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`        The wave length of an incident light is 6000A° if work function of the material is 1ev calculate the velocity of the emitted electron`
one year ago

## Answers : (1)

Arun
18681 Points
```							Dear Pooja E = Wo + Kmax E = 12375/6000 = 2.0625 ev NowKmax = E - Wo = 2.0625 - 1 = 1.0625 ev Now1/2* m* V² = 1.0625 * 1.6 * 10^-19 joules Put m = 9.1 * 10^-31 kg And find V  RegardsArun (askIITians forum expert)
```
one year ago
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