First slide

Photo Electric Effect

Question

When the light of frequency $2 ϑ_{0}$ (where $ϑ_{0}$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is $v_{1}$ . When the frequency of the incident radiation is increased to $5 ϑ_{0}$ , the maximum velocity of electrons emitted from the same plate is $v_{2}$ . The ratio of $v_{1}$ to $v_{2}$ is

Easy

A

1:2
B

1:4
C

4:1
D

2:1

Solution

According to the Einstein's photoelectric equation, $E = W_{0} + \frac{1}{2} m v^{2}$
When frequency of incident light is $2 ϑ_{0}$
$h (2 ϑ_{0}) = h ϑ_{0} + \frac{1}{2} m v_{1}^{2} \Rightarrow h ϑ_{0} = \frac{1}{2} m v_{1}^{2} . . . . . . . . . . . (i)$
When frequency of incident light is $5 ϑ_{0}$
$h (5 ϑ_{0}) = h ϑ_{0} + \frac{1}{2} m v_{2}^{2} \Rightarrow 4 h ϑ_{0} = \frac{1}{2} m v_{2}^{2} . . . . . . . . (i i)$
Dividing (i) by (ii),
$\frac{1}{4} = \frac{v_{1}^{2}}{v_{2}^{2}} or \frac{v_{1}}{v_{2}} = \frac{1}{2}$

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