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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.
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Solution
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.
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