Question

In a photoelectric experiment the stopping potential V is plotted against applied frequency v for three different materials A, B and C as shown. Which of the following materials has largest threshold wavelength ?

• A. A
• B. B
• C. C
• D. All have same

Answer 1

Answer: (C)

C

Answer 2

The photoelectric effect is described by Einstein's equation:

$K_{max} = h\nu - \phi_0$

where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the material.

The stopping potential $V_s$ is related to the maximum kinetic energy by:

$eV_s = K_{max}$

where $e$ is the charge of an electron.

Combining these equations, we get:

$eV_s = h\nu - \phi_0$

Dividing by $e$, we express the stopping potential as a function of frequency:

$V_s = \left(\frac{h}{e}\right)\nu - \left(\frac{\phi_0}{e}\right)$

This equation represents a straight line in a $V_s$ vs $\nu$ graph, with:

• Slope $m = \frac{h}{e}$ (which is a universal constant for all materials).
• Y-intercept $c = -\frac{\phi_0}{e}$.

The threshold frequency ($\nu_0$) is the minimum frequency of incident light required to eject electrons, at which $K_{max} = 0$ and thus $V_s = 0$. Setting $V_s = 0$ in the equation:

$0 = \left(\frac{h}{e}\right)\nu_0 - \left(\frac{\phi_0}{e}\right)$ $\left(\frac{h}{e}\right)\nu_0 = \left(\frac{\phi_0}{e}\right)$ $h\nu_0 = \phi_0$

This means the threshold frequency $\nu_0$ is the x-intercept (the point where the line crosses the $\nu$-axis, i.e., $V_s = 0$) of the graph.

From the given graph:

• The lines for materials A, B, and C are parallel, indicating they have the same slope $\frac{h}{e}$, which is consistent.
• We need to find the threshold frequencies for each material by observing their x-intercepts.
  • For material C, the line intersects the $\nu$-axis at the smallest frequency. So, $\nu_{0C}$ is the smallest.
  • For material B, the line intersects at an intermediate frequency. So, $\nu_{0B}$ is intermediate.
  • For material A, the line intersects at the largest frequency. So, $\nu_{0A}$ is the largest.

Therefore, we have the order of threshold frequencies:

$\nu_{0C} < \nu_{0B} < \nu_{0A}$

The threshold wavelength ($\lambda_0$) is related to the threshold frequency by:

$\lambda_0 = \frac{c}{\nu_0}$

where $c$ is the speed of light.

This inverse relationship means that a smaller threshold frequency corresponds to a larger threshold wavelength. Since $\nu_{0C}$ is the smallest threshold frequency, material C will have the largest threshold wavelength.

Thus, the order of threshold wavelengths is:

$\lambda_{0C} > \lambda_{0B} > \lambda_{0A}$

Material C has the largest threshold wavelength.