Question

A muon has the same charge as an electron but a mass that is 207 times greater. The negatively charged muon can bind to a proton to form a new type of hydrogen atom. How does the binding energy, \(E_{\mathrm{B} \mu}\), of the muon in the ground state of a muonic hydrogen atom compare with the binding energy, \(E_{\mathrm{Be}}\) of an electron in the ground state of a conventional hydrogen atom? a) \(\left|E_{\mathrm{B} \mu}\right|=\mid E_{\mathrm{Be}}\) b) \(\left|E_{\mathrm{B} \mu}\right|=100\left|E_{\mathrm{Be}}\right|\) c) \(\left|E_{\mathrm{B} \mu}\right|=\left|E_{\mathrm{Be}}\right| / 100\) d) \(\left|E_{\mathrm{B} \mu}\right|=200\left|E_{\mathrm{Be}}\right|\) e) \(\left|E_{\mathrm{Bu}}\right|=\left|E_{\mathrm{Be}}\right| / 200\)

Short answer

Based on the calculations, the binding energy of the muonic hydrogen atom is 207 times larger than that of a regular hydrogen atom, but the closest option given is d) |\(E_{\mathrm{B} \mu}\)| = 200|\(E_{\mathrm{Be}}\)|. Therefore, the best answer is option d.

Step-by-step solution

Recall the Rydberg formula for hydrogen-like atoms

The Rydberg formula for the energy levels of hydrogen-like atoms is given by \(E_n=-\frac{Z^2 e^4 m_e}{32\pi^2 n^2 \varepsilon_0^2 h^2}\), where \(Z\) is the atomic number, \(n\) is the principal quantum number, \(m_e\) is the mass of the electron, \(e\) is the elementary charge, \(\varepsilon_0\) is the vacuum permittivity, and \(h\) is Planck's constant. We are comparing the ground state (n = 1) of the muonic and regular hydrogen atom, so we will need to consider how the mass and charge are affected in the muonic hydrogen atom.

Change the mass of the electron to the mass of the muon

In the muonic hydrogen atom, we have a muon bound to a proton. The muon has the same charge as an electron but 207 times the mass: \(m_\mu = 207m_e\). So, the energy levels of the muonic hydrogen atom can be expressed as \(E_{n\mu}=-\frac{Z^2 e^4 m_\mu}{32\pi^2 n^2 \varepsilon_0^2 h^2}\), with \(m_\mu=207m_e\).

Compute the ratio of the binding energy of the muonic hydrogen atom to that of the regular hydrogen atom

We want to find \(\frac{E_{1\mu}}{E_{1e}}\), the ratio of the binding energies of the ground state of the muonic hydrogen atom and the regular hydrogen atom: $$\frac{E_{1\mu}}{E_{1e}} = \frac{-\frac{Z^2 e^4 m_\mu}{32 \pi^2\cdot1^2\varepsilon_0^2 h^2}}{-\frac{Z^2 e^4 m_e}{32\pi^2\cdot1^2\varepsilon_0^2 h^2}} = \frac{m_\mu}{m_e}$$ Since \(m_\mu = 207m_e\), the ratio is: $$\frac{E_{1\mu}}{E_{1e}} = 207$$

Compare the calculated ratio with the given options and determine the answer

We found that the binding energy of the muonic hydrogen atom is 207 times larger than that of a regular hydrogen atom, or \(\left|E_{\mathrm{B} \mu}\right| = 207 \left|E_{\mathrm{Be}}\right|\). The closest option is: d) \(\left|E_{\mathrm{B} \mu}\right| = 200\left|E_{\mathrm{Be}}\right|\) Though not exact, this is the closest approximation. Thus, the answer is option d.