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 stated as:

\[ KE_{\text{max}} = hu - \phi \]

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

The maximum kinetic energy can also be expressed using the stopping potential \( V_0 \) as:

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

where \( e \) is the elementary charge.

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

\[ eV_0 = hu - \phi \]

This equation can be rearranged to:

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

When compared to the standard linear equation \( y = mx + c \), the slope \( m \) is identified as:

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

Therefore, the charge of an electron \( e \) can be determined as:

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

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