Question

Two light rays of wavelength \(\lambda\) and \(\frac{\lambda}{4}\) incident on the surface of a photo sensitive material emit electrons with max kinetic energy \(E\) and \(6E\) respectively. The work function of the material is (\(h\) = Planck’s constant, \(c\) = velocity of light in free space)

• A. \(\frac{hc}{\lambda}\)
• B. \(\frac{hc}{2\lambda}\)
• C. \(\frac{2hc}{5\lambda}\)
• D. \(\frac{3hc}{5\lambda}\)

Hint:

Always solve for the kinetic energy term first and then eliminate it to find the work function. This avoids dealing with multiple fractions for \(\phi\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the energy of incident photons to the work function of the metal and the maximum kinetic energy of the emitted photoelectrons.

Step 2: Key Formula or Approach:
\(\frac{hc}{\lambda} = \phi + K_{max}\), where \(\phi\) is the work function.

Step 3: Detailed Explanation:
1. For wavelength \(\lambda\):
\[ \frac{hc}{\lambda} = \phi + E \implies E = \frac{hc}{\lambda} - \phi \quad \text{--- (i)} \]
2. For wavelength \(\lambda/4\):
\[ \frac{hc}{\lambda/4} = \phi + 6E \implies \frac{4hc}{\lambda} = \phi + 6E \quad \text{--- (ii)} \]
Substitute the value of \(E\) from (i) into (ii):
\[ \frac{4hc}{\lambda} = \phi + 6\left( \frac{hc}{\lambda} - \phi \right) \]
\[ \frac{4hc}{\lambda} = \phi + \frac{6hc}{\lambda} - 6\phi \]
Rearrange to solve for \(\phi\):
\[ 5\phi = \frac{6hc}{\lambda} - \frac{4hc}{\lambda} \]
\[ 5\phi = \frac{2hc}{\lambda} \implies \phi = \frac{2hc}{5\lambda} \]

Step 4: Final Answer:
The work function of the material is \(\frac{2hc}{5\lambda}\).