Q 12-11-134JEE MainJEE Main 2021 (26 Aug, Shift 1)Medium
In a photoelectric experiment, ultraviolet light of wavelength $280$ nm is used with lithium cathode having work function $\phi = 2.5$ eV. If the wavelength of incident light is switched to $400$ nm, find out the change in the stopping potential. ($h = 6.63\times10^{-34}$ J s, $c = 3\times10^8$ m s$^{-1}$)
Answer: (C) $1.3$ V
$eV_s = \dfrac{hc}{\lambda} - \phi$, so $e\,\Delta V_s = hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$.
$hc = 1.989\times10^{-25}$ J m $\approx 1243$ eV nm.
$$\Delta V_s = 1243\left(\frac{1}{280} - \frac{1}{400}\right) = 4.44 - 3.11 \approx 1.3\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics