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 $n^{t h}$ 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)

$B_{n}=\frac{\mu _{0} I_{n}}{2 r_{n}}$
Or $B_{n} \, \propto \frac{I_{n}}{r_{n}} \propto \frac{\left(f_{n}\right)}{2 r_{n}}$
$B_{n} \, \propto \frac{\left(\frac{v_{n}}{r_{n}}\right)}{r_{n}} \, \propto \frac{v_{n}}{\left(r_{n}\right)^{2}}$ $ \propto \frac{\left(\frac{Z}{n}\right)}{\left(\frac{n^{2}}{Z}\right)^{2}} \propto \frac{Z^{3}}{n^{5}}$