Q148 / 160 · TS EAMCET 2003 · Physics
When radiation of the wavelength $\lambda$ is incident on a metallic surface, the stopping potential is $4.8 \mathrm{~V}$. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes $1.6 \mathrm{~V}$. Then, the threshold wavelength for the surface is :
- A$2 \lambda$
- Correct answerB$4 \lambda$
- C$6 \lambda$
- D$8 \lambda$
Solution
Stopping potential.
$V_0=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$
where, $V_0=$ stopping potential
Ist case, $4.8=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$
IInd case, $1.6=\frac{h c}{e}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)$
Dividing Eqs. (i) by (ii),
$3=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)}$
$\frac{3}{2 \lambda}-\frac{3}{\lambda_0}=\frac{1}{\lambda}-\frac{1}{\lambda_0}$
$\frac{1}{\lambda_0}-\frac{3}{\lambda_0}=\frac{1}{\lambda}-\frac{3}{2 \lambda}$
$\frac{-2}{\lambda_0}=\frac{2-3}{2 \lambda}$
$\frac{2}{\lambda_0}=\frac{1}{2 \lambda}$
$\lambda_0=4 \lambda$
Hence, the correct answer is option: 'B'.