Question

In his experiment on photoelectric effect, Robert A. Millikan found the slope of the cut-off voltage versus frequency of incident light plot to be \( 4.12 \times 10^{-15} \, \text{Vs} \). Calculate the value of Planck’s constant from it.

Hint:

The slope of the cut-off voltage versus frequency plot provides a direct way to calculate Planck's constant using the equation \( eV_{\text{cut}} = h f - \phi \).

Answer 1

Robert A. Millikan's photoelectric effect experiment yielded a slope of \( 4.12 \times 10^{-15} \, \text{Vs} \) for the cut-off voltage versus frequency plot. This allows us to determine Planck's constant (\( h \)).

1. Photoelectric Effect Equation:

The photoelectric effect is governed by:

\[ eV_{\text{cut-off}} = h u - \phi \] where:

• \( e \) is the elementary charge ( \( 1.6 \times 10^{-19} \, \text{C} \) ),
• \( V_{\text{cut-off}} \) is the cut-off voltage,
• \( h \) is Planck’s constant,
• \( u \) is the incident light frequency, and
• \( \phi \) is the material's work function.

2. Slope-Planck's Constant Relation:

The plot of \( V_{\text{cut-off}} \) against \( u \) is linear. Rearranging the photoelectric equation yields:

\[ V_{\text{cut-off}} = \frac{h}{e} u - \frac{\phi}{e} \]

This is in the form \( y = mx + c \), with:

• \( y = V_{\text{cut-off}} \),
• \( x = u \),
• \( m = \frac{h}{e} \) (the slope), and
• \( c = -\frac{\phi}{e} \) (the y-intercept).

Therefore, the slope \( m = \frac{h}{e} \), enabling Planck's constant calculation from the given slope.

 

3. Planck's Constant Calculation:

Given slope \( m = 4.12 \times 10^{-15} \, \text{Vs} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \). Using:

\[ \frac{h}{e} = 4.12 \times 10^{-15} \, \text{Vs} \]

Solving for \( h \):

\[ h = 4.12 \times 10^{-15} \times 1.6 \times 10^{-19} \, \text{Js} \]

4. Calculation Performed:

\[ h = 6.592 \times 10^{-34} \, \text{Js} \]

5. Final Result:

Planck's constant \( h \) is approximately \({6.59 \times 10^{-34}} \, \text{Js}\).