Ncert Exercises & Solutions

12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which is a negatively charged muon (µ–) of mass about 207m_e orbits around a proton].

Hint: Apply Bohr's model of an atom.

Step 1: Find the first Bohr's orbit of a muon.

Mass of a negatively charged muon,

m_{μ} = 207 m_{e}

According to Bohr's model,

B o h r r a d i u s, r_{e} \propto \frac{1}{m_{e}}

And, the energy of a ground state electronic hydrogen atom,

E_{e} \propto m_{e}

Also, the energy of a ground state muonic hydrogen atom,

E_{u} \propto m_{u}

We have the value of the first Bohr orbit,

r_{e} = 0.53 A = 0.53 \times 10^{- 10} m

Let

r_{μ}

, be the radius of the muonic hydrogen atom.
At equilibrium, we can write the relation as:

m_{μ} r_{μ} = m_{e} r_{e}

207 m_{e} \times r_{μ} = m_{e} r_{e}

∴ r_{μ} = \frac{0.53 \times 10^{- 10}}{207} = 2.56 \times 10^{- 13} m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is

2.56 \times 10^{- 13} m

Step 2: Find the ground state energy of muon.

We have,

E_{e} = - 13.6 e V

Take the ratio of these energies as:

\frac{E_{e}}{E_{μ}} = \frac{m_{e}}{m_{μ}} = \frac{m_{e}}{207 m_{e}}

E_{μ} = 207 E_{e}

= 207 \times (- 13.6) = - 2.81 k e V

Hence, the ground state energy of a muonic hydrogen atom is -2.81 keV.

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