Following graph shown the variation of stopping potential corresponding to the frequency of incident…

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]
  1. (C)
  2. (B)
  3. (A)
  4. (D)

Solution

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.

Asked in: MHT CET 2022 (05 Aug Shift 2)

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