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 $v_0$. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $v = 5v_0$?
Answer & explanation
Correct answer: option 2
As $v_0$ is the threshold frequency.
∴ Work function, $\phi_0=hv_0$
According to Einstein photoelectric equation
$\frac{1}{2}mv_{max}^2=hv-\phi_0$
Where hv is the incident energy, $\phi_0$ is the work function of the metal and $\frac{1}{2}mv_{max}^2$ is the maximum kinetic energy of the emitted photoelectron.
As per question
$\frac{1}{2}mv_0^2=h(2v_0)-hv_0=hv_0$ (i)
and $\frac{1}{2}mv'^2=5(5v_0)-hv_0=4hv_0$ (ii)
Divide (ii) by (i), we get
$\frac{v'^2}{v_0^2}=\frac{4}{1}$
$v'^2=4v_0^2$ or $v'=2v_0$