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Question
Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.
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Solution
Mass of the electron, m = 9.1×10-31kg
Radius of the ground state, r = 0.53×10-10m
Let f be the frequency of revolution of the electron moving in ground state and A be the area of orbit.
Dipole moment of the electron (μ) is given by
μ = niA = qfA
`= e xx (me^4)/(4∈_0^2h^3n^3)xx(pi^2n^2)`
`= (me^5xx (pir^2n^2))/(4∈_0^2h^3n^3)`
Here,
h = Planck's constant
e = Charge on the electron
`epsilon_0` = Permittivity of free space
n = Principal quantum number
`therefore mu = ((9.1xx10^-13)(1.6xx10^-19)^5xxpi xx(0.53xx10^-10)^2)/(4 xx (8.85xx10^-12)^2xx(6.64xx10^-34)^3xx(1)^3`
= 0.000917 × 10-20
= 9.176 × 10-24 A-m25
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