Question

Obtain the first Bohr radius and ground state energy of a muonic hydrogen atom [i.e. an atom in which a negatively charged muon $\left( {{\mu ^ - }} \right)$ of mass about $207{m_e}$ orbits around a proton].

Answer 1

Hint : In order to solve this question, we are going to find the first Bohr radius which depends on the value of ${m_e}$ .After that, the energy is computed by using the fact that it is directly proportional to ${m_e}$ and using the fact that the energy of the Bohr first orbit is equal to $- 13.6eV$ .
The Bohr radius of an atom is given by the formula
${r_e} = \dfrac{1}{{{m_e}}}$
Energy to mass ratios of the muonic hydrogen atom is
$\dfrac{{{E_e}}}{E} = \dfrac{{{m_e}}}{m}$

Complete Step By Step Answer:
It is given that the charge of the muon is $207{m_e}$
The Bohr radius is given by the relation
${r_e} = \dfrac{1}{{{m_e}}}$
Now as we know that the radius of the first Bohr orbit is equal to $0.53 \times {10^{ - 10}}m$
We know that at equilibrium, $mr = {m_e}{r_e}$
The radius of a muonic hydrogen atom $r = 2.56 \times {10^{ - 13}}m$
Now as the energy of the atom is directly proportional to the mass, this implies
${E_e} \propto {m_e}$
And for the first orbit, the energy equals $- 13.6eV$
Also the ratios of the energy to mass remain constant
Thus,
$\dfrac{{{E_e}}}{E} = \dfrac{{{m_e}}}{m}$
Now finding the value of $E$ from this relation and putting the other values, we get
$E = \dfrac{{{E_e} \times m}}{{{m_e}}} = - 2.81keV$
Hence the ground state energy of a muonic hydrogen atom is $- 2.81keV$ .

Note :
The first Bohr orbit is equal to $0.53 \times {10^{ - 10}}m$ . As the energy of the atom is directly proportional to the mass, this implies;
${E_e} \propto {m_e}$ .
A muonic hydrogen is the atom in which a negatively charged muon revolves around a proton.