Thank you for reporting, we will resolve it shortly
Q.
The work function of metal is $4.2\, e v$. If radiations of $2000^{\circ} A$ fall on the metal then the kinetic energy of the fastest photoelectron is
Solution:
$E_{0}=4.2 eV =4.2 \times 1.60 \times 10^{-19} \,J$
$=6.72 \times 10^{-19} J$
$E=h v=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2000 \times 10^{-10}}$
$=9.94 \times 10^{-19} J$
$\therefore K E$ of electron emitted
$=(9.94-6.72) \times 10^{-19} J=3.22 \times 10^{-19} J$