Question

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $n^{th}$ orbital will therefore be proportional to:

• A. $n^2$
• B. $n$
• C. $n^{1/2}$
• D. $n^{1/4}$

Answer 1

Answer: (C)

According to the question,
$2\pi r=n\lambda=\frac{nh}{p}=\frac{nh}{mv}$
or $mvr =\frac{nh}{2\pi}$ or $mv=\frac{nh}{2\pi r }$
$F=qv_{B}=\frac{mv^{2}}{r}$
or, $q_{B}=\frac{mv}{r}=\frac{nh}{2\pi r.r }$
or $r^{2}=\frac{nh}{2\pi qB }$
or, $r=\sqrt{\frac{nh}{2\pi qb}}$
i.e., $r\,\propto\,n^{1/2}$