is $100\times10^{-4} = 10^{-2}$ W. The photon energy is $h\nu = \dfrac{1240}{660} = 1.88$ eV $= 3.0\times10^{-19}$ J, so $$n_\gamma = \frac{10^{-2}}{3.0\times10^{-19}} = 3.3\times10^{16}\ \text{photons/s}.$$

Electrons emitted per second. $n_e = \dfrac{I}{e} = \dfrac{100\times10^{-6}}{1.6\times10^{-19}} = 6.25\times10^{14}\ \text{s}^{-1}$.

Probability. Each photon can strike any of the atoms, so the probability that a given atom’s encounter with a photon produces an electron is $$P = \frac{n_e}{n_\gamma\,N_{\text{atoms}}} = \frac{6.25\times10^{14}}{3.3\times10^{16}\times2.5\times10^{18}} \approx 7.4\times10^{-21}.$$ It is tiny, so a photon is overwhelmingly unlikely to be absorbed by any given atom, and the fraction of photons that give an electron is $n_e/n_\gamma \approx 0.019$, about two in a hundred.

(The key computes $h\nu$ with $600$ nm, which gives $2.05$ eV, and so reports about $8\times10^{-21}$. The conclusion is the same.)

Question 26

Consider an electron in front of a metallic surface at a distance $d$ (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\dfrac{1}{4}\dfrac{q^2}{4\pi\varepsilon_0 d^2}$. Calculate work in taking the charge to an infinite distance from the plate. Taking $d = 0.1$ nm, find the work done in electron volts. [Such a force law is not valid for $d < 0.1$ nm].

Solution. The force changes with distance, so the work done against it is found by integration. Moving the electron out from $d$ to infinity against the attraction $$W = \int_d^\infty \frac{q^2}{16\pi\varepsilon_0 x^2}\,dx = \frac{q^2}{16\pi\varepsilon_0}\left[-\frac1x\right]_d^\infty = \frac{q^2}{16\pi\varepsilon_0 d}.$$ For $q = e = 1.6\times10^{-19}$ C and $d = 10^{-10}$ m, $$W = \frac{9\times10^{9}\times(1.6\times10^{-19})^2}{4\times10^{-10}} = 5.76\times10^{-19}\ \text{J} = \frac{5.76\times10^{-19}}{1.6\times10^{-19}}\ \text{eV} = 3.6\ \text{eV}.$$ This is of the order of a metal’s work function, which is the point of the question: the attraction of the induced image charge in the metal is what binds the electron to the surface.

Question 27

A student performs an experiment on photoelectric effect, using two materials A and B. A plot of $V_{\text{stop}}$ vs $\nu$ is given in Fig. 11.2. (i) Which material A or B has a higher work function? (ii) Given the electric charge of an electron $= 1.6\times10^{-19}$ C, find the value of $h$ obtained from the experiment for both A and B. Comment on whether it is consistent with Einstein’s theory.

Solution. The figure shows two straight lines of stopping potential against frequency, A on the left and B to its right, with the frequency axis marked at $5$, $10$ and $15\times10^{14}$ Hz and the voltage axis at $1$, $1.5$, $2$, $2.5$ and $3$ V. Line A cuts the frequency axis at $5\times10^{14}$ Hz and line B at $10\times10^{14}$ Hz.

(i) The intercept is the threshold frequency $\nu_0$, and $\phi = h\nu_0$. Line B has the larger intercept, so B has the higher work function.

(ii) Einstein’s equation $eV_{\text{stop}} = h\nu – \phi$ gives $V_{\text{stop}} = \dfrac{h}{e}\nu – \dfrac{\phi}{e}$, a straight line of slope $h/e$. So $h = e\times\text{slope}$.

Reading the lines off the graph, A reaches $2$ V at $10\times10^{14}$ Hz, so its slope is $\dfrac{2}{5\times10^{14}}$ V s and $$h_A = 1.6\times10^{-19}\times\frac{2}{5\times10^{14}} = 6.4\times10^{-34}\ \text{J s}.$$ B reaches $2.5$ V at $15\times10^{14}$ Hz, so its slope is $\dfrac{2.5}{5\times10^{14}}$ V s and $$h_B = 1.6\times10^{-19}\times\frac{2.5}{5\times10^{14}} = 8.0\times10^{-34}\ \text{J s}.$$

Comment. Einstein’s theory says $h$ is a universal constant, so the two lines should be parallel with the same slope $h/e$ and differ only in their intercepts. Here the slopes (and so the $h$ values) differ, so the experiment as plotted is not consistent with Einstein’s theory for both materials. (Each value is within a factor of about 1.2 of the accepted $6.6\times10^{-34}$ J s, so the readings are rough.)

The key prints $6.04\times10^{-34}$ for A, a transposition of $6.4\times10^{-34}$.

Question 28

A particle A with a mass $m_A$ is moving with a velocity $v$ and hits a particle B (mass $m_B$) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.

Solution. For an elastic head-on collision with B at rest, conservation of momentum and of kinetic energy give the velocity of A after the collision as $$v_1 = \frac{m_A – m_B}{m_A + m_B}\,v.$$ The wavelength of A before and after is $$\lambda_i = \frac{h}{m_A v}, \qquad \lambda_f = \frac{h}{m_A|v_1|} = \frac{h\,(m_A+m_B)}{m_A|m_A – m_B|\,v}.$$ The change is $$\Delta\lambda = \lambda_f – \lambda_i = \frac{h}{m_A v}\left[\left|\frac{m_A+m_B}{m_A-m_B}\right| – 1\right] = \lambda_i\left[\left|\frac{m_A+m_B}{m_A-m_B}\right| – 1\right].$$ The bracket is positive, so the wavelength always increases: A slows down. If $m_A = m_B$, A stops and transfers all its momentum to B, so $\lambda_f$ is undefined (infinite). If $m_A > m_B$, A keeps moving forward; if $m_A < m_B$, it rebounds, and the modulus takes care of the sign.

Question 29

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å. (i) Estimate no. of photons emitted by the bulb per second. [Assume no other losses] (ii) Will there be photoelectric emission? (iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)? (iv) How many photons would atomic disk receive within time duration calculated in (iii) above? (v) Can you explain how photoelectric effect was observed instantaneously? [Hint: Time calculated in part (iii) is from classical consideration and you may further take the target of surface area say 1 cm$^2$ and estimate what would happen?]

Solution. (i) The photon energy is $\dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^{8}}{5\times10^{-7}} = 3.98\times10^{-19}\ \text{J} = 2.49$ eV. So $$n = \frac{P\lambda}{hc} = \frac{20}{3.98\times10^{-19}} \approx 5.0\times10^{19}\ \text{photons/s}.$$

(ii) Yes. Each photon carries $2.49$ eV, which is more than the work function $2$ eV, so $h\nu > \phi$ and emission is possible.

(iii) Treating the light classically, the bulb’s power spreads uniformly over a sphere of radius $2$ m, so the power falling on one atomic disk of radius $r = 1.5\times10^{-10}$ m is $$P_{\text{disk}} = P\,\frac{\pi r^2}{4\pi d^2} = 20\times\frac{\pi(1.5\times10^{-10})^2}{4\pi\times2^2} = 2.8\times10^{-20}\ \text{W}.$$ The energy needed is $2\ \text{eV} = 3.2\times10^{-19}$ J, so $$\Delta t = \frac{3.2\times10^{-19}}{2.8\times10^{-20}} \approx 11.4\ \text{s}.$$

(iv) In that time the disk receives $$N = \frac{2\ \text{eV}}{2.49\ \text{eV}} \approx 0.8\ \text{photon},$$ which is less than one. This is an energy check: the energy $2$ eV is delivered by less than one photon on average, so the classical picture of waiting while energy accumulates cannot be right.

(v) In the quantum picture the energy arrives in whole photons, not continuously. The first photon that is absorbed by an electron gives it the energy $2.49$ eV at once, and the electron is emitted with no waiting time. Whether any particular atom gets a photon in the first moment is a matter of chance, but a target of area $1\ \text{cm}^2$ is a vast number of atoms. The number of photons reaching it per second is $$\frac{5.0\times10^{19}\times10^{-4}}{4\pi\times2^2} \approx 1.0\times10^{14}\ \text{s}^{-1},$$ so within any microsecond hundreds of millions of photons arrive, and some atom is bound to absorb one almost at once. Emission therefore appears instantaneous, as observed, and no accumulation of energy is needed.

(The key prints $\Delta t = 28.4$ s and $N = 2$. With the given numbers $\Delta t = 11.4$ s; and $N = 2$ photons would deliver $5$ eV, not the $2$ eV the question asks about. The key also gives no answer to part (v).)

Printed key errors. Q24: the key’s $0.21$ eV does not follow from its own numbers, which give $0.080$ eV. Q29: the key’s $28.4$ s and $N = 2$ do not follow from the given data; they should be $11.4$ s and $0.8$ photon, and part (v) is left unanswered.

Common mistakes

• Taking the wrong mass ratio in Question 14. The wavelength depends on $\sqrt{mq}$, so the $\alpha$-particle’s $m_\alpha q_\alpha = 8\,m_pq_p$, not $4$, and the ratio is $\sqrt8$.
• Counting the work function twice in Question 15. An electron that absorbs two photons pays $\phi_0$ once, giving $2h\nu – \phi_0$, not $2(h\nu – \phi_0)$.
• Writing $1:500$ the wrong way round in Question 18. The longer wavelength has the lower photon energy, so it has more photons; X-rays are the $1$ in $1:500$.
• Applying “doubles” to the photon energy in Question 20. It is the kinetic energy that doubles, so $E_2 – \phi = 2(E_1 – \phi)$, not $E_2 = 2E_1$.
• Equating frequencies in Question 22. Equal intensity means equal energy per second, $n\,h\nu$, not equal photon numbers.
• Giving only one case in Question 23. The momenta can be parallel or opposed, and the answer differs.
• Using $\sin\theta$ for the wrong angle in Question 24. In Bragg’s relation $\theta$ is measured from the planes, which is what the question gives, so $2d\sin\theta = \lambda$ with $\theta = 30^\circ$.
• Mixing up the units in Question 26. Convert joules to electron volts by dividing by $1.6\times10^{-19}$.
• Concluding “not consistent” in Question 27 from different intercepts. Different intercepts only mean different work functions; it is the different slopes that break Einstein’s relation.
• Forgetting the modulus in Question 28. If $m_B > m_A$ the particle A rebounds, and $\lambda$ uses the speed, not the signed velocity.

Practise next

• Exemplar Dual Nature of Radiation and Matter: Multiple Choice Questions — the page for Questions 1 to 13, on the photoelectric effect, de Broglie waves and the Davisson–Germer experiment.
• NCERT textbook Chapter 11 — Dual Nature of Radiation and Matter for the theory of photons, Einstein’s equation and matter waves.
• Exemplar Current Electricity — an earlier unit, whose charges, potentials and fields underlie the accelerating-potential ideas used here.
• A photon of wavelength 400 nm falls on a metal of work function 2.0 eV. What is the stopping potential?

• An electron is accelerated from rest through $150$ V. Find its de Broglie wavelength.