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 $n^{t h}$ 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, $B=\frac{\mu _{0} i}{2 r}$
$B \propto \frac{\mu _{0} Z^{3}}{2 n^{5}}$
$B \propto \frac{1}{n^{5}}$