Question

When a surface of thickness 1 $cm$ is illuminated with light of wavelength $\lambda$, the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$, the stopping potential is $\frac{V}{3}$. Then the threshold wavelength for the surface is:

• A. $\frac{4\lambda }{3}$
• B. $4\lambda$
• C. $6\lambda$
• D. $\frac{8\lambda }{3}$

Answer 1

The correct option is B $4\lambda$

From given conditions, we can write
$\frac{hC}{\lambda }=\varphi +eV$ ...........(i)
$\frac{hC}{2\lambda }=\varphi +\frac{eV}{3}$ ..........(ii)

On solving ,[ 3x(ii) -(i) ], we get
$\Rightarrow (\frac{3}{2}-1)\frac{hc}{\lambda }=2\varphi \Rightarrow \varphi =\frac{hc}{4\lambda }=\frac{hc}{{\lambda }_{th}}$
$\therefore {\lambda }_{th}=4\lambda$