Question

The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the :

• A. Infrared region
• B. Ultraviolet region
• C. X-ray region
• D. Visible region

Answer 1

Answer: (B)

The correct option is C Ultraviolet region

$E_{incident}=W+K_{max};K_{max}=e{V}_0$

$hv=h{v}_0+e{V}_0;stoppingpotential=V_0$

$\frac{hc}{\lambda }=6.2e+5e$

$\lambda =\frac{h\times c}{11.2\times 1.6\times {10}^{-19}}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{11.2\times 1.6\times {10}^{-19}}$

$=1.1\times {10}^{-7}m$

Hence the wavelength of ultraviolet region.