Question

The work function of a 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. X-ray region
• C. Ultraviolet region
• D. Visible region

Answer 1

Answer: (C)

$eV { }_{\circ}= h \vartheta - h\vartheta_{\circ}$
$h \vartheta =\frac{ hc }{\lambda}= eV _{\circ}+ h \vartheta _{\circ}$
$=5 eV +6.2 eV$
$\frac{ hc }{\lambda}=11.2 eV$
$\rightarrow \lambda=\frac{ hc }{11.2 eV }$
$=1.104 \times 10^{-7}$
$\therefore$ Ultraviolet region