A cesium photocell, with a steady potential difference of 60 V across it, is illuminated by a small bright light placed 1 m away. When the same light is placed 2 m away, the electrons crossing the photocell.

each carry one-quarter of their previous momentum

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each carry one-quarter of their previous energy

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are one-quarter as numerous

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are half as numerous

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Solution

The correct option is C are one-quarter as numerous
Number of Photo electrons (N)
As $I \propto \frac{1}{d^{2}}$
We get $\frac{I_{2}}{I_{1}} = \frac{{(1)}^{2}}{{(2)}^{2}} = \frac{1}{4}$

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A cesium photocell, with a steady potential difference of 60 V across it, is illuminated by a small bright light placed 1 m away. When the same light is placed 2 m away, the electrons crossing the photocell.

The correct option is C are one-quarter as numerousNumber of Photo electrons (N)AsI∝1d2We get I2I1=(1)2(2)2=14

The correct option is C are one-quarter as numerous
Number of Photo electrons (N)
As $I \propto \frac{1}{d^{2}}$
We get $\frac{I_{2}}{I_{1}} = \frac{{(1)}^{2}}{{(2)}^{2}} = \frac{1}{4}$