Q138 / 180 · NEET 2013 (Karnataka) · Physics

A source of light is placed at a distance of $50 \mathrm{~cm}$ from a photo cells and the stopping potential is found to be $\mathrm{V}_{0^*}$. If the distance between the light source and photo cells is made $25 \mathrm{~cm}$, the new stopping potential will be:

  • A
    $V_0 / 2$
  • Correct answer
    B
    $V_0$
  • C
    $4 V_0$
  • D
    $2 V_0$

Solution

Stopping potential is independent of distance hence new stopping potential will remain unchanged in required. Stopping potential $=V_0$

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

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