Q155 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics

In a photoelectric effect experiment if the frequency of light is doubled, the stopping potential will

  • A
    be halved
  • Correct answer
    B
    become more than double
  • C
    become less than double
  • D
    be doubled

Solution

The energy equation in photoelectric effect is given as $e V_0=h v-\phi_0 \Rightarrow V_0=\frac{h v}{e}-\frac{\phi_0}{e}$ $\ldots$ (i) When the frequency is doubled, $V_0^{\prime}=\frac{2 h v}{e}-\frac{\phi_0}{e} \Rightarrow V_0^{\prime}=2\left(\frac{h v}{e}-\frac{\phi_0}{e}\right)+\frac{\phi_0}{e}$ $V_0^{\prime}=2 V_0+\frac{\phi_0}{e} \quad$ [using Eq. (i)] $\Rightarrow \quad V_0^{\prime}>2 V_0$ Thus, the stopping potential will become more than double.

Hence, the correct answer is option: 'B'.

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