(i) The momentum of an electron is determined using its kinetic energy. The formula for kinetic energy is given by: \[ E_k = \frac{p^2}{2m} \] where \( p \) represents momentum and \( m \) is the electron's mass. Rearranging to solve for momentum yields: \[ p = \sqrt{2mE_k} \] Given the kinetic energy \( E_k = 80 \, \text{eV} \), we convert this to joules for SI unit consistency: \[ E_k = 80 \times 1.602 \times 10^{-19} \, \text{J} = 1.2816 \times 10^{-17} \, \text{J} \] The electron's mass is \( m = 9.11 \times 10^{-31} \, \text{kg} \). Substituting these values into the momentum equation: \[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.2816 \times 10^{-17}} = 1.13 \times 10^{-24} \, \text{kg m/s} \] Therefore, the electron's momentum is \( 1.13 \times 10^{-24} \, \text{kg m/s} \).
(ii) The de Broglie wavelength \( \lambda \) is calculated using the formula: \[ \lambda = \frac{h}{p} \] where \( h = 6.626 \times 10^{-34} \, \text{J s} \) is Planck's constant. Substituting the momentum value previously calculated: \[ \lambda = \frac{6.626 \times 10^{-34}}{1.13 \times 10^{-24}} = 5.87 \times 10^{-10} \, \text{m} \] Consequently, the de Broglie wavelength is \( 5.87 \times 10^{-10} \, \text{m} \).