Show that the ratio of the magnetic dipole moment to the angular momentum (*l* = *mvr*) is a universal constant for hydrogen-like atoms and ions. Find its value.

#### Solution

Mass of the electron, *m* = 9.1×10^{-31}kg

Radius of the ground state,* r = *0.53×10 -^{10 }m

Let * f* be the frequency of revolution of the electron moving in the ground state and *A* be the area of orbit.

Dipole moment of the hydrogen like elements (*μ*) is given by

*μ* = *ni*A = *qf*A

`= e xx m/(4∈_0^2 h^3n^3 )xx(pir_0^2n^2)`

`= (me^5xx(pir_0^2n^2))/(4∈_0^2h^3n^3)`

Here,

*h* = Planck's constant

*e *= Charge on the electron

ε_{0} = Permittivity of free space

*n* = Principal quantum number

Angular momentum of the electron in the hydrogen like atoms and ions (*L)* is given by

`L = mvr = (nh)/(2pi)`

Ratio of the dipole moment and the angular momentum is given by

`mu/L =( e^5xxmxxpir^2n^2)/(4∈_0h^3n^3)xx (2pi)/(nh)`

`mu/L =((1.6xx10^-19)^5xx(9.10xx10^-31)(3.14)^2xx(0.53xx10xx^-10)^2)/(2(8.85xx10^-12)^2xx(6.63xx10^-34)^3xx1^2`

`mu/L = 3.73 xx 10^10 C // kg`

Ratio of the magnetic dipole moment and the angular momentum do not depends on the atomic number '*Z*'.

Hence, it is a universal constant.