A student draws a graph between stopping potential \( V_0 \) (y-axis) and the reciprocal of the wavelength \( \left(\displaystyle\frac{1}{\lambda}\right) \) of the incident radiation on the x-axis. What is the slope of the graph?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{hc}{e} \)
By Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons is \( eV_0 \), equal to the photon energy minus the work function \( \phi_0 \).
\( \displaystyle eV_0 = \frac{hc}{\lambda} - \phi_0 \)
\( \displaystyle V_0 = \left(\frac{hc}{e}\right)\frac{1}{\lambda} - \frac{\phi_0}{e} \)
This is a straight line of the form \( y = mx + c \) with \( x = 1/\lambda \), so the slope is \( hc/e \).