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Standard XII

Physics

Summary

A muonic atom...

Question

A muonic atom contains a nucleus of charge Ze and a negative muon, $μ^{-}$ , moving about it. The $μ^{-}$ is an elementary charge with -e and a mass that is approximately 200 times as large as an electron mass.
Calculate the binding energy of a muonic atom(in eV) with $Z = 1$ .

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Solution

General equation of binding energy is
$E = - \frac{m k^{2} e^{4}}{2 h^{2}}$

For regular electron model, $E_{b i n d i n g} = - 13.6 e V$
Hence for muonic atom, $E_{b i n d i n g} = - 13.6 \times 200 e V = 2720 e V$
(as $E_{m u o n i c} = - \frac{μ k^{2} e^{4}}{2 h^{2}}$ , where $μ = 200 m_{e}$ )

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A muonic atom contains a nucleus of charge Ze and a negative muon, μ−, moving about it. The μ− is an elementary charge with -e and a mass that is approximately 200 times as large as an electron mass. Calculate the binding energy of a muonic atom(in eV) with Z=1.

General equation of binding energy is E=−mk2e42h2 For regular electron model, Ebinding=−13.6 eV Hence for muonic atom, Ebinding=−13.6×200 eV=2720 eV (as Emuonic=−μk2e42h2, where μ=200me)

A muonic atom contains a nucleus of charge Ze and a negative muon, $μ^{-}$ , moving about it. The $μ^{-}$ is an elementary charge with -e and a mass that is approximately 200 times as large as an electron mass.
Calculate the binding energy of a muonic atom(in eV) with $Z = 1$ .

General equation of binding energy is
$E = - \frac{m k^{2} e^{4}}{2 h^{2}}$

For regular electron model, $E_{b i n d i n g} = - 13.6 e V$
Hence for muonic atom, $E_{b i n d i n g} = - 13.6 \times 200 e V = 2720 e V$
(as $E_{m u o n i c} = - \frac{μ k^{2} e^{4}}{2 h^{2}}$ , where $μ = 200 m_{e}$ )