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Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively.
Ratio of maximum speeds of emitted electrons will be
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- 1 : 4
- 1 : 2
- 1 : 1
- 1 : 5
Correct Option: B
The maximum kinetic energy of emitted electrons is given by
K.E = φ – φ 0
K.E1 = 1 eV – 0.5 eV = 0.5 eV
K.E2 = 2.5 eV – 0.5 eV = 2 eV
| ∴ | = | = | ||||
| KE2 | 2eV | 4 |
| ⇒ | = |  | = | ||||
| v2 | 4 | 2 |
