Question

The threshold frequency for a metallic surface corresponds to an energy of $6.2 \,eV$ and the stopping potential for a radiation incident on this surface is $5 \,V$. 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=hv_{0}+eV_{0} $
$ =6.2+5=11.2\,\,eV $
$ \Rightarrow $ $ \frac{hc}{\lambda }=11.2\,eV $
or $ \lambda =\frac{hc}{11.2\,eV} $
$ =\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,