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Q.

A certain metal when irradiated to light (v = 3.2 × 1016 Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light ( v = 20 × 1016 Hz). The v0 (threshold frequency) of metal is

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a

1.2×1014Hz

b

8×1015Hz

c

1.2×1016Hz

d

4×1012Hz

answer is B.

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Detailed Solution

K.E1=hv1−hv0K.E2=hv2−hv0As(K.E)1=2(K.E)2hv1−hv0=2(hv2−hv0) =h(v1−v0)=2h(v2−v0)v1−v0=2v2−2v02v0−v0=2v2−v1v0=2v2−v1 =2(2.0×1016)−(3.2×1016) =8×1015s−1=8×1015Hz
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