Question

The stopping potential \( V_0 \) measured in a photoelectric experiment for a metal surface is plotted against frequency \( \nu \) of the incident radiation. Let \( m \) be the slope of the straight line so obtained. Then the value of the charge of an electron is given by \( h \) (where \( h \) is the Planck’s constant):

• A. \( \frac{h}{m} \)
• B. \( \frac{m}{h} \)
• C. \( \frac{m}{h} \)
• D. \( \frac{m h}{1} \)

Hint:

In the photoelectric effect, the slope of the graph of stopping potential versus frequency is related to the Planck constant divided by the electron charge.

Answer 1

The Correct Option is A

In the photoelectric effect, the stopping potential \( V_0 \) is related to the frequency \( \nu \) of the incident radiation by the equation: \[ V_0 = \frac{h}{e} \nu - \phi \] where: - \( h \) is the Planck constant, - \( e \) is the charge of the electron, - \( \phi \) is the work function of the metal. From the graph of \( V_0 \) vs. \( \nu \), the slope \( m \) is given by \( \frac{h}{e} \). Thus, the charge of an electron \( e \) is: \[ e = \frac{h}{m} \] Therefore, the correct answer is \( \frac{h}{m} \).