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Question

In a photoemissive cell with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed by 3λ/4, the speed of the fastest emitted electron will be :
  1. v(3/4)1/2
  2. v(4/3)1/2
  3. less than v(4/3)1/2
  4. greater than v(4/3)1/2

A
less than v(4/3)1/2
B
v(4/3)1/2
C
v(3/4)1/2
D
greater than v(4/3)1/2
Solution
Verified by Toppr

From E=W0+12mv2maxvmax=2Em2W0m
where E=hcλ
If wavelength of incident light charges from λ to 3λ4 (Decreases)
Let energy of incident light charges from E and speed of fastest electron changes from v to v then
v=2Em2W0m
As E1λE43E
Hence v=  2(43)Em2W0m
v=(4/3)1/22Em2W0m(4/3)1/2
So, v>(4/3)1/2v

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