Solution to 1992 Problem 30

The Bohr energies for an electron orbiting a nucleus with charge Ze and mass m_n is
\begin{align*}E_n = \frac{- \mu Z^2 e^2}{2 \hbar^2 (4 \pi \epsilon_0)^2 n^2}\end{align*}
where \mu is the reduced mass of the electron and the nucleus. In this case n = 2, \mu = m_e/2. Therefore, the energy is 1/8 of the ground-state energy of hydrogen. Therefore, answer (E) is correct.

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