Ncert Exercises & Solutions

Questions 11.6: In an experiment on the photoelectric effect, the slope of the cut-off voltage versus the frequency of incident light is found to be 4.12 x 10^-15 V s. Calculate the value of Planck's constant.

Hint: The slope of the cut-off voltage (V) versus frequency ($\nu$) of incident light is given as:

$\frac{V}{\nu }$

Step 1: Relate cut-off voltage and frequency.
V is related to frequency by the equation h$\nu$ = eV
Where, e = Charge on an electron = 1.6 x $10^{- 19}$ C and h = Planck's constant

Step 2: Find the value of Planck's constant.
h = e x $\frac{V}{\nu }$= 1.6 x $10^{- 19}$ x 4.12 x $10^{- 15}$ = 6.592 x $10^{- 34}$ Js
Hence, the value of Plank's constant is 6.592 x $10^{- 34}$ Js.

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