Question

If Planck's constant is \( 6.63 \times 10^{-34} \) Js, then the slope of a graph drawn between cut-off voltage and frequency of incident light in a photoelectric experiment is:

• A. \( 4.14 \times 10^{-15} \) Vs
• B. \( 19.776 \times 10^{-15} \) Vs
• C. \( 2.198 \times 10^{-15} \) Vs
• D. \( 1.337 \times 10^{-15} \) Vs

Hint:

The slope of the stopping potential vs frequency graph in a photoelectric experiment gives \( h/e \).

Answer 1

The Correct Option is A

The equation for the photoelectric effect is: \[ eV_s = h f - \phi \] where: - \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \) C), - \( V_s \) is the stopping potential, - \( h \) is Planck’s constant (\( 6.63 \times 10^{-34} \) Js), - \( f \) is the frequency of incident light. The slope of the \( V_s \) vs \( f \) graph is: \[ \frac{h}{e} = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} \] \[ = 4.14 \times 10^{-15} \text{Vs} \]