Question

Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

Answer 1

Answer:

Magnetic moment = $\sqrt{n(n+2)}$  ; where n= no of unpaired electron in an atom.  

                              =$\sqrt{1(1+2)}$ = 1.732 BM

where for H atom no of electron is 1   and BM is Bohr magneton

Explanation:.

Answer 2

Explanation:

Electron has the mass, $m=9.1 \times 10^{-31} \mathrm{kg}$

Ground state has the radius, $r=0.53 \times 10^{-10} \mathrm{m}$

Let the revolution’s frequency of the electron be shown as  f  moving in ground state and the orbit’s area is denoted as A.

Electron’s dipole moment (μ) is shown as $\mu=q f \mathrm{A}=n i \mathrm{A}$

$=m e 5 \times \frac{(\pi r 2 n 2)}{4 \in 02 h 3 n 3}$

where,

h = Constant of Planck  

e = Electron’s charge

ε0 = Free space’s permittivity

n = Principal quantum number

$\therefore \mu=\frac{\left(9.1 \times 10^{-13}\right)\left(1.6 \times 10^{-19}\right)^{5} \times \pi \times\left(0.53 \times 10^{-10}\right)^{2}}{4 \times\left(8.85 \times 10^{-12}\right)^{2} \times\left(6.64 \times 10^{-34}\right)^{3} \times(1)^{3}}$

$=0.000917 \times 10^{-20}$

$=9.176 \times 10^{-24} \mathrm{A}-\mathrm{m}^{25}$