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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: |
equal to the slope e times the slope of cut off voltage versus frequency $e^2$ times the slope (1/e) times the slope |
e times the slope of cut off voltage versus frequency |
The correct answer is Option (2) → e times the slope of cut off voltage versus frequency |
