Question

The magnetic field at the centre (at nucleus) of the hydrogen-like atoms (atomic number = $Z$ ) due to the motion of an electron in $n^{th}$ orbit is proportional to

• A. $\frac{n^3}{Z^5}$
• B. $\frac{n^4}{Z}$
• C. $\frac{Z^2}{n^3}$
• D. $\frac{Z^3}{n^5}$

Answer 1

Answer: (D)

$B_n=\frac{{\mu }_0{I}_n}{2r_n}$
Or $B_n\propto \frac{I_n}{r_n}\propto \frac{\left(f_n\right)}{2r_n}$
$B_n\propto \frac{\left(\frac{v_n}{r_n}\right)}{r_n}\propto \frac{v_n}{{\left(r_n\right)}^2}$ $\propto \frac{\left(\frac{Z}{n}\right)}{{\left(\frac{n^2}{Z}\right)}^2}\propto \frac{Z^3}{n^5}$