Question

The threshold frequency for a metallic surface corresponds to an energy of $6.2eV$ and the stopping potential for a radiation incident on this surface is $5V$. The incident radiation lies in

• A. ultra violet region
• B. infrared region
• C. visible region
• D. X - ray region

Answer 1

Answer: (A)

From Einstein's photoelectric equation
$hv=h{v}_0+e{V}_0$
$=6.2+5=11.2eV$
$\Rightarrow$ $\frac{hc}{\lambda }=11.2eV$
or $\lambda =\frac{hc}{11.2eV}$
$=\frac{6.6\times 10^{-34}\times 3.0\times 10^8}{11.2\times 1.6\times 10^{-19}}$
$=1.1049\times 10^{-7}$
$=1104.9\mathring{A}$
This incident radiation lies in ultra violet region,