91Ó°ÊÓ

To find the momentum, we need to first find the velocity of the electron. We know the kinetic energy formula is: \(K.E. = \frac{1}{2} mv^2\) Where \(K.E.\) is the kinetic energy, \(m\) is the mass of the electron and \(v\) is its velocity. We can rewrite the kinetic energy formula and solve for the momentum (\(p = mv\)) as follows: \(v = \sqrt{\frac{2 \times K.E.}{m}}\) \(p = m \times \sqrt{\frac{2 \times K.E.}{m}}\) Now let's plug in the values for the mass of an electron (\(9.11 \times 10^{-31} \mathrm{kg}\)) and the two given kinetic energies.

Using the formula derived in Step 1, we can now calculate the momentum for the electron with a kinetic energy of \(1.0 \mathrm{eV}\) (note that \(1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\)): \(p_{1.0eV} = (9.11 \times 10^{-31} \mathrm{kg}) \times \sqrt{\dfrac{2 \times (1.0)(1.6 \times 10^{-19} \mathrm{J})}{9.11 \times 10^{-31} \mathrm{kg}}}\) \(p_{1.0eV} = 5.39 \times 10^{-25} \mathrm{kg\,m/s}\)

Similarly, for the \(1.0 \mathrm{keV}\) electron, we have: \(p_{1.0keV} = (9.11 \times 10^{-31} \mathrm{kg}) \times \sqrt{\dfrac{2 \times (1.0)(1.6 \times 10^{-16} \mathrm{J})}{9.11 \times 10^{-31} \mathrm{kg}}}\) \(p_{1.0keV} = 5.39 \times 10^{-22} \mathrm{kg\,m/s}\)

Now that we have the momenta for both electrons, we can calculate their de Broglie wavelengths using the formula: \(\lambda = \dfrac{h}{p}\) For the \(1.0 \mathrm{eV}\) electron: \(\lambda_{1.0eV} = \dfrac{6.626 \times 10^{-34} \mathrm{Js}}{5.39 \times 10^{-25} \mathrm{kg\,m/s}}\) \(\lambda_{1.0eV} = 1.23 \times 10^{-9} \mathrm{m}\) For the \(1.0 \mathrm{keV}\) electron: \(\lambda_{1.0keV} = \dfrac{6.626 \times 10^{-34} \mathrm{Js}}{5.39 \times 10^{-22} \mathrm{kg\,m/s}}\) \(\lambda_{1.0keV} = 1.23 \times 10^{-12} \mathrm{m}\) So, the de Broglie wavelengths for the electrons with kinetic energies of \(1.0 \mathrm{eV}\) and \(1.0 \mathrm{keV}\) are \(1.23 \times 10^{-9} \mathrm{m}\) and \(1.23 \times 10^{-12} \mathrm{m}\), respectively.