In a photoelectric experiment ultraviolet light of wavelength
Question:

In a photoelectric experiment ultraviolet light of wavelength $280 \mathrm{~nm}$ is used with lithium cathode having work function $\phi=2.5 \mathrm{eV}$. If the wavelength of incident light is switched to $400 \mathrm{~nm}$, find out the change in the stopping potential. ( $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$, $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$

1. $1.3 \mathrm{~V}$

2. $1.1 \mathrm{~V}$

3. $1.9 \mathrm{~V}$

4. $0.6 \mathrm{~V}$

Correct Option: 1

Solution:

$\mathrm{KE}_{\max }=\mathrm{eV}_{\mathrm{S}}=\frac{\mathrm{hc}}{\lambda}-\phi$

$\Rightarrow \mathrm{eV}_{\mathrm{S}}=\frac{1240}{280}-2.5=1.93 \mathrm{eV}$

$\rightarrow \mathrm{V}_{\mathrm{S}_{1}}=1.93 \mathrm{~V} \ldots$ (i)

$\rightarrow \mathrm{eV}_{\mathrm{S}_{2}}=\frac{1240}{400}-2.5=0.6 \mathrm{eV}$

$\Rightarrow \mathrm{V}_{\mathrm{S}_{2}}=0.6 \mathrm{~V}$  ……………..(II)

$\Delta \mathrm{V}=\mathrm{V}_{\mathrm{S}_{1}}-\mathrm{V}_{\mathrm{S}_{2}}=1.93-0.6=1.33 \mathrm{~V}$

Option (1)