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

Answer

b

Explanation

`(hc)/(lambda_(A))=K_(A)+phi_(0).......(i)`
and `(hc)/(lambda_(B))=K_(B)+phi_(0).........(i)`
Divinding (i) and (ii), we get
`(lambda_(A))/(lambda_(B))=(K_(B)+phi_(0))/(K_(A)+phi_(0))`
`(2lambda_(B))/(lambda_(B))=(K_(B)+phi_(0))/(K_(A)+phi_(0))` or `2=(K_(B)+phi_(0))/(K_(A)+phi_(0))`
`2K_(A)+2phi_(0)=K_(B)+phi_(0)`
or ` K_(B)=2K_(A)+phi_(0)`
i.e., `K_(B)gt2K_(A)` or `(K_(B))/2gtK_(A)`