Question

If light waves of wavelengths \(\lambda\) and \(\lambda/3\) are incident on the surface of a material, photoelectrons are emitted with maximum kinetic energy E and 4E respectively, then the work function of the material is

• A. \(\frac{3hc}{4\lambda}\)
• B. \(\frac{hc}{3\lambda}\)
• C. \(\frac{hc}{\lambda}\)
• D. \(\frac{hc}{2\lambda}\)
• E. \(\frac{2hc}{\lambda}\)

Hint:

When two sets of data for photoelectric effect are given, subtract the equations to eliminate the unknown kinetic energy and solve for work function.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Using Einstein's photoelectric equation: \(E_k = \frac{hc}{\lambda} - \phi\).

Step 2: Detailed Explanation:
For wavelength \(\lambda\), \(E = \frac{hc}{\lambda} - \phi\) (1)
For wavelength \(\lambda/3\), \(4E = \frac{hc}{\lambda/3} - \phi = \frac{3hc}{\lambda} - \phi\) (2)
Subtract (1) from (2): \(3E = \frac{2hc}{\lambda} \Rightarrow E = \frac{2hc}{3\lambda}\).
Substitute \(E\) in (1): \(\frac{2hc}{3\lambda} = \frac{hc}{\lambda} - \phi \Rightarrow \phi = \frac{hc}{\lambda} - \frac{2hc}{3\lambda} = \frac{hc}{3\lambda}\).

Step 3: Final Answer:
The work function is \(\frac{hc}{3\lambda}\).