In a photoelectric cell, the cathode with work function $W_{1}$ is replaced by another one with work function $W_{2}$ ( $W_{2} > W_{1}) .$ If the current before this change is $I_{1}$ and that after the change is $I_{2}$ and other circumstances remain same and if $h v > W_{2},$ then

I_{1} > I_{2}

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I_{1} < I_{2}

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I_{1} = I_{2}

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I_{1} < I_{2} < 2 I_{1}

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Solution

The correct option is C $I_{1} = I_{2}$
When radiation is incident on a photo cell, the velocity with which the photo electrons are ejected depends on frequency of incident radiation, number of electrons(current) depends on intensity of the incident of radiation. Hence the current would depend on the incident radiation. Both cases have the same incident radiation.
Hence, $I_{1} = I_{2}$
So, the answer is option (C).

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The cathode of a photoelectric cell is changed such that the work function changes from $W_{1} t o W_{2} (W_{2} > W_{1})$ . If the current before and after change are $l_{1} a n d l_{2}$ all other conditions remaining unchanged, then (assuming hn > $W_{2}$ )