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Q.

A certain metal when irradiated with light ( $v$ = $3.2$ X $10^{16}$ Hz) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated with light $(v = 2.0 x 10^{16} H z)$ . The $v_{0}$ (threshold frequency ) of metal is

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a

$4 x 10^{12} H z$

b

$1.2 x 10^{14} H z$

c

$8 x 10^{15} H z$

d

$1.2 x 10^{16} H z$

answer is B.

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Detailed Solution

$K E = h v_{0} - h v_{0}$

$h v_{1} - h v_{0} = 2 (h v_{2} - h v_{0})$

$v_{0} = 2 (v_{2} - v_{1})$

$= 2 (2.0 \times 10^{16}) - (3.2 \times 10^{16})$

$= 8 \times 10^{15} s^{- 1} = 8 \times 10^{15} H z$

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