Q. Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 $\mathring{\mathrm{A}}$ falls on the surface, the velocity of fastest electrons ejected from the surface is
A) $2.5 \times 10^{18}ms^{- 1}$
B) $2.5 \times 10^{13}ms^{- 1}$
C) $6.7 \times 10^{18}ms^{- 1}$
D) None of these
Explanation
(d) $\frac{1}{2}mv^{2} = \frac{hc}{\lambda} - \phi(in\ eV)$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6\ \times 10^{- 19}} - 4.2$
$= 2eV = 2 \times 1.6 \times 10^{- 19}$J
$v = \sqrt{2 \times 2 \times 1.6 \times 10^{- 19}/9.1 \times 10^{- 31}}$
$= \sqrt{6.4/9.1}\ = \ 10^{6}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.