Question

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with the radiation of wavelength 3$\lambda$, the stopping potential is $V/4$. If the threshold wavelength for the metallic surface is n$\lambda$, then the value of n will be_____________

Answer 1

We have from Einstein's photoelectric equation,$\frac{hc}{\lambda }=\varphi +eV$ -------------------$\left(1\right)$ where $\frac{hc}{\lambda }$ is the energy of the incident photon, $\varphi$ is the work function and $V$ is the stopping potential.

1. When the light of 3$\lambda$ is incident the equation becomes, $\frac{hc}{3\lambda }=\varphi +\frac{eV}{4}$-------------$\left(2\right)$
2. Dividing $\left(1\right)$ with $\left(2\right)$,We get, $3=\frac{\varphi +eV}{\varphi +{\displaystyle \frac{eV}{4}}}$
3. Simplifying the equation, $\varphi =\frac{eV}{8}$. Substituting this in eqn(1), $\frac{hc}{\lambda }=\frac{eV}{8}+eV$$=\frac{9eV}{8}$
4. Therefore, $eV=\frac{8hc}{9\lambda }$. Thus eqn (1) will become, $$\begin{gathered} \varphi -\frac{hc}{\lambda }=\frac{-8hc}{9\lambda } \\ \varphi =\frac{hc}{9\lambda } \end{gathered}$$
5. From this the value of$n$will be $9$.