Question

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. 

Answer 1

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

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 = niA = qfA

`= 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

ε₀ = 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.