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Question

Energy of incident radiation increases from 5 eV to 10 eV. If the threshold wavelength for the given metal surface is 310 nm, find the ratio of maximum kinetic energies of the emitted electrons.

A
1:6
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B
1:3
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C
1:5
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D
1:2
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Solution

The correct option is A 1:6
ϕ0=work function=hcλ

=6.63×1034×3×108310×109=4 eV

Now, K.Emax=Eϕ0
As the material is same, ϕ0=4 eV
(K.E)max1(K.E)max2=E1ϕ0E2ϕ0=54104=16

Hence, (A) is the correct answer.
Why this question?This question shows dependance of maximum kinetic energy on energy of incident radiation.

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