Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom
Question:

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].

Solution:

Mass of a negatively charged muon, $m_{\mu}=207 m_{e}$

According to Bohr’s model,

Bohr radius, $r_{e} \propto\left(\frac{1}{m_{e}}\right)$

And, energy of a ground state electronic hydrogen atom, $E_{e} \propto m_{e}$.

Also, energy of a ground state muonic hydrogen atom, $E_{\mu} \propto m_{\mu}$.

We have the value of the first Bohr orbit, $r_{e}=0.53 \mathrm{~A}=0.53 \times 10^{-10} \mathrm{~m}$

Let rμ be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

$m_{\mu} r_{\mu}=m_{e} r_{e}$

$207 m_{e} \times r_{\mu}=m_{e} r_{e}$

$\therefore r_{\mu}=\frac{0.53 \times 10^{-10}}{207}=2.56 \times 10^{-13} \mathrm{~m}$

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

2.56 × 10−13 m.

We have,

Ee= − 13.6 eV

Take the ratio of these energies as:

$\frac{E_{e}}{E_{\mu}}=\frac{m_{e}}{m_{\mu}}=\frac{m_{e}}{207 m_{e}}$

$E_{\mu}=207 E_{e}$

$=207 \times(-13.6)=-2.81 \mathrm{keV}$

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

Administrator

Leave a comment

Please enter comment.
Please enter your name.