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Q.
If the frequency of light incident on metal surface is doubled, then kinetic energy of emitted electron will become
J & K CETJ & K CET 2005
Solution:
From Einstein's photoelectric equation,
we have $E_{k}=h v-W$
where $E_{k}$ is maximum kinetic energy of photoelectrons and $W$ the work function
$\left(=h v_{0}\right)$.
$\therefore E_{k}=h\left(v-v_{0}\right)$ ..(i)
where $v_{0}$ is threshold frequency.
For $h v' =2 h v$
$\therefore 2 E_{k}=2 h\left(v-v_{0}\right)$..(ii)
From Eqs. (i) and (ii),
we observe that kinetic energy of emitted electron will become more than doubled.