Q136 / 150 · MHT CET 2023 (12 May Shift 1) · Physics
When light of wavelength $\lambda$ is incident on a photosensitive surface the stopping potential is ' $\mathrm{V}$ '. When light of wavelength $3 \lambda$ is incident on same surface the stopping potential is $\frac{\mathrm{V}}{6}$. Then the threshold wavelength for the surface is
- A$2 \lambda$
- B$3 \lambda$
- C$4 \lambda$
- Correct answerD$5 \lambda$
Solution
$$
\begin{aligned}
& \lambda_1=\lambda,\left(\mathrm{V}_0\right)_1=\mathrm{V} \\
& \lambda_2=3 \lambda,\left(\mathrm{V}_0\right)_2=\frac{\mathrm{V}}{6}
\end{aligned}
$$
Photo electric equation is given by $\mathrm{eV}_0=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)....(i)$
In first case,
$$
\mathrm{eV}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)
$$
For second case,
$$
\frac{\mathrm{eV}}{6}=\mathrm{hc}\left(\frac{1}{3 \lambda}-\frac{1}{\lambda_0}\right).....(ii)
$$
Dividing equation (i) by equation (ii),
$$
\begin{aligned}
6 & =\frac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{3 \lambda}-\frac{1}{\lambda_0}}=\frac{3 \lambda_0-\lambda}{\lambda_0-3 \lambda} \\
\therefore \quad \lambda_0 & =5 \lambda
\end{aligned}
$$
Hence, the correct answer is option: 'D'.