Solution to 1992 Problem 3

K x-rays are released when an electron transitions to an n = 1 orbital from a higher orbital. The energy released is
\begin{align*}\frac{\mu Z^2 e^4}{2 \hbar^2 \left(4 \pi \epsilon_0 \right)^2}\left(1 - \frac{1}{m^2} \right)\end{align*}
where m is the shell of the higher orbital. We are given that Z = 6 for carbon and Z = 12 for magnesium, so the desired ratio is
\begin{align*}\left(\frac{6}{12}\right)^2 = \boxed{\frac{1}{4}}\end{align*}
Therefore, answer (A) is correct.

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