Advertisement Remove all ads

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. - Physics

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.

Advertisement Remove all ads

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`

Concept: Einstein’S Photoelectric Equation: Energy Quantum of Radiation
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 11 Dual Nature of Radiation and Matter
Q 6 | Page 407
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×