Category: Photoelectric Effect

Q: Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

(A) $\large \lambda_0 = \frac{2 m c \lambda^2}{h} $

(B) $\large \lambda_0 = \frac{2 h}{m c} $

(C) $\large \lambda_0 = \frac{2 m^2 c^2 \lambda^3}{h^2} $

(D) $\large \lambda_0 = \lambda $

Ans: (A)

Sol: Kinetic Energy of Striking Electrons is

$\large K = \frac{p^2}{2 m} $

As , $\large p = \frac{h}{\lambda}$

$\large K = \frac{h^2}{2 m \lambda^2} $

This is the maximum energy of X-ray photons .

Therefore , $\large \frac{h c}{\lambda_0} = \frac{h^2}{2 m \lambda^2} $

$\large \lambda_0 = \frac{2 m c \lambda^2}{h} $

Q: A beam of electron is used in an YDSE experiment. The slit width is d. When the velocity of electron is increased, then

(A) no interference is observed

(B) fringe width increases

(C) fringe width decreases

(D) fringe width remains same

Ans: (C)

Sol: As momentum of electron is increased , the wavelength $\large ( \lambda = \frac{h}{p} ) $ will decrease . Hence fringe width $ (\beta \propto \lambda )$ will decrease .

Q: The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let λ1 be the de-Broglie wavelength of the proton and λ2 be the wavelength of the photon. The ratio λ1/λ2 is proportional to

(A) E⁰

(B) E^1/2

(C) E^–1

(D) E^–2

Ans: (B)

Sol: $\large \frac{\lambda_1}{\lambda_2} = \frac{ \frac{h}{\sqrt{2mE}}}{ \frac{hc}{E}}$

$\large \frac{\lambda_1}{\lambda_2} \propto E^{1/2} $

Q: A particle of mass M at rest decays into two particles of masses m₁ and m₂ having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles λ₁/λ₂ is

(A) m₁ / m₂

(B) m₂ / m₁

(C) 1

(D) $\large \frac{\sqrt{m_2}}{\sqrt{m_1}}$

Ans: (C)

Sol: From Law of Conservation of Momentum;

P₁ = P₂

As , de-broglie wavelength is given by

$\large \lambda = \frac{h}{P}$

Since , Momentum is equal.

Hence , λ₁ = λ₂

Q: 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

Ans: (C)

Sol: $\large \lambda (in A^o) = \frac{12375}{\phi (eV)}$

$\large = \frac{12375}{4} = 3093 A^o$

= 3o9.3 nm ≈ 310 nm