For spontaneous transitions from the $3 d$ state, the largest probability is for $3 d \rightarrow 2 p$. In nonrelativistic approximation, the photon energy is given by

$$
\begin{aligned}
h \nu &=\frac{m_{\mu} Z^{2} e^{4}}{2 \hbar^{2}}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \\
&=\frac{m_{\mu} c^{2} Z^{2}}{2}\left(\frac{e^{2}}{\hbar c}\right)^{2}\left(\frac{5}{36}\right) \\
&=\frac{105.7 \times 13^{2}}{2}\left(\frac{1}{137}\right)^{2}\left(\frac{5}{36}\right)=6.61 \times 10^{-2} \mathrm{MeV} .
\end{aligned}
$$

The corresponding wavelength is

$$
\lambda=\frac{c}{\nu}=\frac{h c}{h \nu}=\frac{4.135 \times 10^{-15} \times 3 \times 10^{10}}{6.61 \times 10^{4}}=1.88 \times 10^{-9} \mathrm{~cm} .
$$The transition probability per unit time is

$$
A \propto \omega^{3}\left|\mathbf{r}_{k k^{\prime}}\right|^{2} .
$$

For hydrogen-like atoms, as

$$
\left|\mathbf{r}_{k k^{\prime}}\right| \propto \frac{1}{Z}, \quad \omega \propto m Z^{2}, \quad \text { and so } \quad A \propto m^{3} Z^{4},
$$

the mean life of the muonic atom in the $3 d$ state is

$$
\begin{aligned}
T_{\mu} &=\left(\frac{A_{0}}{A}\right) T_{0}=\left(\frac{m_{0}}{m_{\mu}}\right)^{3} \frac{T_{0}}{Z^{4}} \\
&=\left(\frac{0.51}{105.7}\right)^{3} \times \frac{1}{13^{4}} \times 1.6 \times 10^{-8}=6.3 \times 10^{-20} \mathrm{~s} .
\end{aligned}
$$
Consider the situation which arises when a negative muon is captured by an aluminum atom (atomic number $Z=13$ ). After the muon gets inside the "electron cloud" it forms a hydrogen-like muonic atom with the aluminum nucleus. The mass of the muon is $105.7 \mathrm{MeV}$.

(Slide rule accuracy; neglect nuclear motion)

Compute the mean life of the above muonic atom in the $3 d$ state, taking into account the fact that the mean life of a hydrogen atom in the $3 d$ state is $1.6 \times 10^{-8}$ sec.
