If the work function of a certain metal is $$3.2$$×$$10-19J$$ and it is a illuminated with light of frequency $$10$$×$$1014Hz$$, the maximum kinetic energy of photo electrons would $$be(h$$ $$=$$ $$6.6$$×$$10-34Js)$$

Correct option is 13.4×10−19J

Questions

If the work function of a certain metal is $3 .2 \times 10^{-19} J$ and it is a illuminated with light of frequency $10 \times 10^{14} Hz,$ the maximum kinetic energy of photo electrons would be $(h = 6 .6 \times 10^{-34} Js)$

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detailed solution

Correct option is A

hv=w+kE kE=hv−w =6.6×10−34×10×1014−3.2×10−19 =6.6×10−19−3.2×10−19 =3.4×10−19J

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If the work function of a certain metal is 3.2×10-19J and it is a illuminated with light of frequency 10×1014Hz, the maximum kinetic energy of photo electrons would be(h = 6.6×10-34Js)

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If the work function of a certain metal is $3 .2 \times 10^{-19} J$ and it is a illuminated with light of frequency $10 \times 10^{14} Hz,$ the maximum kinetic energy of photo electrons would be $(h = 6 .6 \times 10^{-34} Js)$