Q82 / 90 · JEE Main 2019 (11 Jan Shift 2) · Physics
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 :
$\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\right)$
- A$0.5 \mathrm{~V}$
- B$1.5 \mathrm{~V}$
- Correct answerC$1.0 \mathrm{~V}$
- D$2.0 \mathrm{~V}$
Solution
Let $\phi=$ work function of the metal,
$\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\mathrm{eV}_{1}$
$\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\mathrm{eV}_{2} \quad \ldots \ldots$
(ii)
Sutracting (ii) from (i) we get
$\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$
$\Rightarrow V_{1}-V_{2}=\frac{h c}{e}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{nm} \\ \lambda_{2}=400 \mathrm{nm} \\ \frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\end{array}\right]$
$=(1240 \mathrm{nm}-\mathrm{v})\left(\frac{100 \mathrm{nm}}{300 \mathrm{nm} \times 400 \mathrm{nm}}\right)$
$=1,03 \mathrm{~V} \approx 1 \mathrm{~V}$
Hence, the correct answer is option: 'C'.