# NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Q. An electron, an a-particle and a proton have the same kinetic energy. Which of these particles has the shortest de-Broglie wavelength ?

Ans For a particle, de-Broglie wavelength

$$(\lambda)=\frac{h}{p}\\\text{Kinetic energy (K)}=\frac{\text{p}^{2}}{2\text{m}}\\\text{Then,}\space\lambda=\frac{h}{\sqrt{2\text{mk}}}\\\text{For the same kinetic energy K},\\\text{the de-Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.}\\\text{A proton}(^{1}_1\text{H})\text{is 1836 times massive than an electron and an }\alpha-\text{particle is}(^{4}_{2}\text{He})\\\text{four times that of a proton.}\\\text{Hence,}\space\alpha-\text{particle has the shortest de-Broglie wavelength.}$$

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

• Ans. Given, Photoelectric cut-off voltage (V0) = 1.5 V
• The maximum kinetic energy of the emitted photoelectrons is given by
• Kmax = eV0
• Here, charge on an electron (e) = 1.6 × 10– 19 C
• ∴ Kmax = 1.6 × 10– 19 × 1.5
• = 2.4 × 10– 19 J
• Hence, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10– 19 J.

Q. Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons ?

Ans Given, frequency of the incident photon
(v) = 7·21 × 1014 Hz

Maximum speed of the electrons (v) = 6.0 × 105 m/s

Planck’s constant (h) = 6.626 × 10– 34 Js

Mass of an electron (m) = 9.1 × 10– 31 kg

For threshold frequency (v0) kinetic energy is given by

$$\frac{1}{2}\text{mv}^{2}=h(v-v_0)\\v_0=v-\frac{mv^{2}}{2h}\\=7.21×10^{14}-\frac{(9.1×10^{-31})×(6×10^{5})^{2}}{2×(6.626×10^{-34})}\\=7.21×10^{14}-2.472×10^{14}\\=4.738×10^{14}\text{Hz}$$

Hence, the threshold frequency for the photoemission of electrons is  4.738×1014 Hz
Q. The threshold frequency for a certain metal is 3.3×1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Ans Given, threshold frequency of the metal (v0) = 3.3 × 1014 Hz

Frequency of light incident on the metal (v) = 8.2 × 1014 Hz

Charge on an electron (e) = 1.6 × 10– 19 C

Planck’s constant, h = 6.626 × 10– 34 Js

Cut-off voltage for the photoelectric emission from the metal = V0

The equation for the cut-off energy is given by

$$ev_0=h(v-v_0)\\\text{v}_0=\frac{h(v-v_0)}{e}\\=\frac{6.626×10^{-34}×(8.2×10^{14}-3.3×10^{14})}{1.6×10^{-19}}\\=2.0292\space\text{V}$$

Q. 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 photo-electrons is 0.38 V. Find the work function of the material from which the emitter is made.

Ans Given, Wavelength of light produced by the argon laser (l) = 488 nm = 488 × 10– 9 m

Stopping potential of the photoelectrons (V0) = 0.38 V

$$1\text{eV}=1.6×10^{-19}\text{J}\\\therefore(\text{V}_0)=\frac{0.38}{1.6×10^{-19}}\text{eV}\\\text{Planck’s constant (h) =}6.6×10^{-34}\text{Js}\\\text{Charge on an electron =(e)}=1.6×10^{-19}\text{C}\\\text{Speed of light (c)}=3×10^{8}\space\text{m/s}\\\text{From Einstein’s photoelectric effect,}\\\text{we have the relation involving the work function}\space\phi_0\space\text{of the material of the emitter}\\eV_0=\frac{hc}{\lambda}-\phi_0\\\phi_0=\frac{hc}{\lambda}-eV_0\\=\frac{6.6×10^{-34}×3×10^{8}}{1.6×10^{-19}×488×10^{-9}}-\frac{1.6×10^{-19}×0.38}{1.6×10^{-19}}\\=2.54-0.38=2.16\text{eV}$$

Therefore, the material with which the emitter is made has the work function of 2.16 eV.

Q. Find the :

• (i) Maximum frequency, and
• (ii) Minimum wavelength of X-rays produced by 30 kV electrons.

Ans Given, Potential of the electrons (V) = 30 kV = 3 × 104 V

Hence, energy of the electrons (E) = 3 × 104 eV

Charge on an electron (e) = 1.6 × 10– 19 C

(i) Maximum frequency produced by the X-rays = n

The energy of the electrons is given by

(E) = hv

$$\text{Here, h= Planck’s constant =6.626×10}^{-34}\text{Js}\\\therefore v=\frac{\text{E}}{\text{h}}\\=\frac{1.6×10^{-19}×3×10^{4}}{6.626×10^{-34}}\\=7.24×10^{18}\text{Hz}.\\\text{The maximum frequency of X-rays produced is}\space 7.24×10^{18}\text{Hz}.\\\text{(ii)}\space\text{The minimum wavelength produced by the X-rays is given by}\\\lambda=\frac{c}{v}=\frac{3×10^{8}}{7.24×10^{18}}\\=4.14×10^{-11}m\\=0.0414\text{nm}\\\text{Hence, the minimum wavelength of X-rays produced is 0.0414 nm.}$$

Q. The work function of caesium is 2.14 eV. Find :

• (i) the threshold frequency for caesium, and
• (ii) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V.
Ans. (i) For the cut-off or threshold frequency, the energy (hv0) of the incident radiation must be equal to work function (Φ0), so that

$$v_0=\frac{\phi_0}{h}\\=\frac{2.14\text{eV}}{6.63×10^{-34}\text{Js}}\\=\frac{2.14×1.6×10^{-19}\text{J}}{6.63×10^{-34}\text{Js}}\\=5.16×10^{14}\text{Hz}\\\text{Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.}\\\text{(ii) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals}\\\text{ the potential energy}(eV_0)\text{by the retarding potential}(v_0).\\\text{Einstein’s photoelectric equation is}\\eV_0=hv-\phi_0\\=\frac{hc}{\lambda}-\phi_0\\ \text{or}\space\lambda=\frac{hc}{eV_0+\phi_0}\\=\frac{(6.63×10^{-34}\text{Js})×(3×10^{8}\text{m/s})}{(0.60\text{eV}+2.14\text{eV})}\\=\frac{19.89×10^{-26}\text{Jm}}{(2.74\text{eV})}\\\lambda=\frac{19.89×10^{-26}}{2.74×1.6×10^{-19}\text{J}}\\\lambda=454\space\text{nm}$$

Q. In an experiment on photoelectric effect, the slope of the cut off voltage versus frequency of incident light is found to be 4.12 × 10– 15 Vs. Calculate the value of Planck’s constant.

Ans. The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given by

$$\frac{V}{v}=4.12×10^{\normalsize -15}\text{Vs}\\\text{V is related to frequency by}\\\text{hv}=\text{eV}\\\text{Here, Charge on an electron (e) =}1.6×10^{-19}\text{C}\\\text{Planck’s constant (h) = ?}\\\therefore\space\space h=e×\frac{V}{v}\\=1.6×10^{\normalsize -19}×4.12×10^{\normalsize -15}\\=6.592×10^{-34}\text{Js}\\\text{Hence, the value of Planck’s constant is}\space6.592×10^{\normalsize-34}\text{Js}.$$

Q. What is the de-Broglie wavelength associated with (i) an electron moving with a speed of 5.4 × 106 m/s and (ii) a ball of mass 150 g travelling at 30.0 m/s ?

Ans. (i) For the electron :

Mass (m) = 9.11 × 10– 31 kg,

speed (v) = 5.4 × 106 m/s.

Then, momentum p = mv = 9.11 × 10– 31 kg × 5.4 × 106 m/s,

p = 4.92 × 10– 24 kg m/s

$$\text{de-Broglie wavelength} (\lambda)=\frac{h}{p}\\=\frac{6.63×10^{-34}\text{Js}}{4.92×10^{-24}\text{Kgm/s}}\\\lambda=0.135\space\text{nm}\\\textbf{(ii) For the ball :}\\\text{Mass}(m’)=0.150\text{kg},\text{speed}(v’)=30.0\text{m/s}.\\\text{Then momentum p′}=m’v’=0.150\space\text{kg}×30.0\text{m/s},p’=4.50\text{kg\space m/s}\\\text{de-Broglie wavelength}\space\lambda’=\frac{h}{p’}\\=\frac{6.63×10^{-34}\text{Js}}{4.50\text{kgm/s}}\\\lambda’=1.47×10^{-34}\text{m}$$

The de-Broglie wavelength of electron is comparable with X-ray wavelength. However, for the ball it is about 10– 19 times the size of the proton, quite beyond experimental measurement.
Q. The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10– 15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

$$\textbf{Ans.}\space \text{Given, wavelength of a proton or a neutron (l) ≈}10^{-15}\text{m}\\\text{Rest mass energy of an electron (m}_0) = ?\\m_0c^{2}=0.511\text{Mev}\\= 0.511 × 10^{6} × 1.6 × 10^{–19}\\= 0.8176 × 10^{\normalsize– 13} J\\\text{Planck’s constant (h) =}6.6×10^{\normalsize-34}\text{Js}\\\text{Speed of light (c)}=3×10^{8}\text{m/s}\\\text{The momentum of a proton or a neutron is given by}\\\text{(p)}=\frac{h}{\lambda}=\frac{6.6×10^{-34}}{10^{-15}}\\=6.6×10^{-19}\text{kg\space m/s}\\\text{The relativistic relation for energy (E) is given as :}\\\text{E}^{2}=p^{2}c^{2} + m^{2}c^{4}\\=(6.6×10^{-19}×3×10^{8})^{2}+(0.8176×10^{-13})^{2}$$

$$=392.04×10^{-22}+0.6685×10^{-26}\\=392.04×10^{-22}\\\therefore\space\text{E}=1.98×10^{-10}\text{J}\\=\frac{1.98×10^{-10}}{1.6×10^{-19}}\text{eV}\\=1.24×10^{9}\text{eV}=1.24\text{GeV}$$

Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.

Q. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy ? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10– 31 kg).

Ans. An X-ray probe has a greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe (λ) = 1 Å = 10– 10 m

Mass of an electron (me) = 9.11 × 10– 31 kg

Planck’s constant (h) =6.64×10-34 Js

Charge on an electron (e) =1.6×10-19C

The kinetic energy of the electron is given by

$$\text{E}=\frac{1}{2}m_ev^{2}\\m_ev=\sqrt{2Em_e}\\\text{Here},v=\text{Velocity of the electron}\\m_ev=\text{Momentum (p) of the electron}\\\text{According to the de-Broglie principle, the de-Broglie wavelength is given by}\\\lambda=\frac{h}{p}=\frac{h}{m_ev}=\frac{h}{\sqrt{2Em_e}}\\\therefore\space\text{E}=\frac{h^{2}}{2\lambda^{2}m_e}\\=\frac{(6.6×10^{-34})^{2}}{2×(10^{-10})^{2}×9.11×10^{-31}}\\=2.39×10^{-17}\text{J}\\=\frac{2.39×10^{-17}}{1.6×10^{-19}}=149.375eV \text{Energy of a photon (E) =}\frac{hc}{\lambda_0}\\=\frac{6.6×10^{-34}×3×10^{8}}{10^{-10}×1.6×10^{-19}}eV\\=12.375×10^{3}eV\\=12.375keV\\\text{Hence, a photon has a greater energy than an electron for the same wavelength.}$$

Q. Find the typical de-Broglie wavelength associated with a He atom in helium gas at room temperature (27° C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions

Ans. Given, de-Broglie wavelength associated with the He atom = 0.7268 × 10– 10 m

Room temperature (T) = 27°C = 27 + 273 = 300 K

Atmospheric pressure (P) = 1 atm = 1.01 × 105 Pa

Atomic weight of a He atom = 4

Avogadro’s number (NA) = 6.023 × 1023

Boltzmann constant (k) = 1.38 × 10– 23 J mol– 1 K– 1

Average energy of a gas at temperature T, is given by

$$\text{E}=\frac{3}{2}k\text{T}\\\text{de-Broglie wavelength is given by}\\\lambda=\frac{h}{\sqrt{2mE}}\\\text{Here},m=\text{Mass of a He atom}\\=\frac{Atomic weight}{N_A}\\\frac{4}{6.023×10^{23}}\\=6.64×10^{-24}g\\= 6.64 × 10^{\normalsize– 27} kg\\\therefore\space\lambda=\frac{h}{\sqrt{3mkT}}\\=\frac{6.6×10^{-34}}{\sqrt{3×6.64×10^{-27}×1.38×10^{-23}×300}}\\=0.7268×10^{-10}m\\\text{From ideal gas formula :}\\\text{PV}=\text{RT}\\\text{PV}=k\text{NT}\\\frac{V}{N}=\frac{k\text{T}}{\text{P}}\\\text{Here},V=\text{Volume of the gas}\\N=\text{Number of moles of the gas}\\\text{Mean separation between two atoms of the gas is given by}\\r=\bigg(\frac{V}{N}\bigg)^{\frac{1}{3}}=\bigg(\frac{kT}{P}\bigg)^{\frac{1}{3}}\\=\bigg[\frac{1.38×10^{-23}×300}{1.01×10^{5}}\bigg]^{\frac{1}{3}}\\= 3.35 × 10^{– 9} m$$

Thus, the mean separation between the atoms is much greater than the de-Broglie wavelength.

Q. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.

• (i) Find the energy and momentum of each photon in the light beam.
• (ii) 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).
• (iii) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?

Ans. Given, wavelength of the monochromatic light
(λ) = 632.8 nm = 632.8 × 10– 9 m

Power emitted by the laser (P) = 9.42 mW = 9.42 × 10– 3 W

Planck’s constant (h) = 6.626 × 10– 34 Js

Speed of light (c) = 3 × 108 m/s

Mass of a hydrogen atom, m = 1.66 × 10– 27 kg

$$\text{(i)}\space\text{(The energy of each photon is given by)}\\\text{E}=\frac{hc}{\lambda}\\=\frac{6.626×10^{-34}×3×10^{8}}{632.8×10^{-9}}\\=3.14 1×10^{-19}\text{J}\\\text{The momentum of each photon is given by}\\p=\frac{h}{\lambda}=\frac{6.626×10^{-34}}{632.8×10^{-9}}\\=1.047×10^{-27}\text{kg ms}^{-1}\\\text{(ii) Number of photons arriving per second, at a target irradiated by the beam = n}\\\text{Assume that the beam has a uniform crosssection that is less than the target area.}\\\text{Hence, the equation for power can be written by}\\ P = nE\\\therefore\space n=\frac{P}{E}\\=\frac{9.42×10^{-3}}{3.141×10^{-19}}≈ 3 × 10^{16}\space\text{photons}\\\text{(iii)}\text{Momentum of the hydrogen atom is the same as the momentum of the photon,}\\p=1.047×10^{-27}\text{kg ms}^{\normalsize– 1}\\\text{Momentum is given by}\\p=mv\\\text{Here},\space v=\text{Speed of the hydrogen atom}\\\therefore v=\frac{p}{m}.=\frac{1.047×10^{-27}}{1.66×10^{-27}}=0.631\space\text{m/s}$$

Q. Describe de-Broglie concept of matter waves and derive an expression for the momentum and wavelength of the particle.

Ans. According to the de-Broglie concept, all the material particles exhibit wave properties and the wavelength of the wave associated with any material particles is given by,

$$\lambda=\frac{h}{p}\space\space\text{…(i)}\\\text{On the analogy of light radiation, the expression for the wavelength}\\\lambda \text{can be easily derived. The momentum of the photon is,}\\p=\frac{hv}{c}=\frac{h}{\lambda}\space\space\text{…(ii)}\\\lambda\text{or}\frac{h}{p}\text{is called the de-Broglie wavelength.}\\\text{If m is the mass and v the velocity of the material particle, then}\\p=mv\\\text{Hence},\space\lambda=\frac{h}{mv}\space\text{…(iii)}\\\text{If E is the kinetic energy of the material particle, then}\\\text{E}=\frac{1}{2}mv^{2} \Rarr\space\text{E}=\frac{p^{2}}{2m}\space[\text{from equation} (iii)]\text{…(iv)}\\\therefore\text{The de-Broglie wavelength is given by}\\\lambda=\frac{h}{\sqrt{2mE}}\space[\text{from equation (iv)}]\\\text{and for electron,}\space\lambda=\frac{12.27}{\sqrt{V}}\text{Å}.$$

Q. A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :

$$\lambda_1 = 3650Å, \lambda_2 = 4047 Å, \lambda_3 = 4358 Å, \lambda_4 = 5461 Å, \lambda_5 = 6907 Å. \\\text{The stopping voltages, respectively, were measured to be :}\\ V_{01} = 1.28 V, V_{02} = 0.95 V, V_{03} = 0.74 V, V_{04} = 0.16 V, V_{05} = 0 V.\\\text{Determine the value of Planck’s constant h, the threshold frequency and work function for the material.}\\\text{Note : You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C).}\\\text{ Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment)} \\\text{confirmed Einstein’s photoeletric equation and at the same time gave an independent estimate of the value of h.}$$

Ans. Einstein’s photoelectric equation is given by

$$eV_0 = hv – \phi_0\\V_0=\frac{h}{e}V-\frac{\phi_0}{e}\space\text{…(i)}\\\text{Here, V}_0 = \text{Stopping potential}\\\text{h = Planck’s constant}\\\text{e = Charge on an electron}\\\text{n = Frequency of radiation}\\\phi_0=\text{Work function of a material}\\\text{From equation (i) potential} V_0 \text{is directly proportional to frequency v.} \text{Frequency (v)}=\frac{\text{Speed of light}}{\text{Wavelength}(\lambda)}\\\text{This relation can be used to obtain the frequencies of the various lines of the given wavelengths.}\\\text{v}_1=\frac{c}{\lambda_1}=\frac{3×10^{8}}{3650×10^{-10}}=8.219 × 10^{14}\text{Hz}\\\text{v}_2=\frac{c}{\lambda}_2=\frac{3×10^{8}}{4047×10^{-10}}=7.412×10^{14}\text{Hz}\\\text{V}_3=\frac{c}{\lambda}_3=\frac{3×10^{8}}{4358×10^{-10}}=6.884 × 10^{14}\text{Hz}\\\text{V}_4=\frac{c}{\lambda}_4=\frac{3×10^{8}}{5461×10^{-10}}=5.493×10^{14}\text{Hz}\\\text{v}_5=\frac{c}{\lambda}_5=\frac{3×10^{8}}{6907×10^{-10}}=4.343×10^{14}\text{Hz}\\$$

The given quantities can be listed in tabular form as :

 Frequency ×1014Hz 8.219 7.412 6.884 5.493 4.343 Stopping potential V0 1.28 0.95 0.74 0.16 0
The figures shows a graph between v and V0. It can be observed that the obtained curve is a straight line. It intersects the x-axis at 5 × 1014 Hz, which is the threshold frequency (n0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the l5 line, and therefore, no stopping voltage is required to
stop the current.

Slope of the straight line

$$=\frac{\text{AB}}{\text{CB}}=\frac{1.28-0.16}{(8.214-5.493)×10^{14}}\\\text{From equation (i), the slope}\space\frac{h}{e}\space\text{can be written as :}\\\frac{h}{e}=\frac{1.28-0.16}{(8.214-5.493)×10^{14}}\\\therefore\text{h}=\frac{1.12×1.6×10^{-19}}{2.726×10^{14}}\\=6.573×10^{-34}\text{Js}\\\text{The work function of the metal is given as :}\\\phi_0=hv_0\\= 6.573 × 10^{\normalsize– 34} × 5 × 10^{14} \text{J}\\=3.286×10^{-19}\text{J}\\=\frac{3.286×10^{-19}}{1.6×10^{-18}}=2.054\text{eV}$$