Question

An electron moves in a Bohr's orbit. The magnetic field at the centre is proportional to:

• A. $n^{-5}$
• B. $n^{-3}$
• C. $n^{-4}$
• D. $n^{-2}$

Answer 1

Answer: (A)

$n^{-5}$

Let the electron revolves with speed $v_n$ in a circular orbit of radius $r_n$.

$\therefore$ $i=e\nu =e\frac{v_n}{2\pi r_n}$

According to Bohr's model, radius of nth orbit, $r_n=0.529\frac{n^2}{Z}$ $A^o$

$⟹$ $r\propto n^2$

Also velocity of electron in nth orbit, $v_n=2.18\times {10}^6\frac{Z}{n}$ $m/s$

$⟹$ $v\propto n^{-1}$

Magnetic field at the centre, $B=\frac{{\mu }_{o}i}{2r_n}=\frac{{\mu }_o}{4\pi }\frac{e{v}_n}{r_n^2}$

$\therefore$ $B\propto \frac{v}{r^2}\propto \frac{n^{-1}}{(n^2{)}^2}$

$⟹$ $B\propto n^{-5}$