A `mu-meson `("charge -e , mass = 207 m`, where `m` is mass of 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 Å` and `13.6 e V` , repectively. `Take `R = 1.67 xx 109678 cm^(-1)`

Answer

`0.002844 Å, 2530 eV, 653.6 Å`

Explanation

The reduced mass of the system is given by
`mu = ((207m) (1836 m))/((207 m + 1836m)) = 186 m`
where `m = mass` of electron or positron.
The radius of first Bohr's orbit is given by
`r_(1) = (h^(2))/(4 pi^(2) Ke^(2) (186 m))`
`= (1)/(186) xx` Radius of first Bohr's orbit of hydrogen atom
`= (1)/(186) xx 0.529 = 0.002844 Å` (i)
From bohr's theory, the ground state energy for hydrogen-like atom with `Z = 1` is given by
`E_(1) = (2 pi^(2) K^(2) e^(4) mu)/(h^(2)) = (2 pi^(2) K^(2) e^(4) (186 m))/(h^(2)) `
`= - 186 xx 13.6 eV`
`= 2530 eV`
Hence, the binding energy is `= 2530 eV`
The wavelength of Lyman series is given by
`(1)/(lambda) = R_(1) ((1)/(1^(2)) - (1)/(n^(2))) n = 2,3,4,`...
where `R_(mu) =` Rydberg contant for mesic atom.
For first line , `(1)/(lambda) = R_(1) ((1)/(1^(2)) - (1)/(2^(2))) or lambda = (4)/(3 R_(1))`
Now `R_(mu) = (2 pi^(2) K^(2) e^(2) mu)/(ch^(3) = (R)(mu)/(m)`
or ` = 186 R`
or `= 186 xx 10967800 m^(-1)`
Substituting the value of `R_(mu)` in equation (iii), we get
`lambda = (4)/(3 xx 186 xx 109687) m = 653.6 Å`