Question

When one centimeter thick surface 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 $\frac{V}{3}$. Threshold wavelength for metallic 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 $4\lambda$

We know,

$K{E}_{max}=E_{incident}-Thresholdenergy$

For the case 1

Wave length is $\lambda$ and stopping potential is V

So, $eV=\frac{hc}{\lambda }-Thresholdenergy$ -----------------(1)

For the case 2

Wave length is $2\lambda$ and stopping potential is $\frac{V}{3}$

So, $e\frac{V}{3}=\frac{hc}{2\lambda }-Thresholdenergy$ ----------------------(2)

Dividing equation 2 by 1

We get,

$Thresholdenergy=\frac{hc}{4\lambda }=\frac{hc}{{\lambda }_0}$

So we can write

${\lambda }_0=4\lambda$

Hence the threshold wavelength is $4\lambda$