Q. A light of wavelength 4000$\ \mathring{\mathrm{A}}$ is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is $(R = 6.6 \times 10^{- 34}$Js)
A) $1.35 \times 10^{5}\ ms^{- 1}$
B) $2.7 \times 10^{5}\ ms^{- 1}$
C) $6.2 \times 10^{5}\ ms^{- 1}$
D) $8.1 \times 10^{5}\ ms^{- 1}$
Explanation
(c) $\frac{1}{2}mv^{2} = \frac{hc}{\lambda} - \phi\ (in\ eV)$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{4000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 2$
$= 3.1 - 2 = 1.1\ eV = 1.1 \times 1.6 \times 10^{- 19}\ J$
$= 1.76 \times 10^{- 19}\ J$
$v = \frac{1.76 \times 10^{- 19} \times 2}{9 \times 10^{- 13}}$
$= 6.2 \times 10^{5}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.