NCERT Class 12 Physics — Dual Nature of Radiation and Matter, Chapter 11 Exercises. All 11 questions solved.

The eleven exercises at the end of Chapter 11 fall into three groups. Question 11.1 is about X-rays: the most energetic photon an electron beam can produce. Questions 11.2 to 11.9 are about photons and the photoelectric effect — Einstein’s equation, the stopping potential, the threshold frequency, and the energy and momentum of a single photon. Questions 11.10 and 11.11 are about de Broglie’s matter waves. Everything runs on three results:

• Photon energy and momentum — $E = h\nu = \dfrac{hc}{\lambda}$ and $p = \dfrac{h}{\lambda}$.
• Einstein’s photoelectric equation — $K_{\max} = eV_0 = h\nu – \phi_0 = h(\nu – \nu_0)$, where $\phi_0 = h\nu_0$ is the work function, $\nu_0$ the threshold frequency and $V_0$ the stopping (cut-off) potential.
• de Broglie wavelength — $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$.

Unless a question gives its own values, the constants used are $h = 6.63\times10^{-34}\ \text{J s}$, $c = 3\times10^8\ \text{m s}^{-1}$, $e = 1.6\times10^{-19}\ \text{C}$ and $m_e = 9.1\times10^{-31}\ \text{kg}$, so that $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$.

Key insight. Keep the energy accounts in electronvolts. A photon of energy $h\nu$ arriving at a metal pays the work function first, and whatever is left over is the most kinetic energy an electron can leave with. Measured in electronvolts, that leftover energy is numerically the stopping potential in volts, because stopping a charge $e$ takes an energy $eV_0$. So Question 11.3 is only a change of units, Questions 11.2, 11.6 and 11.9 are one subtraction each, and Question 11.7 is a comparison of two numbers. Convert to joules only when a speed or a frequency has to come out (11.2(c), 11.8).

Question 11.1

Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by $30\ \text{kV}$ electrons.

Question 11.1 (a)

Solution. An electron accelerated through $30\ \text{kV}$ reaches the target with kinetic energy $eV = 30\ \text{keV}$. When it is stopped, that energy can come out as X-ray photons. The most energetic photon possible is the one that carries off the electron’s entire energy in a single step, so $h\nu_{\max} = eV$:

$$\nu_{\max} = \frac{eV}{h} = \frac{(1.6\times10^{-19})(30\times10^{3})}{6.63\times10^{-34}} = \frac{4.8\times10^{-15}}{6.63\times10^{-34}} = 7.24\times10^{18}\ \text{Hz}$$

Question 11.1 (b)

Solution. Wavelength and frequency are inversely related, $\lambda = c/\nu$, so the shortest wavelength belongs to that same highest-frequency photon:

$$\lambda_{\min} = \frac{c}{\nu_{\max}} = \frac{3\times10^8}{7.24\times10^{18}} = 4.14\times10^{-11}\ \text{m}$$

Question 11.2

The work function of caesium metal is $2.14\ \text{eV}$. When light of frequency $6\times10^{14}\ \text{Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?

Question 11.2 (a)

Solution. Each electron absorbs one photon. It spends at least the work function getting out of the metal, and the electrons that spend exactly that much leave with the most energy. So $K_{\max} = h\nu – \phi_0$. The photon energy, converted to electronvolts so that it can be compared with the work function, is

$$h\nu = (6.63\times10^{-34})(6\times10^{14}) = 3.98\times10^{-19}\ \text{J} = \frac{3.98\times10^{-19}}{1.6\times10^{-19}}\ \text{eV} = 2.49\ \text{eV}$$

and so

$$K_{\max} = 2.49 – 2.14 = 0.35\ \text{eV} = (0.346)(1.6\times10^{-19}) = 5.5\times10^{-20}\ \text{J}$$

Note on the book’s answer. The key prints $0.34\ \text{eV} = 0.54\times10^{-19}\ \text{J}$. $K_{\max}$ is the small difference of two nearly equal energies, so its second figure depends on the constants used: with $h = 6.626\times10^{-34}\ \text{J s}$ and $e = 1.602\times10^{-19}\ \text{C}$ the photon energy is $2.48\ \text{eV}$ and $K_{\max} = 0.34\ \text{eV}$, the key’s value. This is a difference in constants, not an error.

Question 11.2 (b)

Solution. The stopping potential is the reverse voltage that just turns back the fastest electrons. Pushing a charge $e$ through a potential difference $V_0$ takes energy $eV_0$, so $eV_0 = K_{\max}$. With $K_{\max}$ in electronvolts, $V_0$ in volts is the same number.

Question 11.2 (c)

Solution. Now the energy has to be in joules, because the speed comes from $K_{\max} = \tfrac12 m_e v_{\max}^2$ in SI units:

$$v_{\max} = \sqrt{\frac{2K_{\max}}{m_e}} = \sqrt{\frac{2\times5.54\times10^{-20}}{9.1\times10^{-31}}} = \sqrt{1.22\times10^{11}} = 3.5\times10^5\ \text{m s}^{-1}$$

Note on the book’s answer. The key prints $344\ \text{km/s}$, which is what the rounded $0.54\times10^{-19}\ \text{J}$ gives. Carrying $5.54\times10^{-20}\ \text{J}$ through gives $349\ \text{km/s}$, and precise constants give $347\ \text{km/s}$. All three are $3.5\times10^5\ \text{m s}^{-1}$ to two figures; the spread is rounding.

Question 11.3

The photoelectric cut-off voltage in a certain experiment is $1.5\ \text{V}$. What is the maximum kinetic energy of photoelectrons emitted?

Solution. The cut-off voltage is the retarding potential that just stops the fastest photoelectrons from reaching the collector. To be stopped, an electron must do work $eV_0$ against the field, so its kinetic energy must have been exactly that:

$$K_{\max} = eV_0 = (1.6\times10^{-19})(1.5) = 2.4\times10^{-19}\ \text{J}$$

In electronvolts no arithmetic is needed at all: an electron stopped by $1.5\ \text{V}$ had $1.5\ \text{eV}$.

Question 11.4

Monochromatic light of wavelength $632.8\ \text{nm}$ is produced by a helium-neon laser. The power emitted is $9.42\ \text{mW}$. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Question 11.4 (a)

Solution. Each photon carries energy $E = h\nu = hc/\lambda$. Because it travels at $c$, its momentum is $p = E/c = h/\lambda$:

$$E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^8)}{632.8\times10^{-9}} = 3.14\times10^{-19}\ \text{J} \approx 1.96\ \text{eV}$$

$$p = \frac{h}{\lambda} = \frac{6.63\times10^{-34}}{632.8\times10^{-9}} = 1.05\times10^{-27}\ \text{kg m s}^{-1}$$

Question 11.4 (b)

Solution. The power of the beam is the energy it delivers per second. Every photon in monochromatic light carries the same energy, so the number arriving per second is the power divided by the energy of one photon. The beam is narrower than the target, so every photon lands on it:

$$N = \frac{P}{E} = \frac{9.42\times10^{-3}}{3.14\times10^{-19}} = 3.0\times10^{16}\ \text{photons per second}$$

Question 11.4 (c)

Solution. A hydrogen atom has mass $m_H \approx 1.67\times10^{-27}\ \text{kg}$, about that of a proton. For its momentum $m_H v$ to equal the photon’s:

$$v = \frac{p}{m_H} = \frac{1.05\times10^{-27}}{1.67\times10^{-27}} = 0.63\ \text{m s}^{-1}$$

The answer shows how little momentum a single visible photon carries: a hydrogen atom matches it at walking pace.

Question 11.5

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12\times10^{-15}\ \text{V s}$. Calculate the value of Planck’s constant.

Solution. Einstein’s equation, $eV_0 = h\nu – \phi_0$, divided through by $e$, reads

$$V_0 = \frac{h}{e}\,\nu – \frac{\phi_0}{e}$$

For a given metal $\phi_0/e$ is a constant, so a graph of $V_0$ against $\nu$ is a straight line whose slope is $h/e$, the same for every metal. That is how Millikan measured Planck’s constant, and it is what this question asks us to do:

$$h = e\times\text{slope} = (1.6\times10^{-19})(4.12\times10^{-15}) = 6.59\times10^{-34}\ \text{J s}$$

Question 11.6

The threshold frequency for a certain metal is $3.3\times10^{14}\ \text{Hz}$. If light of frequency $8.2\times10^{14}\ \text{Hz}$ is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Solution. The work function is the energy of a photon at the threshold frequency, $\phi_0 = h\nu_0$, so Einstein’s equation can be written as $eV_0 = h\nu – h\nu_0 = h(\nu – \nu_0)$:

$$V_0 = \frac{h(\nu – \nu_0)}{e} = \frac{(6.63\times10^{-34})(8.2 – 3.3)\times10^{14}}{1.6\times10^{-19}} = \frac{3.25\times10^{-19}}{1.6\times10^{-19}} = 2.03\ \text{V}$$

Question 11.7

The work function for a certain metal is $4.2\ \text{eV}$. Will this metal give photoelectric emission for incident radiation of wavelength $330\ \text{nm}$?

Solution. An electron absorbs one photon at a time, so emission happens only if a single photon has at least the work function’s worth of energy — that is, only if $\nu \geq \nu_0$. The threshold frequency is

$$\nu_0 = \frac{\phi_0}{h} = \frac{4.2\times1.6\times10^{-19}}{6.63\times10^{-34}} = 1.01\times10^{15}\ \text{Hz}$$

and the frequency of the incident radiation is

$$\nu = \frac{c}{\lambda} = \frac{3\times10^8}{330\times10^{-9}} = 9.09\times10^{14}\ \text{Hz}$$

Since $\nu < \nu_0$, no electrons are emitted. The same comparison in energies: each photon carries $hc/\lambda = 3.77\ \text{eV}$, short of the $4.2\ \text{eV}$ needed. The longest wavelength that would work is $hc/\phi_0 = 296\ \text{nm}$. Making the light more intense would not help: it sends more photons, but none of them has enough energy on its own.

Question 11.8

Light of frequency $7.21\times10^{14}\ \text{Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0\times10^5\ \text{m/s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Solution. Einstein’s equation in the form $h\nu = h\nu_0 + \tfrac12 m_e v_{\max}^2$ says the photon’s energy goes partly into freeing the electron and partly into its kinetic energy. The kinetic energy of the fastest electrons is

$$K_{\max} = \tfrac12(9.1\times10^{-31})(6.0\times10^5)^2 = 1.64\times10^{-19}\ \text{J}$$

Dividing Einstein’s equation by $h$ turns every energy into a frequency:

$$\nu_0 = \nu – \frac{K_{\max}}{h} = 7.21\times10^{14} – \frac{1.64\times10^{-19}}{6.63\times10^{-34}} = (7.21 – 2.47)\times10^{14} = 4.74\times10^{14}\ \text{Hz}$$

Note on the book’s answer. The key prints $4.73\times10^{14}\ \text{Hz}$. With precise constants the result is $4.735\times10^{14}\ \text{Hz}$, so the two differ only in how the last figure is rounded.

Question 11.9

Light of wavelength $488\ \text{nm}$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38\ \text{V}$. Find the work function of the material from which the emitter is made.

Solution. Rearranged, Einstein’s equation gives the work function as the photon energy minus the kinetic energy of the fastest electrons: $\phi_0 = h\nu – eV_0$. The photon energy is

$$\frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^8)}{488\times10^{-9}} = 4.08\times10^{-19}\ \text{J} = 2.55\ \text{eV}$$

A stopping potential of $0.38\ \text{V}$ means $K_{\max} = 0.38\ \text{eV}$, so

$$\phi_0 = 2.55 – 0.38 = 2.17\ \text{eV} = 3.47\times10^{-19}\ \text{J}$$

Note on the book’s answer. The key prints $2.16\ \text{eV} = 3.46\times10^{-19}\ \text{J}$. With precise values of $h$, $c$ and $e$ the photon energy is $2.54\ \text{eV}$ and the work function $2.16\ \text{eV}$, so the difference comes from the constants, not from an error.

Question 11.10

What is the de Broglie wavelength of (a) a bullet of mass $0.040\ \text{kg}$ travelling at the speed of $1.0\ \text{km/s}$, (b) a ball of mass $0.060\ \text{kg}$ moving at a speed of $1.0\ \text{m/s}$, and (c) a dust particle of mass $1.0\times10^{-9}\ \text{kg}$ drifting with a speed of $2.2\ \text{m/s}$?

Solution. De Broglie proposed that any particle of momentum $p = mv$ has a wave associated with it, of wavelength $\lambda = h/p = h/mv$. Each part is a direct substitution, with every speed in metres per second.

Question 11.10 (a)

Solution. The speed is $1.0\ \text{km/s} = 1.0\times10^3\ \text{m/s}$, so $p = (0.040)(1.0\times10^3) = 40\ \text{kg m s}^{-1}$:

$$\lambda = \frac{6.63\times10^{-34}}{40} = 1.66\times10^{-35}\ \text{m}$$

Question 11.10 (b)

Solution. Here $p = (0.060)(1.0) = 0.060\ \text{kg m s}^{-1}$:

$$\lambda = \frac{6.63\times10^{-34}}{0.060} = 1.1\times10^{-32}\ \text{m}$$

Question 11.10 (c)

Solution. Here $p = (1.0\times10^{-9})(2.2) = 2.2\times10^{-9}\ \text{kg m s}^{-1}$:

$$\lambda = \frac{6.63\times10^{-34}}{2.2\times10^{-9}} = 3.0\times10^{-25}\ \text{m}$$

All three wavelengths are far smaller than a nucleus, about $10^{-15}\ \text{m}$ across, so there is no slit or obstacle small enough to diffract them. That is why the wave nature of everyday objects, even one as light as a speck of dust, is never seen.

Note on the book’s answer. The key prints $3.0\times10^{-23}\ \text{m}$, a hundred times the value above. Putting it back into $\lambda = h/p$ would need a momentum of $6.63\times10^{-34}/3.0\times10^{-23} = 2.2\times10^{-11}\ \text{kg m s}^{-1}$, but the particle’s momentum is $2.2\times10^{-9}\ \text{kg m s}^{-1}$. The printed power of ten is a misprint; the wavelength is $3.0\times10^{-25}\ \text{m}$.

Question 11.11

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Solution. Electromagnetic radiation of frequency $\nu$ has wavelength $\lambda = c/\nu$. Its quantum, the photon, has energy $E = h\nu$, and since it travels at speed $c$ and has no rest mass, its momentum is $p = E/c = h\nu/c$. The de Broglie wavelength of the photon is therefore

$$\lambda_{\text{dB}} = \frac{h}{p} = \frac{h}{h\nu/c} = \frac{c}{\nu}$$

which is exactly the wavelength of the radiation. De Broglie’s relation, invented for material particles, gives the right answer when it is applied back to light, which is the consistency the idea needs.

Common mistakes

• Question 11.1(b): using the electron’s de Broglie wavelength. The question asks for the wavelength of the X-ray photon, $c/\nu_{\max}$. Working out $h/p$ for the $30\ \text{keV}$ electron instead gives about $0.007\ \text{nm}$, a real wavelength but of the wrong thing.
• Question 11.2(a): subtracting electronvolts from joules. The photon energy comes out of $h\nu$ in joules and the work function is given in electronvolts. Taking $2.14$ away from $3.98\times10^{-19}$ is meaningless; convert one of them first.
• Question 11.2(c): putting electronvolts into $\tfrac12mv^2$. The kinetic energy must be in joules when the mass is in kilograms. Using $0.35$ directly gives a speed of nearly $10^{15}\ \text{m s}^{-1}$, faster than light, which should be a warning.
• Question 11.4(b): forgetting that the power is in milliwatts. $9.42\ \text{mW}$ is $9.42\times10^{-3}\ \text{W}$. Using $9.42\ \text{W}$ gives $3\times10^{19}$ photons per second, a thousand times too many.
• Question 11.5: taking the slope to be $h$ itself. The graph is of cut-off voltage, not energy, against frequency, so its slope is $h/e$. Reading the slope off as Planck’s constant gives $4.12\times10^{-15}$, wrong by the factor $e$.
• Question 11.7: expecting brighter light to cause emission. Each electron absorbs a single photon, so what matters is the energy per photon, $3.77\ \text{eV}$, against the $4.2\ \text{eV}$ needed. More intense light only sends more photons that are each still too weak.
• Question 11.7: comparing the wavelengths the wrong way round. A longer wavelength means a lower frequency and less energy per photon. The threshold wavelength is $296\ \text{nm}$, and $330\ \text{nm}$ is longer, so it falls short; it is not “more than enough”.
• Question 11.10(a): leaving the speed in km/s. Using $v = 1.0$ instead of $1.0\times10^3\ \text{m/s}$ makes the wavelength a thousand times too large.

Practise next

• Chapter 12, Atoms — photons emitted and absorbed as electrons jump between energy levels, and Bohr’s allowed orbits read as whole numbers of de Broglie wavelengths.
• Chapter 10, Wave Optics — interference and the wave picture of light, which the photoelectric effect shows to be only half the story.