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Question

In a photocell, with excitation wavelength λ, the faster electrons has speed v. If the excitation wavelength is changed to 3λ/4, the speed of the fastest electron will be

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

The correct option is D greater than v(4/3)1/2
Initially, using energy conservation
12mv2=hcλW0
Let the speed of the fastest electron be v1, when excitation wavelength is changed to 3λ/4.
12mv21=4hc3λW0
12mv21=43(hcλW0)+W03
12mv21=43(12mv2)+W03 [using Eq. (i)]
v21=4v23+2W03m
v1>43v

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