Question

Monochromatic light of wavelength $198$mm is incident on the surface of a metal, whose work function is $2.5$ eV. Calculate the stopping potential.

Answer 1

Given: $\lambda =198$nm $=198\times {10}^{-9}$m
$W=2.5eV=2.5\times 1.6\times {10}^{-19}$J
$V_0=?$
We know, $\frac{hc}{\lambda }=W+E_k$
$\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{198\times {10}^{-9}}=2.5\times 1.6\times {10}^{-19}+E_k$
$10\times {10}^{-19}=4\times {10}^{-19}+E_k$
$\therefore E_k=6\times {10}^{-19}$J
Now $E_k=e{V}_0$
$6\times {10}^{-19}=1.6\times {10}^{-19}\times V_0$
$\therefore V_0=3.75$Volt.