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.

Step by step solution

01

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.

02

The hyperfine splitting in the ground state of muonic hydrogen.

(a)

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

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

03

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

04

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

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