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my2cts
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$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$) this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.

$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$ this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.

$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$) this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.

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my2cts
  • 26.6k
  • 2
  • 22
  • 73

$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$ this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.