Question

When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $\text{V}$. When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\frac{\text{V}}{4}$. If the threshold wavelength for the metallic surface is $n\lambda .$ The value of $n$ will be

• A. 9
• B. 9.00
• C. 9.0

Answer 1

Answer: (A), (B), (C)

Consider maximum kinetic energy by Einstien equation.
$K{E}_{max}=\left(\frac{hc}{\lambda }\right)-\varphi$

When the radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is $\text{V}$.
Then: $\frac{hc}{\lambda }=\varphi +e\text{V}$ ........(1)

When the same surface is illuminated with radiation of wavelength $3\lambda$, the stopping potential is $\frac{\text{V}}{4}$
Then: $\frac{hc}{3\lambda }=\varphi +\frac{e\text{V}}{4}$ .......(2)

Divide equation (2) by (1):
$3=\frac{\varphi +e\text{V}}{\varphi +\frac{e\text{V}}{4}}$
$3\varphi +\frac{3e\text{V}}{4}=\varphi +e\text{V}$
$2\varphi =\frac{e\text{V}}{4}$
$\varphi =\frac{e\text{V}}{8}$

Substitute the above value in equation (1),
Then: $\frac{hc}{\lambda }=\frac{e\text{V}}{8}+e\text{V}$
$e\text{V}=\frac{8}{9}\cdot \frac{hc}{\lambda }$

So,
$\varphi =\frac{hc}{\lambda }-\frac{8}{9}\cdot \frac{hc}{\lambda }$
$\varphi =\frac{1}{9}\cdot \frac{hc}{\lambda }$
$\frac{hc}{{\lambda }_{th}}=\frac{hc}{9\lambda }$
Then ${\lambda }_{th}=9\lambda$
If the threshold wavelength for the metallic surface is $n\lambda$ then value of $n$ will be $9$.