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.