Question

Two monochromatic sources emit light at wavelengths \(\lambda\) and \(\lambda/2\). The stopping potentials for a photosensitive material using these two sources are found to be \(1\ \text{V}\) and \(3\ \text{V}\), respectively. What is the work function of the material?

• A. \(1\ \text{eV}\)
• B. \(2\ \text{eV}\)
• C. \(1.5\ \text{eV}\)
• D. \(1.25\ \text{eV}\)

Hint:

Halving the wavelength doubles the energy of the incident photons.
Let \(E\) be the initial photon energy.
We have \(E - \phi = 1\ \text{eV}\) and \(2E - \phi = 3\ \text{eV}\).
Subtracting twice the first equation from the second equation directly yields \(\phi = 1\ \text{eV}\) with minimal calculation.

Answer 1

The Correct Option is A

This problem involves understanding the photoelectric effect and how it relates to the stopping potential and work function. Let's solve this step-by-step.

The photoelectric effect equation is given by the formula:

\(E_k = h\nu - \Phi\)

where:

• \(E_k\) is the kinetic energy of the emitted electron,
• \(h\) is Planck's constant,
• \(\nu\) is the frequency of the incident light,
• \(\Phi\) is the work function of the material.

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

\(E_k = eV_0\)

Given Sources:

1. First source with wavelength \(\lambda\) and stopping potential \(V_0 = 1 \, \text{V}\).
2. Second source with wavelength \(\lambda/2\) and stopping potential \(V_0 = 3 \, \text{V}\).

First, convert wavelength to frequency:

\(\nu = \frac{c}{\lambda}\) and \(\nu' = \frac{c}{\lambda/2} = \frac{2c}{\lambda}\)

Now, using the photoelectric equation for each source:

• For wavelength \(\lambda\):
• For wavelength \(\lambda/2\):

We can set up the two equations:

1. \(e \times 1 = \frac{hc}{\lambda} - \Phi\)
2. \(e \times 3 = \frac{2hc}{\lambda} - \Phi\)

Subtract the first equation from the second:

\(e \times 3 - e \times 1 = \frac{2hc}{\lambda} - \Phi - \left( \frac{hc}{\lambda} - \Phi \right)\)

This simplifies to:

\(2e = \frac{hc}{\lambda}\)

Substituting back into one of the original equations, we find:

\(e = \frac{hc}{\lambda} - \Phi\)

Solving for \(\Phi\):

\(\Phi = \frac{hc}{\lambda} - e = 2e - e = 1 \, \text{eV}\)

Thus, the work function of the material is \(1 \, \text{eV}\).

The correct answer is: \(1 \, \text{eV}\)