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

The photoelectric equation is given by: 

\[ KE_{\text{max}} = h\nu - \phi \]

where \( KE_{\text{max}} \) is the maximum kinetic energy of emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

In terms of stopping potential \( V_0 \), the maximum kinetic energy can also be expressed as:

\[ KE_{\text{max}} = eV_0 \]

where \( e \) is the charge of the electron.

Equating the two expressions for \( KE_{\text{max}} \):

\[ eV_0 = h\nu - \phi \]

This can be rearranged to:

\[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]

Comparing with the equation of a straight line \( y = mx + c \), the slope \( m \) is:

\[ m = \frac{h}{e} \]

Thus, the charge of an electron \( e \) is:

\[ e = \frac{h}{m} \]

Hence, the correct answer is \( \frac{h}{m} \).