Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength \(λ\) and \((λ/3)\). If the maximum kinetic energy of the emitted photo electrons in the second case is \(4\) times that in the first case, the work function of the surface of the material is (\(h\) = Planck's constant, \(c\) = speed of light)

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

Hint:

Apply Einstein's equation twice and use the factor of 4 between the kinetic energies.

Answer 1

The Correct Option is B

Step 1: Difference of the Two Equations:
Subtract: $K_2-K_1=\dfrac{2hc}\lambda$. With $K_2=4K_1$, $3K_1=\dfrac{2hc}\lambda$, so $K_1=\dfrac{2hc}{3\lambda}$.

Step 2: Find the Work Function:
$\phi=\dfrac{hc}\lambda-K_1=\dfrac{hc}\lambda-\dfrac{2hc}{3\lambda}=\dfrac{hc}{3\lambda}$.

Step 3: Answer:
Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } \frac{hc}{3\lambda}} \]