Question

By appropriate modification of the hydrogen formula, determine the hyperfine splitting in the ground state of

(a) muonic hydrogen (in which a muon-same charge and g-factor as the electron, but 207times the mass-substitutes for the electron),

(b) positronium (in which a positron-same mass and g-factor as the electron, but opposite charge-substitutes for the proton), and

(c) muonium (in which an anti-muon-same mass and g-factor as a muon, but opposite charge-substitutes for the proton). Hint: Don't forget to use the reduced mass (Problem 5.1) in calculating the "Bohr radius" of these exotic "atoms." Incidentally, the answer you get for positronium $\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{82}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{eV}\mathbf{)}$is quite far from the experimental value; $\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{eV}\mathbf{)}$the large discrepancy is due to pair annihilation $\mathbf{(}{\mathbf{e}}^{\mathbf{+}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{\gamma }\mathbf{+}\mathbf{\gamma }\mathbf{)}$, which contributes an extra localid="1656057412048" $\mathbf{}\mathbf{(}\mathbf{3}\mathbf{/}\mathbf{4}\mathbf{)}\mathbf{\Delta E}\mathbf{,}\mathbf{}$and does not occur (of course) in ordinary hydrogen, muonic hydrogen, or muoniun.

Short answer

a) ${\mathrm{\Delta E}}_{{}_{\mathrm{muonichydrogen}}}=0.183\mathrm{eV}$.

b) role="math" localid="1656056468638" ${\mathrm{\Delta E}}_{\mathrm{positronium}}=4.82\cdot {10}^{-4}\mathrm{eV}.$

c) ${\mathrm{\Delta E}}_{\mathrm{muonium}}=1.84\cdot {10}^{-5}\mathrm{eV}.$

Step-by-step solution

Definition of hyperfine spliting.

The interaction of the magnetic moments of the electron and proton causes hyperfine splitting, which results in a slightly variable magnetic energy for each spin state.

The hyperfine splitting in the ground state of muonic hydrogen.

(a)

For muonic hydrogen$:{\mathrm{m}}_{\mathrm{e}}\to {\mathrm{m}}_{\mathrm{\mu }}=207{\mathrm{m}}_{\mathrm{e}},\mathrm{and}\mathrm{a}\to {\mathrm{a}}_{\mathrm{\mu }-}.$

$$\begin{gathered} \frac{\mathrm{a}}{{\mathrm{a}}_{\mathrm{\mu }}}=\frac{{\mathrm{m}}_{\mathrm{\mu },\mathrm{reduced}}}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{{\mathrm{m}}_{\mathrm{\mu }}{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{\mu }}+{\mathrm{m}}_{\mathrm{p}}}\frac{1}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{207{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{p}}}{207{\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{p}}}\frac{1}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{207{\mathrm{m}}_{\mathrm{p}}}{207{\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{p}}} \\ =\frac{207}{1+207.{\displaystyle \frac{9.11.{10}^{-31}}{1.67.{10}^{-27}}}} \\ \cong 186 \\ ∆\mathrm{E}=5.88.{10}^{-6}\mathrm{eV}.\frac{1}{207}.{186}^3 \\ =0.183\mathrm{eV} \end{gathered}$$

The hyperfine splitting in the ground state of positronium.

(b)

For positronium $\mathrm{g}=2\mathrm{and}{\mathrm{m}}_{\mathrm{p}}\to {\mathrm{m}}_{\mathrm{\mu }}$

role="math" localid="1656057236437" $$\begin{gathered} \frac{\mathrm{a}}{{\mathrm{a}}_{{}_{\mathrm{positronium}}}}=\frac{{\mathrm{m}}_{{}_{\mathrm{positronium}}}}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{{\mathrm{m}}_{\mathrm{e}}^2}{{\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{e}}}\frac{1}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{1}{2} \\ ∆\mathrm{E}=5.88.{10}^{-6}\mathrm{eV}\left(\frac{2}{5.59}\right).\frac{1.67.{10}^{-27}}{9.11.{10}^{-31}}{\left(\frac{1}{2}\right)}^3 \\ =4.82.{10}^{-4}\mathrm{eV} \end{gathered}$$

Step 4:The hyperfine splitting in the ground state of muonium.

(c)

For muonium $\mathrm{g}=2,{\mathrm{m}}_{\mathrm{p}}\to {\mathrm{m}}_{\mathrm{\mu }}$

role="math" localid="1656057368856" $$\begin{gathered} \frac{\mathrm{a}}{{\mathrm{a}}_{\mathrm{m}}}=\frac{{\mathrm{m}}_{\mathrm{m}}}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{\mu }}}{{\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{\mu }}}\frac{1}{{\mathrm{m}}_{\mathrm{e}}} \\ =\frac{207}{208} \\ ∆\mathrm{E}=5.88.{10}^{-6}\mathrm{eV}\frac{2}{5.59}.\frac{1.67.{10}^{-27}}{207.9.11.{10}^{-31}}{\left(\frac{207}{208}\right)}^3 \\ =1.84.{10}^{-5}\mathrm{eV} \end{gathered}$$