Solution to 2008 Problem 66

The energy levels of hydrogen are

\\begin{align*}E_n = - \\frac{\\mu e^4}{2 \\hbar^2 (4 \\pi \\epsilon_0)^2 n^2}  = \\frac{-\\left|E_0\\right|}{n^2}\\end{align*}where $\\mu = m_e m_p/(m_e + m_p)$ is the reduced mass of electron proton system. When the electron is replacing with a muon, $\\mu' = m_{\\mu} m_p/(m_{\\mu} + m_p)$. So, the energy levels of this new form of hydrogen are

\\begin{align*}E_n' &= - \\frac{\\mu' e^4}{2 \\hbar^2 (4 \\pi \\epsilon_0)^2 n^2} \\\\&= E_n \\frac{\\mu'}{\\mu} \\\\&=  E_n \\frac{m_{\\mu} (m_e + m_p)}{m_e(m_{\\mu} + m_p)} \\\\&= \\boxed{\\frac{-\\left|E_0\\right|}{n^2} \\frac{m_{\\mu} (m_e + m_p)}{m_e(m_{\\mu} + m_p)} }\\end{align*}Therefore, answer (D) is correct.

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