Question

Consider a muonic hydrogen atom, in which the electron is replaced by a muon of mass \(105.66 \mathrm{MeV} / \mathrm{c}^{2}\) that orbits the proton. What are the first three energy levels of the muon in this type of atom?

Short answer

**Question:** Calculate the first three energy levels of a muonic hydrogen atom, which consists of a muon orbiting a proton. **Answer:** The first three energy levels of the muon in the muonic hydrogen atom are approximately: - For \(n = 1\): \(E_1 = -2.052\times10^{-11} \mathrm{J}\) - For \(n = 2\): \(E_2 = -5.130\times10^{-12} \mathrm{J}\) - For \(n = 3\): \(E_3 = -2.282\times10^{-12} \mathrm{J}\)

Step-by-step solution

Identify the constants and variables

In this exercise, we need to know the following constants: - Mass of muon, \(m_\mu = 105.66 \mathrm{MeV}/\mathrm{c}^2\) - Mass of proton, \(m_p = 938.272 \mathrm{MeV}/\mathrm{c}^2\) - Charge of electron (or muon), \(е = 1.602 \times 10^{-19} \mathrm{C}\) - Planck's constant, \(h = 6.626 \times 10^{-34} \mathrm{Js}\) - Speed of light in vacuum, \(c = 2.998 \times 10^8 \mathrm{m}/\mathrm{s}\) - Permittivity of free space, \(\varepsilon_0 = 8.854 \times 10^{-12} \mathrm{F/m}\) We need to find out the first three energy levels, \(E_n\) where \(n\) is the principal quantum number (\(n = 1, 2, 3\)).

Calculate the reduced mass of the muonic hydrogen system

To account for the fact that both the muon and proton are in motion, we will use the reduced mass of the system, denoted by \(\mu\). The reduced mass is given by: $$\mu = \frac{m_\mu m_p}{m_\mu + m_p}$$ Plugging in the values for the masses: $$\mu = \frac{(105.66)(938.272)}{(105.66 + 938.272)} \mathrm{MeV}/\mathrm{c}^2$$ Calculating the value for \(\mu\): $$\mu = 102.775 \mathrm{MeV}/\mathrm{c}^2$$

Calculate the energy levels using the Bohr formula

The energy levels of a hydrogen-like system can be given by the Bohr formula: $$E_n = -\frac{\mu e^4}{(4\pi \varepsilon_0)^2 2h^2 n^2}$$ Where \(n\) is the principal quantum number. Now, convert the reduced mass \(\mu\) from \(\mathrm{MeV}/\mathrm{c}^2\) to kg. $$\mu = 102.775\frac{\mathrm{MeV}}{\mathrm{c}^2}(1.783\mathrm{e}^{-36}\frac{\mathrm{kg}}{\mathrm{MeV}}) = 1.831\times 10^{-28} \mathrm{kg}$$ Calculate the first three energy levels (\(n = 1, 2, 3\)) using the Bohr formula and the reduced mass value: $$E_1 = -\frac{1.831\times10^{-28}(1.602\times10^{-19})^4}{(4\pi(8.854\times10^{-12})^2)2(6.626\times10^{-34})^2(1)^2} = -2.052\times10^{-11} \mathrm{J}$$ $$E_2 = -\frac{1.831\times10^{-28}(1.602\times10^{-19})^4}{(4\pi(8.854\times10^{-12})^2)2(6.626\times10^{-34})^2(2)^2} = -5.130\times10^{-12} \mathrm{J}$$ $$E_3 = -\frac{1.831\times10^{-28}(1.602\times10^{-19})^4}{(4\pi(8.854\times10^{-12})^2)2(6.626\times10^{-34})^2(3)^2} = -2.282\times10^{-12} \mathrm{J}$$

Present the final results

The first three energy levels of the muon in the muonic hydrogen atom are approximately: - For \(n = 1\): \(E_1 = -2.052\times10^{-11} \mathrm{J}\) - For \(n = 2\): \(E_2 = -5.130\times10^{-12} \mathrm{J}\) - For \(n = 3\): \(E_3 = -2.282\times10^{-12} \mathrm{J}\)