In a photoelectric experiment, the slope of the graph drawn between stopping potential along y-axis and frequency of incident radiation along x-axis is (Planck's constant = $6.6\times 10^{-34}$ Js)
Show Hint
The slope of the graph of stopping potential ($V_s$) versus incident frequency ($\nu$) in the photoelectric effect experiment is a universal constant equal to $h/e$. This is a key result that allows for the experimental determination of Planck's constant.
Step 1: State Einstein's Photoelectric Equation.
The maximum kinetic energy ($KE_{max}$) of the emitted photoelectron is related to the energy of the incident photon ($h\nu$) and the work function ($\phi_0$) of the metal by:
\[
KE_{max} = h\nu - \phi_0.
\] Step 2: Relate kinetic energy to the stopping potential.
The maximum kinetic energy is also related to the stopping potential ($V_s$) by the equation:
\[
KE_{max} = eV_s.
\]
where $e$ is the elementary charge. Step 3: Combine the equations and find the relation between $V_s$ and $\nu$.
Substituting $KE_{max} = eV_s$ into the photoelectric equation:
\[
eV_s = h\nu - \phi_0.
\]
Solving for the stopping potential $V_s$:
\[
V_s = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}.
\] Step 4: Determine the slope of the $V_s$ versus $\nu$ graph.
The graph is plotted with stopping potential ($V_s$) on the y-axis and frequency ($\nu$) on the x-axis.
Comparing the equation from Step 3 to the equation of a straight line, $y = m x + c$, the slope ($m$) of the graph is:
\[
\text{Slope} = \frac{h}{e}.
\] Step 5: Calculate the numerical value of the slope.
Given Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$.
The elementary charge is $e = 1.6 \times 10^{-19} \text{ C}$.
\[
\text{Slope} = \frac{6.6 \times 10^{-34} \text{ Js}}{1.6 \times 10^{-19} \text{ C}} = \left(\frac{6.6}{1.6}\right) \times 10^{(-34 - (-19))} \text{ JsC}^{-1}.
\]
\[
\text{Slope} = 4.125 \times 10^{-15} \text{ JsC}^{-1}.
\]
Was this answer helpful?
0
Top Questions on Dual nature of radiation and matter