Question

Find the maximum magnitude of velocity and linear momentum of a photoelectron emitted when light of wavelength 4000 Ã… falls on a metal having work function 2.5 eV. 
 

Hint:

The kinetic energy of a photoelectron is the energy of the photon minus the work function. The maximum velocity and linear momentum of the photoelectron can be determined using this energy.

Answer 1

Given:
- Wavelength \( \lambda = 4000 \, \text{Ã…} = 4 \times 10^{-7} \, \text{m} \)
- Work function \( \phi = 2.5 \, \text{eV} = 2.5 \times 1.6 \times 10^{-19} \, \text{J} \)
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \) The energy of the incoming photon is: \[ E_{\text{photon}} = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \, \text{J}. \] The kinetic energy of the emitted photoelectron is: \[ E_k = E_{\text{photon}} - \phi = 4.97 \times 10^{-19} - 2.5 \times 1.6 \times 10^{-19} = 2.57 \times 10^{-19} \, \text{J}. \] The maximum velocity of the photoelectron is given by: \[ v = \sqrt{\frac{2E_k}{m_e}} = \sqrt{\frac{2 \times 2.57 \times 10^{-19}}{9.11 \times 10^{-31}}} = 7.52 \times 10^5 \, \text{m/s}. \] The linear momentum \( p \) of the photoelectron is: \[ p = m_e v = 9.11 \times 10^{-31} \times 7.52 \times 10^5 = 6.85 \times 10^{-25} \, \text{kg} \cdot \text{m/s}. \]