Question

The maximum kinetic energies of photoelectrons emitted are \(K_1\) and \(K_2\) when light of wavelength \(λ_1\) and \(λ_2\) respectively are incident on a metallic surface. If \(λ_1 = 3λ_2\) then

• A. \(K_1 = \frac{K_2}{3}\)
• B. \(K_1 < \frac{K_2}{3}\)
• C. \(K_1 = 3K_2\)
• D. \(K_1 = \frac{2}{3}K_2\)

Hint:

Write Einstein's equation for both wavelengths and eliminate the work function.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Einstein's photoelectric equation gives the maximum kinetic energy of the emitted electrons as the photon energy minus the work function \(\phi\).

Step 2: Key Formula or Approach:
\[ K = \frac{hc}{\lambda} - \phi \]

Step 3: Detailed Explanation:
For the two wavelengths:
\[ K_1 = \frac{hc}{\lambda_1} - \phi, \qquad K_2 = \frac{hc}{\lambda_2} - \phi \]
With \(\lambda_1 = 3\lambda_2\):
\[ K_1 = \frac{hc}{3\lambda_2} - \phi = \frac13\left(K_2 + \phi\right) - \phi \]
\[ K_1 = \frac{K_2}{3} + \frac\phi3 - \phi = \frac{K_2}{3} - \frac{2\phi}{3} \]
Since the work function is positive, \(\dfrac{2\phi}{3} > 0\), so
\[ K_1 < \frac{K_2}{3} \]
Option (A) \(K_1 = K_2/3\) would be true only for zero work function. Option (C) and (D) do not hold, because \(K_1\) is smaller than \(K_2\) (longer wavelength, lower photon energy) by more than a factor of 3.

Final Answer:
\(K_1 < \dfrac{K_2}{3}\), option (B). \[ \boxed{K_1<\frac{K_2}{3} \text{ (B)}} \]