The threshold frequency of metal is . When the light of frequency is incident on the metal plate, the maximum velocity of photoelectron is . [2023]

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Q.
The threshold frequency of metal is $f_{0}$ . When the light of frequency $2 f_{0}$ is incident on the metal plate, the maximum velocity of photoelectron is $v_{1}$ . When the frequency of incident radiation is increased to $5 f_{0}$ , the maximum velocity of photoelectrons emitted is $v_{2}$ . The ratio of $v_{1}$ to $v_{2}$ is [2023]

1 $\frac{v_{1}}{v_{2}} = \frac{1}{2}$

2 $\frac{v_{1}}{v_{2}} = \frac{1}{8}$

3 $\frac{v_{1}}{v_{2}} = \frac{1}{16}$

4 $\frac{v_{1}}{v_{2}} = \frac{1}{4}$

Ans.
(1)

$K_{\max} = h f - h f_{0}$

For $f = 2 f_{0}$

$\frac{1}{2} m v_{1}^{2} = 2 h f_{0} - h f_{0} = h f_{0} \dots (i)$

For $f = 5 f_{0}$

$\frac{1}{2} m v_{2}^{2} = 5 h f_{0} - h f_{0} = 4 h f_{0} \dots (i i)$

From (i) and (ii),

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{2}$

Solution

1 $\frac{v_{1}}{v_{2}} = \frac{1}{2}$

2 $\frac{v_{1}}{v_{2}} = \frac{1}{8}$

3 $\frac{v_{1}}{v_{2}} = \frac{1}{16}$

4 $\frac{v_{1}}{v_{2}} = \frac{1}{4}$

Ans.
(1)

$K_{\max} = h f - h f_{0}$

For $f = 2 f_{0}$

$\frac{1}{2} m v_{1}^{2} = 2 h f_{0} - h f_{0} = h f_{0} \dots (i)$

For $f = 5 f_{0}$

$\frac{1}{2} m v_{2}^{2} = 5 h f_{0} - h f_{0} = 4 h f_{0} \dots (i i)$

From (i) and (ii),

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{2}$

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Solution

Q. The threshold frequency of metal is f0. When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectron is v1. When the frequency of incident radiation is increased to 5f0, the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is [2023]

1 v1v2=12

2 v1v2=18

3 v1v2=116

4 v1v2=14

Ans. (1) Kmax=hf-hf0 For f=2f0 12mv12=2hf0-hf0=hf0 ⋯(i) For f=5f0 12mv22=5hf0-hf0=4hf0 ⋯(ii) From (i) and (ii), ⇒v1v2=12

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1 $\frac{v_{1}}{v_{2}} = \frac{1}{2}$

2 $\frac{v_{1}}{v_{2}} = \frac{1}{8}$

3 $\frac{v_{1}}{v_{2}} = \frac{1}{16}$

4 $\frac{v_{1}}{v_{2}} = \frac{1}{4}$

Ans.
(1)

$K_{\max} = h f - h f_{0}$

For $f = 2 f_{0}$

$\frac{1}{2} m v_{1}^{2} = 2 h f_{0} - h f_{0} = h f_{0} \dots (i)$

For $f = 5 f_{0}$

$\frac{1}{2} m v_{2}^{2} = 5 h f_{0} - h f_{0} = 4 h f_{0} \dots (i i)$

From (i) and (ii),

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{2}$