Question

Two photons of wavelength \( \lambda \) and \(2\lambda\) are incident on a metal surface and emit photoelectrons of maximum kinetic energies \(3k\) and \(k\). Find work function of the metal.

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

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Einstein's Photoelectric Equation states that the energy of the incident photon equals the work function plus the maximum kinetic energy of the emitted electron.
Step 2: Key Formula or Approach:
The photoelectric equation is $E = W + K_{max}$.
Substituting photon energy, we get $\frac{hc}{\lambda} = W + K$.
Step 3: Detailed Explanation:
For the first case (incident wavelength $\lambda$):
\[ \frac{hc}{\lambda} = W + 3k \quad \dots(1) \]
For the second case (incident wavelength $2\lambda$):
\[ \frac{hc}{2\lambda} = W + k \quad \dots(2) \]
We need to eliminate $k$ to find $W$. Multiply equation (2) by 3:
\[ \frac{3hc}{2\lambda} = 3W + 3k \quad \dots(3) \]
Now, subtract equation (1) from equation (3):
\[ \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = (3W + 3k) - (W + 3k) \]
\[ \frac{3hc - 2hc}{2\lambda} = 2W \]
\[ \frac{hc}{2\lambda} = 2W \implies W = \frac{hc}{4\lambda} \]
Step 4: Final Answer:
The work function is $\frac{hc}{4\lambda}$.