Question

The threshold frequency for a certain photosensitive metal is $v_{0}$. When it is illuminated by light of frequency $v=2 v_{0}$, the maximum velocity of photoelectrons is $v_{0}$. The maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $v=5 v_{0}$ is $x v_{0}$. Find $x$.

Answer 1

As $v_{0}$ is the threshold frequency,
$\therefore $ Work function, $\phi_{0}=h v_{0}$
According to Einstein's photoelectric equation,
$\frac{1}{2} m v_{\max }^{2}=h v-\phi_{0}$
where $h v$ is the incident energy, $\phi_{0}$ is the work function of the metal and $\frac{1}{2} m v_{\max }^{2}$ is the maximum kinetic energy of the emitted photoelectrons.
As per the question,
$\frac{1}{2} m v_{0}^{2}=h\left(2 v_{0}\right)-h v_{0}=h v_{0}$
and $\frac{1}{2} m v^{\prime 2}=h\left(5 v_{0}\right)-h v_{0}=4 h v_{0}$
(ii) $\div( i )$, we get
$\frac{v_{0}^{2}}{v^{'2}}=\frac{4}{1} $
$\Rightarrow v^{'2}=4 v_{0}^{2}$
or $v'=2 v_{0}$