A certain metal when irradiated with light (ν=3.2×1016Hz) emits photoelectrons with twice kinetic energy of photoelectrons that are emitted when the same metal is irradiated by light (ν=2.0×1016Hz). Calculate νo of electron.
A
1.2×1014Hz
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B
8×1015Hz
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C
1.2×1016Hz
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D
4×1012Hz
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Solution
The correct option is B8×1015Hz Applying Einstein's Photoelectric Equation, K.E=E−Wo K.E=hν−hνo
In first case, K.E1=h(3.2×1016−νo)
In second case, K.E2=h(2.0×1016−νo)
It is given in that (K.E1)=2(K.E2)
Hence, [h(3.2×1016−νo)]=2[h(2.0×1016−νo)]
On solving, νo=8×1015Hz