Question

According to Bohr's model, the magnetic field at the centre (at the nucleus) of a hydrogen atom due to motion of an electron in the $n^{th}$ orbit is proportional to

• A. $\frac{1}{n^3}$
• B. $\frac{1}{n^5}$
• C. $n^5$
• D. $n^3$

Answer 1

Answer: (B)

For an electron revolving around the nucleus,
$i=ef$
$f\propto \frac{z^2}{n^3}$
$i\propto e\frac{z^2}{n^3}$
Again we know that,
$r\propto \frac{n^2}{Z}$
Magnetic field at the center of the hydrogen atom, $B=\frac{{\mu }_{0}i}{2r}$
$B\propto \frac{{\mu }_0{Z}^3}{2n^5}$
$B\propto \frac{1}{n^5}$