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: Substitute the photon energy of the first beam:
Let $E_2 = hc/\lambda_2$ be the photon energy of the second beam, so $K_2 = E_2 - \phi$. The first beam has $E_1 = E_2/3$.

Step 2: Compute K1:
$K_1 = E_1 - \phi = \dfrac{E_2}{3} - \phi = \dfrac{K_2 + \phi}{3} - \phi$.

Step 3: Compare with K2/3:
$K_1 - \dfrac{K_2}{3} = \dfrac{\phi}{3} - \phi = -\dfrac{2\phi}{3}$, which is negative.
So $K_1$ is less than one third of $K_2$.

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