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Q.

A  certain  metal when irradiated  by light v=3.2×1016Hz emits  photoelectrons  with  twice of K.E as did photoelectrons  when  the same  metal is irradiated  by light v=2.0×1016Hz. The v0 of the  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.E.)1=hv1hv0
(K.E.)2=hv2hv0
As (KE.)1=2×(KE.)2
 hv1hv0=2hv2hv0
or v0=2v2v1=2×2×10163.2×1016
=0.8×1016Hz or 8×1015Hz

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