Question

The wavelength of light incident on a metal surface of work function 2.14 eV is 5000 \(\overset{\text{\small o}}{\text{A}}\). Find the maximum kinetic energy and the maximum velocity of the emitted photoelectrons.

Hint:

The maximum kinetic energy of emitted photoelectrons is the difference between the energy of the incident photon and the work function of the metal.

Answer 1

Step 1: Photoelectric Equation.
The energy of the incident photon is given by: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] where: - \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck’s constant, - \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light, - \( \lambda = 5000 \, \overset{\text{\small o}}{\text{A}} = 5000 \times 10^{-10} \, \text{m} \) is the wavelength of light. \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = 3.976 \times 10^{-19} \, \text{J} \]
Step 2: Maximum Kinetic Energy of Photoelectrons.
The maximum kinetic energy of the emitted photoelectrons is given by: \[ K_{\text{max}} = E_{\text{photon}} - \phi \] where \( \phi = 2.14 \, \text{eV} = 2.14 \times 1.602 \times 10^{-19} \, \text{J} = 3.429 \times 10^{-19} \, \text{J} \) is the work function. \[ K_{\text{max}} = 3.976 \times 10^{-19} - 3.429 \times 10^{-19} = 5.47 \times 10^{-20} \, \text{J} \]
Step 3: Maximum Velocity of Photoelectrons.
The maximum kinetic energy of the photoelectrons is related to their velocity by the equation: \[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \] where \( m = 9.11 \times 10^{-31} \, \text{kg} \) is the mass of the electron. Solving for \( v_{\text{max}} \): \[ v_{\text{max}} = \sqrt{\frac{2 K_{\text{max}}}{m}} = \sqrt{\frac{2 \times 5.47 \times 10^{-20}}{9.11 \times 10^{-31}}} = 3.57 \times 10^5 \, \text{m/s} \]