JEE Physics Dual Nature Of Radiation And Matter MCQs Set 07

Multiple Choice Questions (MCQs) for JEE Physics: Dual Nature Of Radiation and Matter

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Multiple Choice Questions

Question 1: An electron of mass m and a photon have the same energy E. The ratio of de Broglie wavelengths (electron to photon) is
(a) \( \dfrac1c\left(\dfrac{E}{2m}\right)^{1/2} \)
(b) \( \left(\dfrac{E}{2m}\right)^{1/2} \)
(c) \( c(2mE)^{1/2} \)
(d) \( \dfrac1{xc}\left(\dfrac{2m}{E}\right)^{1/2} \)
Show Answer & Explanation

Answer: (a) \( \dfrac1c\left(\dfrac{E}{2m}\right)^{1/2} \)

Explanation:
1. \( \lambda_e=h/\sqrt{2mE} \) and \( \lambda_{ph}=hc/E \), so the ratio is \( \dfrac1c\sqrt{E/2m} \).

Teacher's Note:
1. Divide the two.

Question 2: The stopping potential is V for radiation of wavelength λ and V/4 for 2λ. The threshold wavelength is
(a) \( 4\lambda \)
(b) \( 5\lambda \)
(c) \( 5\lambda/2 \)
(d) \( 3\lambda \)
Show Answer & Explanation

Answer: (d) \( 3\lambda \)

Explanation:
1. \( hc/\lambda=\phi+eV \), \( hc/2\lambda=\phi+eV/4 \) give \( eV=2hc/3\lambda \) and \( \phi=hc/3\lambda \).
2. \( \lambda_0=3\lambda \).

Teacher's Note:
1. Subtract to find eV.

Question 3: Electromagnetic wave of intensity 1400 W/m2 falls on a metal surface of area 1.5 m2 and is completely absorbed. The force exerted by the beam is
(a) \( 14\times10^{-5}\ \text{N} \)
(b) \( 14\times10^{-6}\ \text{N} \)
(c) \( 7\times10^{-5}\ \text{N} \)
(d) \( 7\times10^{-6}\ \text{N} \)
Show Answer & Explanation

Answer: (d) \( 7\times10^{-6}\ \text{N} \)

Explanation:
1. \( F=IA/c=1400\times1.5/3\times10^{8}=7\times10^{-6} \) N.

Teacher's Note:
1. Absorption gives I A / c.

Question 4: The de Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature T (kelvin) and mass m is
(a) \( h/\sqrt{3mkT} \)
(b) \( 2h/\sqrt{3mkT} \)
(c) \( 2h/\sqrt{mkT} \)
(d) \( h/\sqrt{mkT} \)
Show Answer & Explanation

Answer: (a) \( h/\sqrt{3mkT} \)

Explanation:
1. Mean energy is \( \tfrac32kT \), so \( \lambda=h/\sqrt{3mkT} \).

Teacher's Note:
1. Thermal neutron.

Question 5: The photoelectric threshold wavelength of silver is 3250 x 10^-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 x 10^-10 m is (h = 4.14 x 10^-15 eV s, c = 3 x 10^8 m/s)
(a) \( \approx0.6\times10^{4}\ \text{m/s} \)
(b) \( \approx61\times10^{3}\ \text{m/s} \)
(c) \( \approx0.3\times10^{6}\ \text{m/s} \)
(d) \( \approx6\times10^{5}\ \text{m/s} \)
Show Answer & Explanation

Answer: (d) \( \approx6\times10^{5}\ \text{m/s} \)

Explanation:
1. \( K=hc(1/2536-1/3250)\times10^{10}=4.90-3.82=1.08 \) eV.
2. \( v=\sqrt{2K/m}\approx6\times10^{5} \) m/s.

Teacher's Note:
1. Use K in joules.

Question 6: An electron of mass m with initial velocity \( V_0\hat i \) enters an electric field \( \vec E=-E_0\hat i \). If \( \lambda_0 \) is its initial de Broglie wavelength, the wavelength at time t is
(a) \( \dfrac{\lambda_0}{1+\frac{eE_0}{mV_0}t} \)
(b) \( \lambda_0\left(1+\dfrac{eE_0}{mV_0}t\right) \)
(c) \( \lambda_0t \)
(d) \( \lambda_0 \)
Show Answer & Explanation

Answer: (a) \( \dfrac{\lambda_0}{1+\frac{eE_0}{mV_0}t} \)

Explanation:
1. The force on the electron is along +x, so \( v=V_0+eE_0t/m \).
2. \( \lambda=h/mv=\lambda_0/(1+eE_0t/mV_0) \).

Teacher's Note:
1. Electron accelerates here.

Question 7: Light of frequency \( 2\nu_0 \) falls on a metal plate and the maximum velocity of the electrons is \( v_1 \). With frequency \( 5\nu_0 \) it is \( v_2 \). The ratio \( v_1:v_2 \) is
(a) 1 : 2
(b) 1 : 4
(c) 4 : 1
(d) 2 : 1
Show Answer & Explanation

Answer: (a) 1 : 2

Explanation:
1. \( K_1=h\nu_0 \) and \( K_2=4h\nu_0 \), so \( v_1:v_2=1:2 \).

Teacher's Note:
1. v is proportional to sqrt(K).

Question 8: Light of wavelength \( \lambda_1 \) on a metal gives fastest electron speed \( v_1 \). With wavelength \( \lambda_2 \) on the same metal, \( v_2 \). The value of \( v_2^2-v_1^2 \) will be
(a) \( \dfrac{2hc}{m_e}(\lambda_2-\lambda_1) \)
(b) \( \dfrac{2hc}{m_e}(\lambda_2-\lambda_1)^2 \)
(c) \( \dfrac{2hc}{m_e}\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right) \)
(d) \( \dfrac{2hc}{m_e}\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right)^2 \)
Show Answer & Explanation

Answer: (c) \( \dfrac{2hc}{m_e}\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right) \)

Explanation:
1. \( \tfrac12mv^2=hc/\lambda-\phi \).
2. Subtracting the two equations for the same metal gives \( v_2^2-v_1^2=\dfrac{2hc}{m}\left(\dfrac1{\lambda_2}-\dfrac1{\lambda_1}\right) \).

Teacher's Note:
1. Same metal means the same work function.

Question 9: A 500 W bulb is at the centre of a perfectly black sphere of radius 1 m. The approximate pressure on the walls (all photons absorbed) is (take 4π = 12.6)
(a) \( 1.3\times10^{-7}\ \text{N/m}^2 \)
(b) \( 2.5\times10^{-7}\ \text{N/m}^2 \)
(c) \( 6.3\times10^{-7}\ \text{N/m}^2 \)
(d) \( 3.14\times10^{-7}\ \text{N/m}^2 \)
Show Answer & Explanation

Answer: (a) \( 1.3\times10^{-7}\ \text{N/m}^2 \)

Explanation:
1. \( I=P/4\pi R^2=500/12.6 \) W/m2; pressure \( =I/c=1.3\times10^{-7} \) N/m2.

Teacher's Note:
1. Absorption gives I / c.

Question 10: Light of frequency 10^15 Hz falls on a metal of work function 2.5 eV. The stopping potential is
(a) 1.6 V
(b) 2.5 V
(c) 4.1 V
(d) 6.6 V
Show Answer & Explanation

Answer: (a) 1.6 V

Explanation:
1. \( E=4.14 \) eV, \( V_0=4.14-2.5=1.6 \) V.

Teacher's Note:
1. h nu = 4.14 eV.

Question 11: The work function of potassium is 2.30 eV. Light of frequency \( 8\times10^{14} \) Hz falls on it. The stopping potential is
(a) 0.1 V
(b) 1.0 V
(c) 2.3 V
(d) 3.3 V
Show Answer & Explanation

Answer: (b) 1.0 V

Explanation:
1. \( E=8\times10^{14}\times4.14\times10^{-15}=3.31 \) eV, so \( V_0=1.0 \) V.

Teacher's Note:
1. V0 = E/e - phi/e.

Question 12: The threshold frequency of a metal is \( \nu_0 \). With light of frequency \( 2\nu_0 \) the maximum velocity of photoelectrons is \( v_0 \). With \( 5\nu_0 \) it is
(a) \( \sqrt2v_0 \)
(b) \( 2v_0 \)
(c) \( 2\sqrt2v_0 \)
(d) \( 4v_0 \)
Show Answer & Explanation

Answer: (b) \( 2v_0 \)

Explanation:
1. \( K_1=h\nu_0 \) and \( K_2=4h\nu_0 \), so the speed doubles.

Teacher's Note:
1. Same as before.

Question 13: A proton moves on a circular path of radius 6.6 x 10^-3 m in a perpendicular magnetic field of 0.625 T. Its de Broglie wavelength is
(a) 1 Å
(b) 0.1 Å
(c) 0.01 Å
(d) 0.001 Å
Show Answer & Explanation

Answer: (c) 0.01 Å

Explanation:
1. \( p=qBr=1.6\times10^{-19}\times0.625\times6.6\times10^{-3}=6.6\times10^{-22} \).
2. \( \lambda=h/p=10^{-12} \) m = 0.01 Å.

Teacher's Note:
1. p = q B r.

Question 14: The velocity at which the mass of a particle becomes twice its rest mass is
(a) \( 2c/3 \)
(b) \( c/2 \)
(c) \( c\sqrt3/2 \)
(d) \( 3c/4 \)
Show Answer & Explanation

Answer: (c) \( c\sqrt3/2 \)

Explanation:
1. \( 1/\sqrt{1-v^2/c^2}=2 \) gives \( v=c\sqrt3/2 \).

Teacher's Note:
1. Relativistic mass increase.

Question 15: An electron with rest mass m0 moves with a speed of 0.8c. Its mass is
(a) \( m_0 \)
(b) \( m_0/6 \)
(c) \( 5m_0/3 \)
(d) \( 3m_0/5 \)
Show Answer & Explanation

Answer: (c) \( 5m_0/3 \)

Explanation:
1. \( m=m_0/\sqrt{1-0.64}=m_0/0.6=5m_0/3 \).

Teacher's Note:
1. Lorentz factor 5/3.

Question 16: To see inside an atom of diameter 10 pm the width to resolve is about 10 pm. The minimum electron energy required in an electron microscope is about
(a) 15 keV
(b) 1.5 keV
(c) 150 keV
(d) 1.5 MeV
Show Answer & Explanation

Answer: (a) 15 keV

Explanation:
1. \( \lambda=0.1 \) Å and \( E=150/\lambda^2\ \text{eV}=15 \) keV.

Teacher's Note:
1. Use 150 / lambda squared in eV.

Question 17: An electron moving at \( 1.5\times10^{8} \) m/s has the same de Broglie wavelength as a photon. The ratio of the kinetic energy of the electron to the energy of the photon is
(a) 1/4
(b) 1/2
(c) 2
(d) 4
Show Answer & Explanation

Answer: (a) 1/4

Explanation:
1. Photon energy \( =pc=mvc \); \( K=\tfrac12mv^2 \).
2. Ratio \( =v/2c=1/4 \).

Teacher's Note:
1. p is the same for both.

Question 18:
Assertion (A): Photoelectric effect is seen only when the electron is bound
Reason (R): A free electron cannot absorb the whole energy of a photon while conserving momentum and energy
(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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A free electron cannot absorb a photon, so the effect needs a bound electron.

Teacher's Note:
1. Bound electron takes up the momentum.

Question 19:
Assertion (A): When photons of frequency 4.3 x 10^15 Hz are incident on a surface of threshold frequency 5 x 10^15 Hz, no photoelectron is emitted
Reason (R): The number of electrons ejected depends on intensity
(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: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The frequency is below the threshold, so there is no emission.
2. The reason is true but does not explain the assertion.

Teacher's Note:
1. Frequency decides whether emission happens.

Question 20:
Assertion (A): On increasing the frequency of incident light on the metal surface, the kinetic energy of the ejected particle increases
Reason (R): The energy of a photon is proportional to the frequency of 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: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( K=h\nu-\phi \) increases because the photon energy increases with \( \nu \).

Teacher's Note:
1. Energy proportional to frequency.

Question 21:
Assertion (A): The Heisenberg principle is not meaningful for a moving tennis ball
Reason (R): Wave nature is not associated with a moving tennis ball
(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. The assertion is true because the uncertainty is negligible.
2. Wave nature is associated with every moving body, but the wavelength is too small to observe, so the reason is false.

Teacher's Note:
1. Tiny wavelength.

Question 22:
Assertion (A): Ultraviolet rays of low intensity cannot emit photoelectrons
Reason (R): Photoelectric emission requires higher intensity
(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: (d) A is false, but R is true.

Explanation:
1. Emission depends on frequency and not on intensity.
2. Both statements are false.

Teacher's Note:
1. Intensity does not set the threshold.

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