Q122 / 180 · KCET 2008 · Physics
When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$, then stopping potential is $3 \mathrm{~V}_{\mathrm{s}}$. When same surface is illuminated by light of wavelength $2 \lambda$, then stopping potential becomes $\mathrm{V}_{\mathrm{s}}$. The value of threshold wavelength for photoelectric emission will be
- Correct answerA$4 \lambda$
- B$8 \lambda$
- C$\frac{4}{3} \lambda$
- D$6 \lambda$
Solution
According to Einstein's photoelectric equation
$$
\mathrm{eV}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
$$
Ist case
$$
3 \mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right]
$$
IInd case
$$
\mathrm{eV}_{\mathrm{s}}=\mathrm{hc}\left[\frac{1}{2 \lambda}-\frac{1}{\lambda_{0}}\right]
$$
Dividing Eq. (i) by Eq. (ii), we get
$$
\lambda_{0}=4 \lambda
$$
Hence, the correct answer is option: 'A'.