When photons of energy `hv` fall on an aluminium plate (of work function `E_(0)`), photoelectrons of maximum kinetic energy `K` are ejected . If the frequency of the radiation is doubled , the maximum kinetic energy of the ejected photoelectrons will be

Options

A. `K+E_(0)`
B. `2K`
C. `K`
D. `K+hv`

Answer

D

Explanation

According to Einstein's photoelectric effect. Energy of photon=KE of photoelectron+work function of metal
i.e. `hv=(1)/(2)mv^(2)+E_(0)`
or `hv=K_(max)+E_(0)` . .. (i)
now, we have given,
`v'=2v`
therefore, `K_(max)'=2hv-E_(0)` . .. (ii)
from Eqs. (i) and (ii), we have
`K_(max)'=2(K_(max)+E_(0))-E_(0)=2K_(max)+E_(0)`
`=K_(max)+(K_(max)+E_(0))=K_(max)+hv`
Putting `K_(max)=K" "therefore K_(max)'=K+hv`