. In a photoemissive cell with exciting wavelength lambda the fastest electron has speed v
If the exciting wavelength is changed by 3λ /4 , the speed of the fastest emitted electron will be :
A: v (3/4)^1/2
B: v (4/3)^1/2
C: less than v (4/3)^1/2
D: greater than v (4/3)^½
Best Answer
Explanation:
When light of sufficient energy incident on metal, then some of its energy is used to eject an electron
from the surface of metal and rest of its energy is taken by moving electron.
Hence,
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Final answer:
Hence, the correct option is (D).
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