In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.
The ratio between the cut-off voltage and the frequency of light is given as: frequency of lightCut off voltage=νV=4.12×10−15Vs Now, the energy of the photoeletron is given as: E=hν=eV Therefore, h=e\dfrac{V}{\nu} ⇒1.6×10−19×4.12×10−15 So, h=6.592×10−34Js−1
