Question

The stopping potential for photoelectric emission from a metal surface is plotted along Y-axis and frequency v of incident light along X-axis. A straight line is obtained as shown. Planck's constant is given by}

• A. slope of the line
• B. product of slope of the line and charge on the electron
• C. intercept along Y-axis divided by charge on the electron
• D. product of intercept along X-axis and mass of the electron

Hint:

The slope of the $V_s$ vs $\nu$ graph gives $h/e$.

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relation between stopping potential, frequency of incident light, and Planck's constant. The photoelectric effect is described by Einstein's photoelectric equation:

\(K.E._\text{max} = h v - \phi\),

where:

• \(K.E._\text{max}\) is the maximum kinetic energy of the emitted electrons.
• \(h\) is Planck's constant.
• \(v\) is the frequency of the incident light.
• \(\phi\) is the work function of the metal.

The stopping potential \(V_0\) is related to the maximum kinetic energy by:

\(e V_0 = K.E._\text{max}\),

where \(e\) is the charge of the electron. Substituting \(K.E._\text{max}\) from Einstein's equation:

\(e V_0 = h v - \phi\)

Rearranging, we get:

\(V_0 = \frac{h}{e} v - \frac{\phi}{e}\)

This is in the form of a straight line \(y = mx + c\), where:

• \(y = V_0\) (stopping potential).
• \(x = v\) (frequency).
• \(m = \frac{h}{e}\) (slope of the line).
• \(c = - \frac{\phi}{e}\) (Y-intercept).

Thus, Planck's constant \(h\) is given by:

\(h = e \times (\text{slope of the line})\)

Therefore, the correct answer is: product of slope of the line and charge on the electron.