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Question

A certain metal when irradiated with light (ν=3.2×1016 Hz) emits photoelectrons with twice kinetic energy of photoelectrons that are emitted when the same metal is irradiated by light (ν=2.0×1016 Hz). Calculate νo of electron.

A
1.2 × 1014 Hz
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B
8 × 1015 Hz
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C
1.2 × 1016 Hz
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D
4 × 1012 Hz
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Solution

The correct option is B 8 × 1015 Hz
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 × 1015 Hz

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