Question

Two identical metal plates show photoelectric effect. Light of wavelength λ_A falls on plate A and λ_B falls on plate B. λ_A = 2λ_B. The maximum K.E. of the photoelectrons are K_Aand K_B respectively. Which one of the following is true ?

• A. 2K_A = K_B
• B. K_A = 2K_B
• C. K_A< K_B/2
• D. K_A> 2K_B

Answer 1

Answer: (C)

K_A< K_B/2

Answer 2

K_A = – φ = – φ  .... (i)

K_B = – φ ⇒ = K_B + φ .... (ii)

From eqⁿ (i) & (ii)

K_Α = $\frac { 1 } { 2 }$(K_B + φ) = $\frac { 1 } { 2 }$ K_B – $\frac { \phi } { 2 }$

K_A < $\frac { 1 } { 2 }$ K_B