Question
When photons of energy hv fall on an aluminium plate (of work function E0), 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
-
K+E0
-
2K
-
K
-
K+hv
Solution
D.
K+hv
The energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron.
According to Einstein's photoelectric effect energy of photon = KE of photoelectron + work function of metal
i.e., 
or 
Now, we have given,
Therefore, 
...(ii)
From Eqs. (i) and (ii), we have
Putting 
