When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $V$. When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\dfrac{V}{4}$. If the threshold wavelength for the metallic surface is $n\lambda$ then value of $n$ will be ______.
Numerical value type. Enter your answer.
Answer: 9
Einstein's equation for the two cases ($\phi$ = work function):
$$\frac{hc}{\lambda} - \phi = eV, \qquad \frac{hc}{3\lambda} - \phi = \frac{eV}{4}$$
Multiply the second by $4$ and equate with the first: $\dfrac{hc}{\lambda} - \phi = \dfrac{4hc}{3\lambda} - 4\phi$, so $3\phi = \dfrac{hc}{3\lambda}$ and $\phi = \dfrac{hc}{9\lambda}$.
Threshold wavelength $\lambda_0 = \dfrac{hc}{\phi} = 9\lambda$, so $n = 9$.
Solution by Sreeraj P, M.Sc Physics