Q. 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
A) $K$
B) $K + hv$
C) $K + E_{0}$
D) $2K$
Explanation
(b) Let $K$ and $K'$ be the maximum kinetic energy of photoelectrons for incident light of frequency $v$ and $2v$ respectively.
According to Einstein’s photoelectric equation,
$K = hv - E_{0}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ …(i)
and $K^{'} = h(2v) - E_{0}$ …(ii)
$= 2hv - E_{0} = hv + hv - E_{0}$
$K^{'} = hv + K\ \ \ \ \ \ $ [using (i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.