Exercises — Dual Nature of Radiation and Matter

1. 12 marksNCERT Exercises, Chapter 11

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

   Hint. The most energetic X-ray photon is produced when an electron gives up all its kinetic energy in a single collision.

   An electron accelerated through a potential difference $V$ gains kinetic energy $eV$. The most energetic X-ray photon possible is produced when an electron surrenders all of that energy in one collision, so:

   $h{\nu }_{\max }=eV$

   (a) Substituting $V=30$ kV:

   ${\nu }_{\max }=\frac{eV}{h}=\frac{1.6\times 10^{-19}\times 30\times 10^3}{6.63\times 10^{-34}}=7.24\times 10^{18}\text{Hz}$

   (b) The corresponding wavelength is the shortest present:

   ${\lambda }_{\min }=\frac{c}{{\nu }_{\max }}=\frac{3\times 10^8}{7.24\times 10^{18}}=4.14\times 10^{-11}\text{m}$

   This is about $0.041$ nm. The sharp short-wavelength cut-off is a purely quantum effect, since classical theory places no limit on the frequency radiated by a decelerating charge.

   ✦ (a) 7.24 x 10^18 Hz (b) 4.14 x 10^-11 m, about 0.041 nm

2. 23 marksNCERT Exercises, Chapter 11

   The work function of caesium metal is 2.14 eV. When light of frequency $6\times 10^{14}$ 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?

   Hint. Apply Einstein's photoelectric equation, then convert the kinetic energy into a stopping potential and a speed.

   The photon energy is:

   $E=h\nu =6.63\times 10^{-34}\times 6\times 10^{14}=3.98\times 10^{-19}\text{J}=2.49\text{eV}$

   (a) Einstein's photoelectric equation gives the maximum kinetic energy as the photon energy less the work function:

   $K_{\max }=h\nu -{\varphi }_0=2.49-2.14=0.35\text{eV}$

   (b) The stopping potential is the potential that just halts the fastest electrons, so numerically:

   $e{V}_0=K_{\max }⟹V_0=0.35\text{V}$

   Expressed in volts it always equals the maximum kinetic energy expressed in electronvolts, which is why the electronvolt is the convenient unit here.

   (c) Converting to joules, $K_{\max }=0.346\times 1.6\times 10^{-19}=5.54\times 10^{-20}$ J, and from $K=\frac{1}{2}m{v}^2$:

   $v_{\max }=\sqrt{\frac{2K_{\max }}{m}}=\sqrt{\frac{2\times 5.54\times 10^{-20}}{9.1\times 10^{-31}}}=3.49\times 10^5{\text{m s}}^{-1}$

   That is about 349 km s${}^{-1}$, and only the fastest electrons travel this quickly, since electrons emitted from below the surface lose energy on the way out.

   ✦ (a) 0.35 eV (b) 0.35 V (c) 3.49 x 10^5 m/s, about 349 km/s

3. 31 markNCERT Exercises, Chapter 11

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

   Hint. The cut-off voltage is defined as the voltage that just stops the most energetic electrons.

   The cut-off, or stopping, voltage is the reverse potential that just prevents the fastest photoelectrons from reaching the collector. At that point their entire kinetic energy has been spent against the retarding field:

   $K_{\max }=e{V}_0$

   Substituting $V_0=1.5$ V:

   $K_{\max }=1.6\times 10^{-19}\times 1.5=2.4\times 10^{-19}\text{J}$

   Equivalently this is simply $1.5$ eV, because one electronvolt is by definition the energy an electron gains across one volt.

   ✦ K_max = 2.4 x 10^-19 J, that is 1.5 eV

4. 44 marksNCERT Exercises, Chapter 11

   Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 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? (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

   Hint. Energy per photon comes from hc/lambda and momentum from h/lambda; the photon rate is the power divided by the energy per photon.

   (a) The energy of each photon is:

   $E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{632.8\times 10^{-9}}=3.14\times 10^{-19}\text{J}=1.96\text{eV}$

   Its momentum is:

   $p=\frac{h}{\lambda }=\frac{6.63\times 10^{-34}}{632.8\times 10^{-9}}=1.05\times 10^{-27}{\text{kg m s}}^{-1}$

   A photon has momentum despite having no rest mass, because momentum for a massless particle is $E/c$ rather than $mv$.

   (b) Power is energy delivered per second, so the number of photons arriving per second is:

   $N=\frac{P}{E}=\frac{9.42\times 10^{-3}}{3.14\times 10^{-19}}=3.0\times 10^{16}{\text{photons s}}^{-1}$

   The number is enormous, which is why a laser beam appears perfectly continuous rather than granular.

   (c) For a hydrogen atom of mass $1.67\times 10^{-27}$ kg to carry the same momentum:

   $v=\frac{p}{m}=\frac{1.05\times 10^{-27}}{1.67\times 10^{-27}}=0.63{\text{m s}}^{-1}$

   So a hydrogen atom strolling at under a metre per second matches the momentum of a visible-light photon.

   ✦ (a) E = 3.14 x 10^-19 J, that is 1.96 eV; p = 1.05 x 10^-27 kg m/s (b) 3.0 x 10^16 photons per second (c) 0.63 m/s

5. 52 marksNCERT Exercises, Chapter 11

   In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12\times 10^{-15}$ V s. Calculate the value of Planck's constant.

   Hint. Rearrange Einstein's equation into the form of a straight line in V0 against frequency and identify the slope.

   Einstein's photoelectric equation is $e{V}_0=h\nu -{\varphi }_0$. Dividing through by $e$ puts it in the form of a straight line:

   $V_0=\left(\frac{h}{e}\right)\nu -\frac{{\varphi }_0}{e}$

   So a graph of stopping potential against frequency is a straight line whose slope is $h/e$, and whose intercept gives the work function.

   Therefore:

   $h=e\times \text{slope}=1.6\times 10^{-19}\times 4.12\times 10^{-15}=6.59\times 10^{-34}\text{J s}$

   This agrees closely with the accepted value of $6.63\times 10^{-34}$ J s. What makes the result remarkable is that the slope is the same for every metal, since it contains no material constant, and Millikan's careful measurement of it was what confirmed Einstein's equation experimentally.

   ✦ h = 6.59 x 10^-34 J s

6. 62 marksNCERT Exercises, Chapter 11

   The threshold frequency for a certain metal is $3.3\times 10^{14}$ Hz. If light of frequency $8.2\times 10^{14}$ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

   Hint. Write Einstein's equation using the threshold frequency in place of the work function.

   Since the work function equals $h{\nu }_0$, Einstein's equation can be written entirely in terms of frequencies:

   $e{V}_0=h(\nu -{\nu }_0)$

   Substituting $\nu =8.2\times 10^{14}$ Hz and ${\nu }_0=3.3\times 10^{14}$ Hz:

   $V_0=\frac{6.63\times 10^{-34}\times (8.2-3.3)\times 10^{14}}{1.6\times 10^{-19}}=\frac{6.63\times 10^{-34}\times 4.9\times 10^{14}}{1.6\times 10^{-19}}$

   $V_0=2.03\text{V}$

   Only the difference of the frequencies matters, so no separate value of the work function is needed here.

   ✦ Cut-off voltage = 2.03 V

7. 72 marksNCERT Exercises, Chapter 11

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

   Hint. Compare the photon energy with the work function; emission occurs only if the photon energy is the larger.

   Photoemission requires each individual photon to carry at least the work function, since one photon is absorbed by one electron and energy cannot be accumulated from several.

   The photon energy at 330 nm is:

   $E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{330\times 10^{-9}}=6.03\times 10^{-19}\text{J}$

   Converting to electronvolts:

   $E=\frac{6.03\times 10^{-19}}{1.6\times 10^{-19}}=3.77\text{eV}$

   Since $3.77\text{eV}<4.2\text{eV}$, no photoemission occurs.

   Increasing the intensity would not help either, because that merely sends more photons, each still too weak individually. Only radiation of shorter wavelength, below the threshold of $hc/{\varphi }_0=295$ nm, would eject electrons.

   ✦ No. The photon energy is 3.77 eV, which is less than the 4.2 eV work function, so no emission occurs at any intensity

8. 83 marksNCERT Exercises, Chapter 11

   Light of frequency $7.21\times 10^{14}$ Hz is incident on a metal surface. Electrons with a maximum speed of $6.0\times 10^5$ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

   Hint. Find the maximum kinetic energy from the speed, then work backwards through Einstein's equation.

   First find the maximum kinetic energy from the measured speed:

   $K_{\max }=\frac{1}{2}m{v}^2=\frac{1}{2}\times 9.1\times 10^{-31}\times (6.0\times 10^5{)}^2=1.64\times 10^{-19}\text{J}$

   Einstein's equation in frequency form is $h\nu =h{\nu }_0+K_{\max }$, so:

   ${\nu }_0=\nu -\frac{K_{\max }}{h}=7.21\times 10^{14}-\frac{1.64\times 10^{-19}}{6.63\times 10^{-34}}$

   ${\nu }_0=7.21\times 10^{14}-2.47\times 10^{14}=4.74\times 10^{14}\text{Hz}$

   The threshold frequency is a property of the metal alone, so this same value would be obtained whatever frequency of light had been used in the experiment.

   ✦ Threshold frequency = 4.74 x 10^14 Hz

9. 93 marksNCERT Exercises, Chapter 11

   Light of wavelength 488 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 V. Find the work function of the material from which the emitter is made.

   Hint. Find the photon energy from the wavelength, then subtract the energy corresponding to the stopping potential.

   The photon energy at 488 nm is:

   $E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{488\times 10^{-9}}=4.08\times 10^{-19}\text{J}$

   In electronvolts:

   $E=\frac{4.08\times 10^{-19}}{1.6\times 10^{-19}}=2.55\text{eV}$

   Einstein's equation rearranged for the work function gives:

   ${\varphi }_0=h\nu -e{V}_0=2.55-0.38=2.17\text{eV}$

   Working in electronvolts throughout makes the subtraction trivial, since the stopping potential in volts is numerically the maximum kinetic energy in electronvolts.

   ✦ Work function = 2.17 eV

10. 103 marksNCERT Exercises, Chapter 11

    What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass $1.0\times 10^{-9}$ kg drifting with a speed of 2.2 m/s?

    Hint. Apply the de Broglie relation to each, and compare the results with the size of an atom.

    The de Broglie wavelength of any particle is:

    $\lambda =\frac{h}{mv}$

    (a) Bullet: $\lambda =\frac{6.63\times 10^{-34}}{0.040\times 1000}=1.7\times 10^{-35}$ m

    (b) Ball: $\lambda =\frac{6.63\times 10^{-34}}{0.060\times 1.0}=1.1\times 10^{-32}$ m

    (c) Dust particle: $\lambda =\frac{6.63\times 10^{-34}}{1.0\times 10^{-9}\times 2.2}=3.0\times 10^{-25}$ m

    Why none of these can ever be observed. An atomic nucleus is about $10^{-15}$ m across and an atom about $10^{-10}$ m. Even the largest of these wavelengths is ten orders of magnitude smaller than a nucleus, so no aperture or crystal lattice exists that could diffract them.

    Wave behaviour becomes measurable only when the mass is tiny, which is why electrons, with wavelengths comparable to atomic spacings, do show diffraction while everyday objects never do.

    ✦ (a) 1.7 x 10^-35 m (b) 1.1 x 10^-32 m (c) 3.0 x 10^-25 m, all far too small to be detected

11. 113 marksNCERT Exercises, Chapter 11

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

    Hint. Write the photon's momentum in terms of its energy, then substitute into the de Broglie relation.

    The de Broglie wavelength of any particle of momentum $p$ is:

    ${\lambda }_{dB}=\frac{h}{p}$

    A photon has zero rest mass, so its momentum cannot be written as $mv$. Instead the relativistic relation for a massless particle gives:

    $p=\frac{E}{c}$

    The energy of a photon of radiation of frequency $\nu$ is $E=h\nu$, so:

    $p=\frac{h\nu }{c}$

    Substituting this momentum into the de Broglie relation:

    ${\lambda }_{dB}=\frac{h}{p}=\frac{h}{h\nu /c}=\frac{c}{\nu }$

    But $c/\nu$ is precisely the wavelength $\lambda$ of the electromagnetic radiation itself. Hence:

    ${\lambda }_{dB}=\lambda$

    The two descriptions agree exactly, which is the consistency check that makes wave-particle duality a single coherent picture rather than two rival ones. The same $h$ appears in both the wave and particle descriptions of light.

    ✦ lambda_dB = h/p = h/(h nu/c) = c/nu = lambda, so the two wavelengths are identical

Solutions written by the tuition.in editorial team and checked against leph203.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.