The threshold frequency of a metal is f 0. When the light of frequency 2 f 0 is incident on the metal plate, the maximum velocity of photoelectrons is v1. When the frequency of incident radiation is increased to 5 f 0, the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is :

Question

The threshold frequency of a metal is f 0. When the light of frequency 2 f 0 is incident on the metal plate, the maximum velocity of photoelectrons is v1. When the frequency of incident radiation is increased to 5 f 0, the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is : (A) (v1/v2)=(1/8) (B) (v1/v2)=(1/4) (C) (v1/v2)=(1/16) (D) (v1/v2)=(1/2)

Tardigrade Admin · Accepted Answer

K max = hf - hf 0 text For f =2 f 0 (1/2) m V 12=2 h 0- hf 0= hf 0 text For f =5 f 0 (1/2) m V 22=5 hf0 - hf f 0=4 hf f 0 ( V 1/ V 2)=(1/2)

$\frac{v _{1}}{v _{2}} = \frac{1}{8}$

$\frac{v _{1}}{v _{2}} = \frac{1}{4}$

$\frac{v _{1}}{v _{2}} = \frac{1}{16}$

$\frac{v _{1}}{v _{2}} = \frac{1}{2}$

$K_{m a x} = h f - h f_{0}$
$For f = 2 f_{0}$
$\frac{1}{2} m V_{1}^{2} = 2 h_{0} - h f_{0} = h f_{0}$
$For f = 5 f_{0}$
$\frac{1}{2} m V_{2}^{2} = 5 h f_{0} - h f f_{0} = 4 h f f_{0}$
$\frac{V _{1}}{V _{2}} = \frac{1}{2}$

v2​v1​​=81​

v2​v1​​=41​

v2​v1​​=161​

v2​v1​​=21​

Kmax​=hf−hf0​ For f=2f0​ 21​mV12​=2h0​−hf0​=hf0​ For f=5f0​ 21​mV22​=5hf0​−hff0​=4hff0​ V2​V1​​=21​

$\frac{v _{1}}{v _{2}} = \frac{1}{8}$

$\frac{v _{1}}{v _{2}} = \frac{1}{4}$

$\frac{v _{1}}{v _{2}} = \frac{1}{16}$

$\frac{v _{1}}{v _{2}} = \frac{1}{2}$

$K_{m a x} = h f - h f_{0}$
$For f = 2 f_{0}$
$\frac{1}{2} m V_{1}^{2} = 2 h_{0} - h f_{0} = h f_{0}$
$For f = 5 f_{0}$
$\frac{1}{2} m V_{2}^{2} = 5 h f_{0} - h f f_{0} = 4 h f f_{0}$
$\frac{V _{1}}{V _{2}} = \frac{1}{2}$