When the light source is kept $20 c m$ away from a photo cell, stopping the potential of $0.6 V$ is obtained. When the source is kept $40 c m$ away, the stopping potential will be:

0.3 V

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0.6 V

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1.2 V

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2.4 V

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Solution

The correct option is B $0.6 V$
Stopping Potential
when distance of the source is
increased the intensity of light
falling on the photo cell is reduced.
Hence, maximum K.E of Emitted
Photo electrons will be same as
before
Stopping Potential and maximum K.E
are related by $K . E_{m a x} = e V$
Since $K . E_{m a x}$ is not changing
stopping will not change
Photo electrons Emitted will be reduced
and saturation current will reduce
as the intensity varies inversely
with square of the distance
$0.6 V \Rightarrow$ required answer

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When the light source is kept 20cm away from a photo cell, stopping the potential of 0.6V is obtained. When the source is kept 40cm away, the stopping potential will be:

When the light source is kept $20 c m$ away from a photo cell, stopping the potential of $0.6 V$ is obtained. When the source is kept $40 c m$ away, the stopping potential will be: