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 $207m_e$ 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
 $r_e={\displaystyle \frac{1}{m_e}}$
Energy to mass ratios of the muonic hydrogen atom is
 ${\displaystyle \frac{E_e}{E}}={\displaystyle \frac{m_e}{m}}$

Complete Step By Step Answer:
It is given that the charge of the muon is $207m_e$
The Bohr radius is given by the relation
 $r_e={\displaystyle \frac{1}{m_e}}$
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,
 ${\displaystyle \frac{E_e}{E}}={\displaystyle \frac{m_e}{m}}$
Now finding the value of $E$ from this relation and putting the other values, we get
 $E={\displaystyle \frac{E_e\times m}{m_e}}=-2.81keV$
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.