Two identical metal plates show photoelectric effect by a light of wavelength `lambda_(A)` falls on plate `A` and `lambda_(B)` on plate `B(lambda_(A) = 2 lambda_(B))`. The maximum kinetic energy is

Options

A. `2K_(A) = K_(B)`
B. `K_(A) lt K_(B)//2`
C. `K_(A) = 2 K_(B)`
D. `K_(A) = K_(B)//2`

Answer

B

Explanation

`(hc)/(lambda) = W_(0) + K_(max) rArr (hc)/(lambda_(A)) = W_(0) + K_(A)` ....(i)
and `(hc)/(lambda_(B)) = W_(0) + K_(B)` ....(ii)
Subtracting (i) from (ii) , `hc[(1)/(lambda_(B)) - (1)/(lambda_(A))] = K_(B) -K_(A)`
`rArr (hc)/(2 lambda_(B)) = K_(B) - K_(A)`
From (ii) and (iii) ,
`2K_(B) - 2K_(A) = W_(0) + K_(B)`
`rArr K_(B) - 2K_(A) = W_(0)`
`rArr K_(A) = (K_(B))/(2) - (W_(0))/(2)` which gives `K_(A) lt (K_(B))/(2)`.