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When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is :
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- 4λ
- 5λ
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5 λ 2 - 3λ
Correct Option: D
According to Einstein's photoelectric effect,
| eV = | - | .....(i) | ||
| λ | λ0 |
| = | - | ....(ii) | ||||
| 4 | 2λ | λ0 |
Dividing equation (i) by (ii) by
| ⇒ 4 = | - | ||
| λ0 | |||
| - | |||
| 2λ | λ0 | ||
on solving we get,
λ0 = 3λ
