In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 \mathrm{~nm}$ to $400 \mathrm{~nm}$. The decrease in the stopping potential is close to:
$\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{~nm}-\mathrm{V}\right)$
Correct Option: 3,
(3) Let $\phi=$ work function of the metal,
$\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\mathrm{eV}_{1}$ ...(1)
$\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}$ ...(2)
Sutracting (ii) from (i) we get
$\operatorname{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$
$\Rightarrow \mathrm{V}_{1}-\mathrm{V}_{2}=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{~nm} \\ \lambda_{2}=400 \mathrm{~nm} \\ \frac{h_{c}}{\mathrm{e}}=1240 \mathrm{~nm}-\mathrm{V}\end{array}\right]$
$=(1240 \mathrm{~nm}-\mathrm{v})\left(\frac{100 \mathrm{~nm}}{300 \mathrm{~nm} \times 400 \mathrm{~nm}}\right)$
$=1.03 \mathrm{~V} \approx 1 \mathrm{~V}$