JEE Physics Dual Nature Of Radiation And Matter MCQs Set 25

Physics Objective Questions and Answers: Dual Nature Of Radiation and Matter

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Multiple Choice Questions

Question 1: The kinetic energy of an electron which is accelerated through a potential difference of 100 V is
(a) \( 1.6\times10^{-17} \) J
(b) \( 1.6\times10^{-14} \) J
(c) \( 1.6\times10^{-10} \) J
(d) \( 1.6\times10^{-8} \) J
Show Answer & Explanation

Answer: (a) \( 1.6\times10^{-17} \) J

Explanation:
1. \( K=eV=1.6\times10^{-19}\times100=1.6\times10^{-17} \) J.

Teacher's Note:
1. 1 eV \( =1.6\times10^{-19} \) J.

Question 2: If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor
(a) \( \tfrac12 \)
(b) 2
(c) \( \dfrac1{\sqrt2} \)
(d) \( \sqrt2 \)
Show Answer & Explanation

Answer: (c) \( \dfrac1{\sqrt2} \)

Explanation:
1. \( \lambda\propto K^{-1/2} \), so doubling \( K \) multiplies \( \lambda \) by \( 1/\sqrt2 \).

Teacher's Note:
1. Wavelength falls as the kinetic energy rises.

Question 3: The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is
(a) four times the initial energy
(b) equal to the initial energy
(c) twice the initial energy
(d) thrice the initial energy
Show Answer & Explanation

Answer: (d) thrice the initial energy

Explanation:
1. Halving \( \lambda \) quadruples \( K \) (\( K\propto1/\lambda^2 \)), so the added energy is \( 4K-K=3K \).

Teacher's Note:
1. Added energy is the difference of the kinetic energies.

Question 4: An electron and a neutron can have the same (i) kinetic energy, (ii) momentum, or (iii) speed. Which particle has the shorter de-Broglie wavelength in each case, respectively?
(a) neutron, same, neutron
(b) neutron, electron, same
(c) electron, same, neutron
(d) electron, neutron, electron
Show Answer & Explanation

Answer: (a) neutron, same, neutron

Explanation:
1. Same \( K \): \( \lambda=h/\sqrt{2mK} \) is smaller for the heavier neutron.
2. Same momentum: \( \lambda=h/p \) is the same. Same speed: \( \lambda=h/mv \) is smaller for the neutron.

Teacher's Note:
1. Compare using the formula that matches the quantity held equal.

Question 5: A particle of mass \( M \) at rest decays into two masses \( m_1 \) and \( m_2 \) with non-zero velocities. The ratio of the de-Broglie wavelengths of the particles \( \lambda_1/\lambda_2 \) is
(a) \( \dfrac{m_2}{m_1} \)
(b) \( \dfrac{m_1}{m_2} \)
(c) \( \dfrac{\sqrt{m_1}}{\sqrt{m_2}} \)
(d) 1 : 1
Show Answer & Explanation

Answer: (d) 1 : 1

Explanation:
1. By momentum conservation the two fragments have equal and opposite momenta, so \( \lambda=h/p \) is the same.

Teacher's Note:
1. Total momentum was zero.

Question 6: For an electron in the second orbit of Bohr's hydrogen atom, the moment of linear momentum (angular momentum) is
(a) \( \pi h \)
(b) \( 2\pi h \)
(c) \( \dfrac h\pi \)
(d) \( \dfrac{2h}\pi \)
Show Answer & Explanation

Answer: (c) \( \dfrac h\pi \)

Explanation:
1. \( L=\dfrac{nh}{2\pi}=\dfrac{2h}{2\pi}=\dfrac h\pi \).

Teacher's Note:
1. Bohr's quantisation condition.

Question 7: A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are \( u_1 \) and \( u_2 \) respectively. If the ratio \( u_1:u_2=2:1 \) and \( hc=1240 \) eV nm, the work function of the metal is nearly
(a) 3.7 eV
(b) 3.2 eV
(c) 2.8 eV
(d) 2.5 eV
Show Answer & Explanation

Answer: (a) 3.7 eV

Explanation:
1. Photon energies: 5.0 eV and 4.0 eV. \( K_1=4K_2 \).
2. \( 5-\phi=4(4-\phi)\Rightarrow3\phi=11\Rightarrow\phi=3.7 \) eV.

Teacher's Note:
1. Kinetic energy is proportional to the speed squared.

Question 8: The radiation corresponding to the \( 3\to2 \) transition of a hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of \( 3\times10^{-4} \) T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to
(a) 1.8 eV
(b) 1.1 eV
(c) 0.8 eV
(d) 1.6 eV
Show Answer & Explanation

Answer: (b) 1.1 eV

Explanation:
1. Photon energy \( =13.6\left(\tfrac14-\tfrac19\right)=1.89 \) eV.
2. \( K_{max}=\dfrac{(eBr)^2}{2m}=\dfrac{(1.6\times10^{-19}\times3\times10^{-4}\times0.01)^2}{2\times9.1\times10^{-31}}=1.27\times10^{-19} \) J \( =0.79 \) eV.
3. \( \phi=1.89-0.79\approx1.1 \) eV.

Teacher's Note:
1. The fastest electrons follow the largest circle: \( p=eBr \).

Question 9: Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 nm is incident on it, a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop the photoelectrons when light of wavelength 200 nm is incident on it?
(a) 15.4 V
(b) 2.35 V
(c) 3.85 V
(d) 1.5 V
Show Answer & Explanation

Answer: (d) 1.5 V

Explanation:
1. Photon energy at 200 nm \( =\dfrac{1240}{200}=6.2 \) eV.
2. \( eV_0=6.2-4.7=1.5 \) eV, so \( V_0=1.5 \) V.

Teacher's Note:
1. \( eV_0=h\nu-\phi \).

Question 10: The maximum kinetic energy of the photoelectrons depends only on
(a) potential
(b) frequency
(c) incident angle
(d) pressure
Show Answer & Explanation

Answer: (b) frequency

Explanation:
1. \( K_{max}=h\nu-\phi \) depends on the frequency of the incident light (and the metal), not on intensity.

Teacher's Note:
1. Einstein's photoelectric equation.

Question 11: The ionisation energy of hydrogen is 13.6 eV. The energy of the photon released when an electron jumps from the first excited state (\( n=2 \)) to the ground state of a hydrogen atom is
(a) 3.4 eV
(b) 4.53 eV
(c) 10.2 eV
(d) 13.6 eV
Show Answer & Explanation

Answer: (c) 10.2 eV

Explanation:
1. \( E=13.6\left(1-\tfrac14\right)=10.2 \) eV.

Teacher's Note:
1. \( E_n=-13.6/n^2 \).

Question 12: Find the maximum wavelength of the light that will excite an electron in the valence band of diamond to the conduction band. The energy gap is 5.5 eV.
(a) 225 nm
(b) 315 nm
(c) 352 nm
(d) 412 nm
Show Answer & Explanation

Answer: (a) 225 nm

Explanation:
1. \( \lambda=\dfrac{1240}{5.5}=225 \) nm.

Teacher's Note:
1. \( \lambda(\text{nm})=1240/E(\text{eV}) \).

Question 13: Which of the following devices is an application of the photoelectric effect?
(a) light emitting diode
(b) diode
(c) photocell
(d) transistor
Show Answer & Explanation

Answer: (c) photocell

Explanation:
1. A photocell works on the photoelectric effect.

Teacher's Note:
1. It converts light into electric current.

Question 14: A monochromatic light of frequency \( 3\times10^{14} \) Hz is produced by a laser. It emits a power of \( 3\times10^{-3} \) W. The number of photons emitted per second is (\( h=6.63\times10^{-34} \) J s)
(a) \( 1.5\times10^{16} \)
(b) \( 2.5\times10^{16} \)
(c) \( 4.5\times10^{16} \)
(d) \( 8.5\times10^{16} \)
Show Answer & Explanation

Answer: (a) \( 1.5\times10^{16} \)

Explanation:
1. \( n=\dfrac P{h\nu}=\dfrac{3\times10^{-3}}{6.63\times10^{-34}\times3\times10^{14}}=1.5\times10^{16} \) s\(^{-1}\).

Teacher's Note:
1. Divide the power by the photon energy.

Question 15: In a photoelectric effect measurement, the stopping potential for a given metal is found to be \( V_0 \) volt when radiation of wavelength \( \lambda_0 \) is used. If radiation of wavelength \( 2\lambda_0 \) is used with the same metal, then the stopping potential (in volt) will be
(a) \( \dfrac{V_0}2 \)
(b) \( 2V_0 \)
(c) \( V_0+\dfrac{hc}{2e\lambda_0} \)
(d) \( V_0-\dfrac{hc}{2e\lambda_0} \)
Show Answer & Explanation

Answer: (d) \( V_0-\dfrac{hc}{2e\lambda_0} \)

Explanation:
1. \( eV_0=\dfrac{hc}{\lambda_0}-\phi \) and \( eV'=\dfrac{hc}{2\lambda_0}-\phi \); subtracting gives \( V'=V_0-\dfrac{hc}{2e\lambda_0} \).

Teacher's Note:
1. Subtract the two equations to eliminate \( \phi \).

Question 16: What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 Å?
(a) 700 m/s
(b) 1000 m/s
(c) 1400 m/s
(d) 2800 m/s
Show Answer & Explanation

Answer: (c) 1400 m/s

Explanation:
1. \( mv=\dfrac h\lambda\Rightarrow v=\dfrac{6.63\times10^{-34}}{9.1\times10^{-31}\times5.2\times10^{-7}}=1.4\times10^3 \) m/s.

Teacher's Note:
1. Equate \( p=mv \) to \( h/\lambda \).

Question 17: The work function for aluminium is 4.125 eV. The cut-off wavelength for photoelectric effect for aluminium will be
(a) 420 nm
(b) 350 nm
(c) 300 nm
(d) 200 nm
Show Answer & Explanation

Answer: (c) 300 nm

Explanation:
1. \( \lambda_0=\dfrac{1240}{4.125}=300 \) nm.

Teacher's Note:
1. \( \lambda_0=hc/\phi \).

Question 18: When a certain metal surface is illuminated with light of frequency \( \nu \), the stopping potential for photoelectric current is \( V_0 \). When the same surface is illuminated by light of frequency \( \nu/2 \), the stopping potential is \( V_0/4 \). The threshold frequency for photoelectric emission is
(a) \( \dfrac\nu6 \)
(b) \( \dfrac\nu3 \)
(c) \( \dfrac{2\nu}3 \)
(d) \( \dfrac{4\nu}3 \)
Show Answer & Explanation

Answer: (b) \( \dfrac\nu3 \)

Explanation:
1. \( eV_0=h\nu-\phi \) and \( \dfrac{eV_0}4=\dfrac{h\nu}2-\phi \).
2. Eliminating \( eV_0 \): \( h\nu-\phi=4\left(\dfrac{h\nu}2-\phi\right)\Rightarrow\phi=\dfrac{h\nu}3 \), so \( \nu_0=\dfrac\nu3 \).

Teacher's Note:
1. Use the photoelectric equation twice.

Question 19: The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately
(a) 540 nm
(b) 400 nm
(c) 310 nm
(d) 220 nm
Show Answer & Explanation

Answer: (c) 310 nm

Explanation:
1. \( \lambda_{max}=\dfrac{1240}{4.0}=310 \) nm.

Teacher's Note:
1. This is the threshold wavelength.

Question 20: Calculate the linear momentum of a 3 MeV photon
(a) 0.01 eV-s/m
(b) 0.02 eV-s/m
(c) 0.03 eV-s/m
(d) 0.0 eV-s/m
Show Answer & Explanation

Answer: (a) 0.01 eV-s/m

Explanation:
1. \( p=\dfrac Ec=\dfrac{3\times10^6\ \text{eV}}{3\times10^8\ \text{m/s}}=0.01 \) eV-s/m.

Teacher's Note:
1. Photon momentum \( =E/c \).

Question 21: A monochromatic source of light emits photons of frequency \( 6\times10^{14} \) Hz. The power emitted by the source is \( 8\times10^{-3} \) W. Calculate the number of photons emitted per second (take \( h=6.63\times10^{-34} \) J s).
(a) \( 6\times10^{16} \)
(b) \( 4\times10^{15} \)
(c) \( 2\times10^{16} \)
(d) \( 1\times10^{17} \)
Show Answer & Explanation

Answer: (c) \( 2\times10^{16} \)

Explanation:
1. \( n=\dfrac{8\times10^{-3}}{6.63\times10^{-34}\times6\times10^{14}}=2\times10^{16} \) s\(^{-1}\).

Teacher's Note:
1. Power divided by \( h\nu \).

Question 22: A photosensitive metallic surface has work function \( \phi \). If a photon of energy \( 3\phi \) falls on this surface, the electron comes out with a maximum velocity of \( 6\times10^6 \) m/s. When the photon energy is increased to \( 9\phi \), then the maximum velocity of the photoelectron will be
(a) \( 12\times10^6 \) m/s
(b) \( 6\times10^6 \) m/s
(c) \( 3\times10^6 \) m/s
(d) \( 24\times10^6 \) m/s
Show Answer & Explanation

Answer: (a) \( 12\times10^6 \) m/s

Explanation:
1. \( K_1=3\phi-\phi=2\phi \) and \( K_2=9\phi-\phi=8\phi \).
2. \( v\propto\sqrt K \): \( v_2=6\times10^6\sqrt{4}=12\times10^6 \) m/s.

Teacher's Note:
1. Kinetic energy is photon energy minus work function.

Question 23: The photoelectric threshold wavelength for silver is \( \lambda_0 \). The energy of the electron ejected from the surface of silver by an incident wavelength \( \lambda \) (\( \lambda<\lambda_0 \)) will be
(a) \( hc\left(\dfrac{\lambda_0-\lambda}{\lambda\lambda_0}\right) \)
(b) \( \dfrac hc\left(\dfrac{\lambda_0-\lambda}{\lambda\lambda_0}\right) \)
(c) \( \dfrac{hc}{\lambda_0-\lambda} \)
(d) \( h c(\lambda_0-\lambda) \)
Show Answer & Explanation

Answer: (a) \( hc\left(\dfrac{\lambda_0-\lambda}{\lambda\lambda_0}\right) \)

Explanation:
1. \( K=\dfrac{hc}\lambda-\dfrac{hc}{\lambda_0}=hc\,\dfrac{\lambda_0-\lambda}{\lambda\lambda_0} \).

Teacher's Note:
1. Photon energy minus work function.

Question 24: If \( e/m \) of an electron is \( 1.76\times10^{11} \) C kg\(^{-1}\) and the stopping potential is 0.71 V, then the maximum velocity of the photoelectron is
(a) 150 km/s
(b) 200 km/s
(c) 500 km/s
(d) 250 km/s
Show Answer & Explanation

Answer: (c) 500 km/s

Explanation:
1. \( \tfrac12mv^2=eV_0\Rightarrow v=\sqrt{2\tfrac emV_0}=\sqrt{2\times1.76\times10^{11}\times0.71}=5\times10^5 \) m/s \( =500 \) km/s.

Teacher's Note:
1. Use the specific charge directly.

Question 25: The frequency of a photon having energy 100 eV is (take \( h=6.67\times10^{-34} \) J s, 1 eV \( =1.6\times10^{-19} \) J)
(a) \( 2.4\times10^{-16} \)
(b) \( 2.4\times10^{16} \) Hz
(c) \( 2.4\times10^{17} \)
(d) \( 10.54\times10^{16} \)
Show Answer & Explanation

Answer: (b) \( 2.4\times10^{16} \) Hz

Explanation:
1. \( \nu=\dfrac Eh=\dfrac{1.6\times10^{-17}}{6.67\times10^{-34}}=2.4\times10^{16} \) Hz.

Teacher's Note:
1. Convert eV to joules.

Question 26:
Assertion (A): When ultraviolet light is incident on a photocell, its stopping potential is \( V_0 \) and the maximum kinetic energy of the photoelectrons is \( K_{max} \). When the ultraviolet light is replaced by X-rays, both \( V_0 \) and \( K_{max} \) increase.
Reason (R): Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. X-rays have higher frequency, so \( K_{max}=h\nu-\phi \) and \( V_0 \) increase: A is true.
2. The spread of electron speeds (zero to maximum) comes from energy lost in collisions inside the metal, not from a range of frequencies, so R is false.

Teacher's Note:
1. For monochromatic light the electrons still have a range of energies.
2. \( K_{max} \) corresponds to electrons emitted from the surface.

Question 27: According to Einstein's photoelectric equation, the graph of the kinetic energy of the photoelectrons emitted from the metal versus the frequency of the incident radiation gives a straight line graph whose slope
(a) depends on the intensity of the incident radiation
(b) depends on the nature of the metal and also on the intensity of incident radiation
(c) is the same for all metals and independent of the intensity of the incident radiation
(d) depends on the nature of the metal
Show Answer & Explanation

Answer: (c) is the same for all metals and independent of the intensity of the incident radiation

Explanation:
1. \( K_{max}=h\nu-\phi \): the slope is Planck's constant \( h \) for every metal.

Teacher's Note:
1. The intercept depends on the metal; the slope does not.

Question 28: What is the work function (in eV) of a substance, if photoelectrons are just ejected for a monochromatic light of wavelength \( \lambda=3300 \) Å?
(a) 3.75
(b) 3.25
(c) 1.63
(d) 0.75
Show Answer & Explanation

Answer: (a) 3.75

Explanation:
1. \( \phi=\dfrac{12400}{3300}=3.76\approx3.75 \) eV.

Teacher's Note:
1. \( \phi(\text{eV})=12400/\lambda(\text{Å}) \).

Question 29: The photoelectric effect represents that
(a) light has a particle nature
(b) electron has a wave nature
(c) proton has a wave nature
(d) none of the above
Show Answer & Explanation

Answer: (a) light has a particle nature

Explanation:
1. The photoelectric effect is explained only by photons, so it shows the particle nature of light.

Teacher's Note:
1. Wave theory fails to explain the threshold frequency.

Question 30: An atom of mass \( M \), which is in the state of rest, emits a photon of wavelength \( \lambda \). As a result, the atom will recoil with the kinetic energy equal to (\( h \) is Planck's constant)
(a) \( \dfrac{h^2}{M\lambda^2} \)
(b) \( \dfrac12\dfrac{h^2}{M\lambda^2} \)
(c) \( \dfrac h{M\lambda} \)
(d) \( \dfrac12\dfrac h{M\lambda} \)
Show Answer & Explanation

Answer: (b) \( \dfrac12\dfrac{h^2}{M\lambda^2} \)

Explanation:
1. The atom recoils with momentum \( p=h/\lambda \), so \( K=\dfrac{p^2}{2M}=\dfrac{h^2}{2M\lambda^2} \).

Teacher's Note:
1. Momentum conservation.

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