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Chapter 6: Problem 111

(a) What is the energy of an electron whose de Broglie wavelength is that of a photon of yellow light with wavelength 590 nm? (b) What is the de Broglie wavelength of an electron whose energy is that of the photon of yellow light?

Short Answer

Expert verified
(a) The energy of an electron with the same de Broglie wavelength as a photon of yellow light with wavelength 590 nm is approximately \(6.965 \times 10^{-19} \, \text{J}\). (b) The de Broglie wavelength of an electron whose energy is that of the photon of yellow light is approximately \(1.710 \times 10^{-10} \, \text{m}\).

Step by step solution

01

Find the de Broglie wavelength of the electron

We are given the de Broglie wavelength (\(\lambda\)) of the electron as the same as that of yellow light, which is 590 nm. Since the de Broglie wavelength is given in nm, let's convert it to meters: \(\lambda = 590 \times 10^{-9} \, \text{m}\)
02

Use the de Broglie wavelength formula to find the momentum of the electron

The de Broglie wavelength formula relates the wavelength (\(\lambda\)) of a particle to its momentum (\(p\)): \(\lambda = \dfrac{h}{p}\) where \(h\) is the Planck constant \(h = 6.626 \times 10^{-34} \, \text{Js}\). We can rearrange the formula to solve for the momentum of the electron: \(p = \dfrac{h}{\lambda}\) Plug in the given values: \(p = \dfrac{6.626 \times 10^{-34} \, \text{Js}}{590 \times 10^{-9} \, \text{m}}\) Calculate the momentum: \(p \approx 1.123 \times 10^{-24} \, \text{kg m/s}\)
03

Use the kinetic energy formula to find the energy of the electron

The kinetic energy formula for a particle with momentum (\(p\)) is: \(E = \dfrac{p^2}{2m}\) where \(E\) is the energy and \(m\) is the mass of the particle. Since we have an electron, the mass of the electron is \(m = 9.109 \times 10^{-31} \, \text{kg}\). Plug in the values: \(E = \dfrac{(1.123 \times 10^{-24} \, \text{kg m/s})^2}{2(9.109 \times 10^{-31} \, \text{kg})}\) Calculate the energy: \(E \approx 6.965 \times 10^{-19} \, \text{J}\) So the energy of the electron is approximately \(6.965 \times 10^{-19} \, \text{J}\).
04

Use the energy to find the momentum of the electron with the same energy as the photon of yellow light

We are given that the energy of the electron in part (b) is the same as the energy found in part (a) (\(E \approx 6.965 \times 10^{-19} \, \text{J}\)). We can use the kinetic energy formula to find the momentum of the electron: \(E = \dfrac{p^2}{2m}\) Rearrange the formula to solve for the momentum (\(p\)): \(p = \sqrt{2mE}\) Plug in the given values: \(p = \sqrt{2(9.109 \times 10^{-31} \, \text{kg})(6.965 \times 10^{-19} \, \text{J})}\) Calculate the momentum: \(p \approx 3.873 \times 10^{-24} \, \text{kg m/s}\)
05

Use the de Broglie wavelength formula to find the de Broglie wavelength of the electron

Now that we have the momentum of the electron, we can use the de Broglie wavelength formula to find the de Broglie wavelength of the electron: \(\lambda = \dfrac{h}{p}\) Plug in the values we calculated: \(\lambda = \dfrac{6.626 \times 10^{-34} \, \text{Js}}{3.873 \times 10^{-24} \, \text{kg m/s}}\) Calculate the de Broglie wavelength: \(\lambda \approx 1.710 \times 10^{-10} \, \text{m}\) So the de Broglie wavelength of an electron whose energy is that of the photon of yellow light is approximately \(1.710 \times 10^{-10} \, \text{m}\).

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