Question

The threshold frequency for a certain photosensitive metal is $v_0$. When it is illuminated by light of frequency $v=2v_0$, the maximum velocity of photoelectrons is $v_0$. The maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $v=5v_0$ is $x{v}_0$. Find $x$.

Answer 1

Answer: 2

As $v_0$ is the threshold frequency,
$\therefore$ Work function, ${\varphi }_0=h{v}_0$
According to Einstein's photoelectric equation,
$\frac{1}{2}m{v}_{\max }^2=hv-{\varphi }_0$
where $hv$ is the incident energy, ${\varphi }_0$ 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,
$\frac{1}{2}m{v}_0^2=h\left(2v_0\right)-h{v}_0=h{v}_0$
and $\frac{1}{2}m{v}^{\prime 2}=h\left(5v_0\right)-h{v}_0=4h{v}_0$
(ii) $÷(i)$, we get
$\frac{v_0^2}{v^{{}^{\prime }2}}=\frac{4}{1}$
$\Rightarrow v^{{}^{\prime }2}=4v_0^2$
or $v^{\prime }=2v_0$