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The stopping potential for a certain photosensitive metal is \mathrm{V_0} when the frequency of incident radiation is\mathrm{V_0} . When the frequency of the incident radiations is doubled, what will be the stopping potential?

Option: 1

\mathrm{V_0}


Option: 2

\mathrm{2V_0}


Option: 3

\mathrm{4V_0}


Option: 4

none of the above


Answers (1)

best_answer

\mathrm{h v-h v_o=e v_0}                                       .....[1]

\mathrm{\text { If } h v^{\prime}-h v_o=e v_o^{\prime}}                                 .....[2]

\mathrm{\text { If } v^{\prime}=2 v}                                                  .....[3]

\mathrm{\text { From given information, we can't have } v_0^{\prime} \text { in terms of } v_0 \text {. }}

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Ritika Kankaria

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