Question

The work function $(W_0)$ of metals A, B and C is 2.25, 2.42 and 3.7 eV respectively. These metals were irradiated with light of wavelength 400 nm. Identify the metals from which photoelectrons are emitted $(h = 6.6 \times 10^{-34} Js; c = 3 \times 10^8 ms^{-1}; 1 eV = 1.6 \times 10^{-19} J)$

• A. A & B only
• B. A, B & C
• C. A & C only
• D. B & C only

Hint:

A convenient formula is \[ E(\text{eV})=\frac{1240}{\lambda(\text{nm})} \] which quickly gives photon energy in electron volts.

Answer 1

The Correct Option is A

Concept: Photoelectric emission occurs only when \[ hf \ge W_0 \] or equivalently, \[ E=\frac{hc}{\lambda} \] must be greater than the work function.

Step 1: Calculate the energy of incident photon.
\[ E=\frac{1240}{400} \] \[ E=3.1eV \]

Step 2: Compare with work functions.
For metal A: \[ W_A=2.25eV<3.1eV \] Photoelectrons are emitted. For metal B: \[ W_B=2.42eV<3.1eV \] Photoelectrons are emitted. For metal C: \[ W_C=3.7eV>3.1eV \] Photoelectrons are not emitted.

Step 3: State the answer.
\[ \boxed{\text{A and B only}} \]