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. `2 K`
C. `K`
D. `K + hv`

Answer

D

Explanation

According to Einstein's photoelectric effect energy of photon ` = KE "of photoelectron + word function of metal"`
That is , `hv = (1)/(2) mv^(2) + E_(0)`
or `hv = K_(max) + E_(0)` …..(i)
Now , we have given `K`_(max) = h( 2v) - 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` [ From Eq. (i)]
Putting `K_(max) = K rArr K_(max) = K + hv`