Q. Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy of the faster electron emitted is approximately
A) $3.2 \times 10^{- 15}$ J
B) $3.2 \times 10^{- 17}$ J
C) $3.2 \times 10^{- 19}$ J
D) $3.2 \times 10^{- 21}$ J
Explanation
(c) KE of emitted electron is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{K} = hv - W$
$= \ 6.2\ eV\ - \ 4.2\ eV\ = \ 2.0\ eV$
$= \ 2 \times 1.6 \times 10^{- 19}\ J$
$= \ 3.2 \times 10^{- 19}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.