Question

Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Hint:

Shortcut:

Convert photon energy directly to eV
\( K_{\max} = h\nu - \phi \)
Stopping potential (in volts) = kinetic energy (in eV)

Answer 1

Concept: Photoelectric equation: \[ E = h\nu = \phi_0 + K_{\max} \] Stopping potential: \[ K_{\max} = eV_0 \] Constants: \[ h = 6.63 \times 10^{-34} \, \text{J·s}, \quad 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Step 1: Energy of photon. \[ E = h\nu = 6.63 \times 10^{-34} \times 6.4 \times 10^{14} \] \[ E = 4.24 \times 10^{-19} \, \text{J} \] Convert to eV: \[ E = \frac{4.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.65 \, \text{eV} \]
Step 2: Maximum kinetic energy. \[ K_{\max} = E - \phi_0 = 2.65 - 1.96 = 0.69 \, \text{eV} \]
Step 3: Stopping potential. \[ K_{\max} = eV_0 \] Since kinetic energy is in eV: \[ V_0 = 0.69 \, \text{V} \] Final Answers:

[(a)] Energy of photon = \( 2.65 \, \text{eV} \)
[(b)] Maximum kinetic energy = \( 0.69 \, \text{eV} \)
[(c)] Stopping potential = \( 0.69 \, \text{V} \)