Question

A $\mu -meson$ (charge $-e$, mass $=207m$, where $m$ is the mass of the electron) can be captured by a proton to form a hydrogen-like "mesic" atom. Calculate the radius of the first Bohr orbit, the binding energy, and the wavelength of the line in the Lyman series for such an atom. The mass of the proton is $1836$ times the mass of the electron. The radius of the first Bohr orbit and the binding energy of Hydrogen are $0.529\mathring{A}$ and $13.6eV$, respectively. Take $R=109678c{m}^{-1}$

Answer 1

Bohrs radius is given by

$r=\frac{n^2{h}^2}{4{\pi }^{2}mkz{e}^2}e=-1.6\times {10}^{19}Cn=1h=6.63\times {10}^{-34}㎡㎏/sm=207\times 9.11\times {10}^{-31}kgz=1g=9\times {10}^9{r}_1=\frac{{\left(1\right)}^2{(6.63\times {10}^{-34})}^2\times 7\times 7}{4\times 22\times 22\times 207\times 9.11\times {10}^{-31}\times 9\times {10}^9\times 1\times 1.6\times {10}^{19}\times 1.6\times {10}^{19}}=2.55\times {10}^{-13}m=0.00255$

Binding energy is the energy of electron in first order is given by

$E_1=\frac{2{\pi }^{2}m{k}^2{z}^2{e}^4}{n^2{h}^2}$

For $n=1m=207\times 9.11\times {10}^{-31}kgz=1$

For Hydrogen Binding energy is

$E_1^{{}^{\prime }}=\frac{-2{\pi }^{2}m{k}^2{z}^2{e}^4}{n^2{h}^2}=13.6ev$

$E_1=207\times 13.6=2.8keV$

For lyman series for hydrogen atom

$R=\frac{2{\pi }^{2}m{k}^2{e}^4}{c{h}^3}=109678{㎝}^{-1}=10967800m^{-1}{R}^{{}^{\prime }}=\frac{2{\pi }^2{m}_c{k}^2{e}^4}{c{h}^3}=207\times R=207\times 10967800=22703346m^{-1}$

Radius of First Bohrs orbit=0.00255

Binding Energy =2.8keV

Wavelength of Lymon =$=R^{{}^{\prime }}[\frac{1}{1^2}-\frac{1}{n^2}]$