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Question

The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency 2×1014Hz is 6.63×1020J . The threshold frequency of the metal is: (Take the value of h= 6.63×1034Jsec)

A
1×1014Hz
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B
2×1014Hz
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C
1×1015Hz
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D
2×1015Hz
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Solution

The correct option is A 1×1014Hz
Given:
Kinetic energy =6.63×1019J
Frequency of radiation (ν)=2×1014HzPlancks constant(h)=6.63×1034Js
Therefore,
Absorbed energy = hν=6.63×1034×2×1014=13.26×1020
As we know that,
Absorbed energy = Threshold energy + Kinetic energy
Threshold energy = Absorbed energy − Kinetic energy
hν0=hνK.EThreshold frequency (ν0)=13.26×10206.63×1020h=1×1014Hz
Hence the threshold frequency of the metal will be 1×1014Hz

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