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Following graph shown the variation of stopping potential corresponding to the frequency of incident radiation $(v)$ for a given metal. The correct variation is shown in graph [ $v_0=$ threshold frequency]
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The correct answer is:
(A)
Concept: Stopping potential $V$ is related to the maximum kinetic energy $K_{\max }$ of the photoelectrons, via Einstien's photo-electric effect relation:
$e V=K_{\max }=h v-h v_0$
Therefore,
$V=\left(\frac{h}{e}\right) v-\left(\frac{h}{e}\right) v_0$
Therefore, $V$ vs $v$ is straighline with positive slope and at $v=v_0$ the stopping potential is zero.
$e V=K_{\max }=h v-h v_0$
Therefore,
$V=\left(\frac{h}{e}\right) v-\left(\frac{h}{e}\right) v_0$
Therefore, $V$ vs $v$ is straighline with positive slope and at $v=v_0$ the stopping potential is zero.
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