Question
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 Vs. Calculate the value of Planck’s constant.
Solution
The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as:
`V/v = 4.12 xx 10^(-15) Vs`
V is related to frequency by the equation:
`hv = eV`
Where
e = Charge on an electron = 1.6 × 10−19 C
h = Planck’s constant
`:. h = e xx V/v`
`= 1.6 xx 10^(-19) xx 4.12 xx 10^(-15) = 6.592 xx 10^(-34) Js`
Therefore, the value of Planck’s constant is `6.592 xx 10^(-34) Js`
Is there an error in this question or solution?
Solution In an Experiment on 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. Concept: Einstein’S Photoelectric Equation: Energy Quantum of Radiation.
