Q. The work function of a metal is 1eV. Light of wavelength 3000 $\mathring{\mathrm{A}}$ is incident on this metal surface. The velocity of emitted photoelectrons will be
A) $10\ ms^{- 1}$
B) $10^{3}\ ms^{- 1}$
C) $10^{4}\ ms^{- 1}$
D) $10^{6}\ ms^{- 1}$
Explanation
(d) $\frac{1}{2}mv^{2} = \frac{hc}{\lambda} - \phi_{0}$
$= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{(3 \times 10^{- 7}) \times 1.6 \times 10^{- 19}} - 1$
$= 4.14 - 1 = 3.14eV$
or $v = \sqrt{\frac{2 \times 3.14 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} = 10^{6}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.