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 :
v(3/4)1/2
v(4/3)1/2
less than v(4/3)1/2
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
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Solution
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From E=W0+12mv2max⇒vmax=√2Em−2W0m
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′=√2E′m−2W0m
As E∝1λ⇒E′43E
Hence v′=
⎷2(43)Em−2W0m
⇒v′=(4/3)1/2√2Em−2W0m(4/3)1/2
So, v′>(4/3)1/2v
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