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CUET · PHYSICS · PYQ PAPER 2023

The slope of the cut off voltage versus frequency of incident light is found to be \(4.12 \times 10^{-15} Vs\).
The value of Planck's constant is :

  1. A equal to the slope
  2. B e times the slope of cut off voltage versus frequency
  3. C \(e ^2\) times the slope
  4. D (1/e) times the slope
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Answer & Solution

Correct Answer

(B) e times the slope of cut off voltage versus frequency

Step-by-step Solution

Detailed explanation

\(eV_0 = hf - \phi\) \(V_0 = \frac{h}{e}f - \frac{\phi}{e}\) \(\text{Slope} = \frac{h}{e}\) \(h = e \times \text{Slope}\) e times the slope of cut off voltage versus frequency
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