The threshold frequency ν0 for a metal is 7.0×1014s−1. Calculate the kinetic energy of an electron emitted when a radiation of frequency ν=1.0×1015s−1 hits the metal.
A
1.988×10−19J
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B
4.567×10−19J
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C
1.988×10−16J
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D
4.567×10−16J
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Solution
The correct option is A1.988×10−19J According to Einstein's equation,
Kinetic energy =12mev2=h(ν−ν0) =(6.626×10−34J s)(1.0×1015s−1−7.0×1014s−1) =(6.626×10−34Js)(10.0×1014s−1−7.0×1014s−1) =(6.626×10−34Js)(3.0×1014s−1) =1.988×10−19J