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 :
- A equal to the slope
- B e times the slope of cut off voltage versus frequency
- C \(e ^2\) times the slope
- D (1/e) times the slope
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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