# In a photoelectric experiment,

Question:

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:

$\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{~nm}-\mathrm{V}\right)$

1. (1) $0.5 \mathrm{~V}$

2. (2) $1.5 \mathrm{~V}$

3. (3) $1.0 \mathrm{~V}$

4. (4) $2.0 \mathrm{~V}$

Correct Option: 3,

Solution:

(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

$\operatorname{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$

$\Rightarrow \mathrm{V}_{1}-\mathrm{V}_{2}=\frac{\mathrm{hc}}{\mathrm{e}}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{~nm} \\ \lambda_{2}=400 \mathrm{~nm} \\ \frac{h_{c}}{\mathrm{e}}=1240 \mathrm{~nm}-\mathrm{V}\end{array}\right]$

$=(1240 \mathrm{~nm}-\mathrm{v})\left(\frac{100 \mathrm{~nm}}{300 \mathrm{~nm} \times 400 \mathrm{~nm}}\right)$

$=1.03 \mathrm{~V} \approx 1 \mathrm{~V}$