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 KA and KB respectively. Which one of the following is true ?
Answer & explanation
Correct answer: option 3
$K_A=\frac{hc}{λ_A}-\phi=\frac{hc}{2λ_B}-\phi$ .... (i)
$K_A=\frac{hc}{λ_B}-\phi⇒\frac{hc}{λ_B}-\phi=K_B+\phi$ .... (ii)
From eqn (i) & (ii)
$K_A=\frac{1}{2}(K_B+\phi)=\frac{1}{2}K_B-\frac{\phi}{2}$
$K_A<\frac{1}{2}K_B$