Question

Calculate the wavelength of a) a \(2.00-\mathrm{eV}\) photon, and b) an electron with a kinetic energy of \(2.00 \mathrm{eV}\).

Short answer

Answer: The wavelength of the 2.00-eV photon is 623 nm, and the wavelength of the electron with a kinetic energy of 2.00 eV is 1.45 nm.

Step-by-step solution

Calculate the wavelength of a 2.00-eV photon

We'll start by converting the photon energy from electron volts to joules, using the energy conversion factor \(1 \mathrm{eV} = 1.6 \times 10^{-19} \mathrm{J}\). This gives us: Photon energy \(E = 2.00 \, \mathrm{eV} \times \frac{1.6 \times 10^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}} = 3.2 \times 10^{-19} \mathrm{J}\) Next, using the energy-wavelength relationship for photons, we have \(E=hf\) where \(h\) is the Planck constant (\(h = 6.626 \times 10^{-34} \, \mathrm{Js}\)) and \(f\) is the frequency of the photon. We'll also use the speed of light \(c\) and the relationship \(c = f\lambda\) to find the wavelength \(\lambda\). Therefore, \(E = \frac{hc}{\lambda}\). Now we can solve for the wavelength \(\lambda\): \(\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \, \mathrm{Js} \times 3.00 \times 10^8 \, \mathrm{m/s}}{3.2 \times 10^{-19} \, \mathrm{J}} = 6.23 \times 10^{-7} \, \mathrm{m} = 623 \, \mathrm{nm}\)

Calculate the wavelength of an electron with a kinetic energy of 2.00 eV

First, we'll convert the electron kinetic energy from electron volts to joules, using the same energy conversion factor as before: Electron kinetic energy \(K = 2.00 \, \mathrm{eV} \times \frac{1.6 \times 10^{-19} \, \mathrm{J}}{1 \, \mathrm{eV}} = 3.2 \times 10^{-19} \mathrm{J}\) Now, we will use the de Broglie wavelength equation: \(\lambda = \frac{h}{p}\), where \(h\) is the Planck constant and \(p\) is the momentum of the electron. Since the electron's momentum is related to its kinetic energy by \(K = \frac{p^2}{2m}\), where \(m\) is the mass of the electron (\(m = 9.11 \times 10^{-31} \, \mathrm{kg}\)), we can express the wavelength as: \(\lambda = \frac{h}{\sqrt{2mK}} = \frac{6.626 \times 10^{-34} \, \mathrm{Js}}{\sqrt{2 \times 9.11 \times 10^{-31} \, \mathrm{kg} \times 3.2 \times 10^{-19} \, \mathrm{J}}} = 1.45 \times 10^{-9} \, \mathrm{m} = 1.45 \, \mathrm{nm}\) So the wavelength of the photon is 623 nm, and the wavelength of the electron is 1.45 nm.