Q52 / 75 · JEE Main 2026 (24 January Shift 2) · Physics
When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is $\_\_\_\_$ m. $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}. \mathrm{s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
- Correct answerA$2.5 \times 10^{-7}$
- B$2.2 \times 10^{-8}$
- C$3.1 \times 10^{-7}$
- D$2.9 \times 10^{-8}$
Solution
Using Einstein's photoelectric equation with stopping potentials:
For first light, $h\nu_1 = \phi + eV_{s1}$.
For second light with $\lambda_2 = 2\lambda_1$ (so $\nu_2 = \nu_1/2$): $h\nu_1/2 = \phi + eV_{s2}$.
From given values: $h\nu_1 - 3.2e = h\nu_1/2 - 0.7e$ gives $h\nu_1/2 = 2.5e$.
Thus $\lambda_1 = \frac{hc}{5e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \approx 2.5 \times 10^{-7}$ m.
For first light, $h\nu_1 = \phi + eV_{s1}$.
For second light with $\lambda_2 = 2\lambda_1$ (so $\nu_2 = \nu_1/2$): $h\nu_1/2 = \phi + eV_{s2}$.
From given values: $h\nu_1 - 3.2e = h\nu_1/2 - 0.7e$ gives $h\nu_1/2 = 2.5e$.
Thus $\lambda_1 = \frac{hc}{5e} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{5 \times 1.6 \times 10^{-19}} \approx 2.5 \times 10^{-7}$ m.
Hence, the correct answer is option: 'A'.