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In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4 , the speed of the fastest emitted electron will be
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- (3/4)1/2. v
- (4/3)1/2. v
- less than (4/3)1/2. v
- greater than (4/3)1/2. v
Correct Option: D
mv2 = | - W0 | |||
2 | λ |
or | = | mv2 + W0 | ||
λ | 2 |
and | mv21 = | - W0 | ||
2 | (3λ/4) |
mv2 + W0 | - W0 | ||||||
4 | 2 |
So, v1 is greater than
v1 = | 1/2 | 3 |