Q. Ultraviolet radiations of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV)
A) $3.2 \times 10^{- 21}J$
B) $3.2 \times 10^{- 19}J$
C) $7 \times 10^{- 25}\ J$
D) $9 \times 10^{- 32}\ J$
Explanation
(b) KE of fastest electron
$= E = \phi_{0} = 6.2 - 4.2 = 2.0\ eV$
$= 2 \times 1.6 \times 10^{- 19} = 3.2 \times 10^{- 19}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.