Question

Calculate: (i) the momentum and (ii) de Broglie wavelength, of an electron with kinetic energy of 80 eV.

Hint:

The momentum of an electron can be found from its kinetic energy using the relation \( p = \sqrt{2mE_k} \), and the de Broglie wavelength is \( \lambda = \frac{h}{p} \).

Answer 1

% Option (i) The momentum of an electron can be found using the kinetic energy formula: \[ E_k = \frac{p^2}{2m} \] where \( p \) is the momentum and \( m \) is the mass of the electron. Rearranging the formula to solve for momentum: \[ p = \sqrt{2mE_k} \] The kinetic energy \( E_k = 80 \, \text{eV} \). To use SI units, we need to convert this to joules: \[ E_k = 80 \times 1.602 \times 10^{-19} \, \text{J} = 1.2816 \times 10^{-17} \, \text{J} \] The mass of the electron is \( m = 9.11 \times 10^{-31} \, \text{kg} \). Now, substituting the values: \[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.2816 \times 10^{-17}} = 1.13 \times 10^{-24} \, \text{kg m/s} \] Thus, the momentum is \( 1.13 \times 10^{-24} \, \text{kg m/s} \). % Option (ii) The de Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \] where \( h = 6.626 \times 10^{-34} \, \text{J s} \) is Planck's constant. Substituting the values: \[ \lambda = \frac{6.626 \times 10^{-34}}{1.13 \times 10^{-24}} = 5.87 \times 10^{-10} \, \text{m} \] Thus, the de Broglie wavelength is \( 5.87 \times 10^{-10} \, \text{m} \).