Question

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)

• A. $2.42\times 10^{15}$ JsC$^{-1}$
• B. $10.56\times 10^{-15}$ JsC$^{-1}$
• C. $4.125\times 10^{-15}$ JsC$^{-1}$
• D. $6.25\times 10^{-20}$ JsC$^{-1}$

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.

Answer 1

The Correct Option is C

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 $e = 1.6 \times 10^{-19} \text{ C}$. \[ \text{Slope} = \frac{6.6 \times 10^{-34} \text{ Js}}{1.6 \times 10^{-19} \text{ C}} = \left(\frac{6.6}{1.6}\right) \times 10^{(-34 - (-19))} \text{ JsC}^{-1}. \] \[ \text{Slope} = 4.125 \times 10^{-15} \text{ JsC}^{-1}. \]