Questions: b. What is the de Broglie wavelength of the electron striking the Cu surface?

Solution Steps

The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \(p\) is the momentum of the particle.

The momentum \(p\) of an electron can be calculated using its mass \(m\) and velocity \(v\): \[ p = mv \] The mass of an electron \(m\) is \(9.109 \times 10^{-31} \, \text{kg}\). If the velocity \(v\) of the electron is given, we can directly calculate the momentum. If not, we need to find the velocity using the kinetic energy.

If the kinetic energy \(E_k\) of the electron is known, we can find the velocity using the relation: \[ E_k = \frac{1}{2}mv^2 \] Solving for \(v\): \[ v = \sqrt{\frac{2E_k}{m}} \]

Assuming we have the kinetic energy \(E_k\) of the electron, we can find the velocity and then the momentum. Finally, we substitute the momentum into the de Broglie wavelength formula.

For example, if the kinetic energy \(E_k\) of the electron is \(1.602 \times 10^{-19} \, \text{J}\) (which corresponds to 1 eV): \[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} = 5.931 \times 10^5 \, \text{m/s} \] \[ p = (9.109 \times 10^{-31} \, \text{kg}) \times (5.931 \times 10^5 \, \text{m/s}) = 5.400 \times 10^{-25} \, \text{kg m/s} \] \[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{5.400 \times 10^{-25} \, \text{kg m/s}} = 1.227 \times 10^{-9} \, \text{m} = 1.227 \, \text{nm} \]

Final Answer