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Home » Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−) of mass about 207me orbits around a proton].
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−) of mass about 207me orbits around a proton].
Atoms | CBSE | Class 12 | NCERT | Physics
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ−) of mass about 207me orbits around a proton].

Answer –

We are given that the mass of a negatively charged muon is  mμ = 207 me

Now, According to Bohr’s model bohr’s radius is given by –

    \[{{r}_{e}}\alpha \left( \frac{1}{{{m}_{e}}} \right)\]

The energy of a ground state muonic hydrogen atom is such that Eμ α mμ

We are given the value of the first Bohr orbit,

re  = 0.53 A = 0.53 × 10-10 m

Let rμ represent the radius of muonic hydrogen atom.

Now, writing the equation at the equilibrium –

    \[{{m}_{\mu }}{{r}_{\mu }}={{m}_{e}}{{r}_{e}}\]

    \[207{{m}_{e}}{{r}_{\mu }}={{m}_{e}}{{r}_{e}}\]

    \[{{r}_{\mu }}=\frac{0.53\times {{10}^{-10}}}{207}=2.56\times {{10}^{-13}}m\]

Therefore, 2.56 × 10−13 m is the value of first Bohr radius for a muonic hydrogen atom.

We know that,

Ee= − 13.6 eV

Applying the ratio of the energies we get the relation –

    \[\frac{{{E}_{e}}}{{{E}_{\mu }}}=\frac{{{m}_{e}}}{{{m}_{\mu }}}=\frac{{{m}_{e}}}{207{{m}_{e}}}\]

    \[{{E}_{\mu }}=207{{E}_{e}}=207\times \left( -13.6 \right)\]

    \[{{E}_{\mu }}=-2.82keV\]

The ground state energy of a muonic hydrogen atom is -2.81keV.

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