The 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) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right|\) d) \(\left|E_{\mathrm{B} \mu}\right| \approx 200 \mid E_{\mathrm{Be}}\) b) \(\left|E_{\mathrm{B} \mu}\right| \approx 100\left|E_{\mathrm{Be}}\right|\) e) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 200\) c) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 100\)

Step by step solution

01

Calculate the reduced mass for the muon-proton system

To start, we need to calculate the reduced mass (\(\mu\)) for the muon-proton system. This is given by the formula: $$\mu = \frac{m_\mu m_p}{m_\mu + m_p}$$ where \(m_\mu\) is the mass of the muon, and \(m_p\) is the mass of the proton. Given that the muon is 207 times heavier than an electron, we can rewrite this formula as: $$\mu = \frac{207m_e m_p}{207m_e + m_p}$$

02

Replace proton mass with electron mass in the formula

Since the mass of the proton is much larger than the mass of the electron (approximately 1836 times larger), we can simplify this expression to: $$\mu \approx \frac{207m_e m_p}{m_p}$$ $$\mu \approx 207m_e$$

03

Calculate the muonic hydrogen binding energy

Now we can calculate the binding energy for the ground state of the muonic hydrogen atom using the Rydberg Formula for reduced mass: $$E_{B\mu} = -\frac{13.6\,\text{eV} \times \mu}{m_e}$$ Therefore, we have: $$E_{B\mu} = -\frac{13.6\,\text{eV} \times 207m_e}{m_e}$$

04

Simplify the expression and compare binding energies

Now, we can simplify the expression as: $$E_{B\mu} = -13.6\,\text{eV} \times 207$$ We know that the binding energy of an electron in a conventional hydrogen atom is \(E_{Be} = -13.6\,\text{eV}\). Comparing these energies, we have: $$\left|E_{B\mu}\right| \approx 207\left|E_{Be}\right|$$ This answer is not provided in the options, since the factor should be within a range 200-210. Based on this, the closest answer is option (d): d) \(\left|E_{\mathrm{B} \mu}\right| \approx 200 \mid E_{\mathrm{Be}}\)

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