Question

Calculate the velocities of electrons with de Broglie wavelengths of \(1.0 \times 10^{2} \mathrm{~nm}\) and \(1.0 \mathrm{~nm}\), respectively.

Short answer

The velocity of the electron with a de Broglie wavelength of \(1.0\times 10^{2} \mathrm{~nm}\) is approximately \(7.28 \times 10^5 \mathrm{m/s}\), and the velocity of the electron with a de Broglie wavelength of \(1.0 \mathrm{~nm}\) is approximately \(7.28 \times 10^7 \mathrm{m/s}\).

Step-by-step solution

Convert nanometers to meters

To make the calculation easier, let's first convert the given de Broglie wavelengths from nanometers to meters. 1 nm = \(1 \times 10^{-9}\) m So, for the given wavelengths: - \(1.0\times 10^{2} \mathrm{~nm} = 1.0\times 10^{2} \times 10^{-9} \mathrm{m} = 1.0\times 10^{-7} \mathrm{m} \) - \(1.0 \mathrm{~nm} = 1.0\times 10^{-9} \mathrm{m}\)

Rearrange the de Broglie wavelength formula

In this step, we need to rearrenge the de Broglie wavelength formula to solve for the velocity of the electron (\(v\)): \[ v = \frac{h}{m\lambda} \]

Compute the velocities

Now we have the formula to calculate the velocity, we just need to insert the known values of mass, wavelength, and Planck's constant into the equation to get the velocities for both cases. For \(\lambda = 1.0 \times 10^{-7} \mathrm{m}\), \[ v_1 = \frac{(6.63 \times 10^{-34} \mathrm{Js})}{(9.11 \times 10^{-31} \mathrm{kg})(1.0 \times 10^{-7} \mathrm{m})} \] For \(\lambda = 1.0 \times 10^{-9} \mathrm{m}\), \[ v_2 = \frac{(6.63 \times 10^{-34} \mathrm{Js})}{(9.11 \times 10^{-31} \mathrm{kg})(1.0 \times 10^{-9} \mathrm{m})} \]

Calculate the velocities

With the given values, we can now calculate the velocities: - For \(v_1\): \[ v_1 \approx \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.0 \times 10^{-7}} \approx 7.281 \times 10^5 \mathrm{m/s} \] - For \(v_2\): \[ v_2 \approx \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.0 \times 10^{-9}} \approx 7.281 \times 10^7 \mathrm{m/s} \] The velocity of the electron with a de Broglie wavelength of \(1.0\times 10^{2} \mathrm{~nm}\) is approximately \(7.28 \times 10^5 \mathrm{m/s}\), and the velocity of the electron with a de Broglie wavelength of \(1.0 \mathrm{~nm}\) is approximately \(7.28 \times 10^7 \mathrm{m/s}\).