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.