Question

A proton, a deutron and an $\alpha$ particle of same kinetic energy enter in a uniform magnetic field perpendicularly. The ratio of their circular path will be

• A. $1:1:2$
• B. $1:\sqrt{2}:1$
• C. $1:\sqrt{2}:2$
• D. $1:2:1$

Answer 1

The correct option is B $1:\sqrt{2}:1$

$F=qvB=\frac{m{v}^2}{r}⟹r=\frac{mv}{qB}$ where $r=$radius of circular path.

$P=mv$ linear momentum and K.E $\frac{P^2}{2m}$

$P=\sqrt{2m\times K.E}$

$\frac{\sqrt{2m\times K.E}}{qB}$

$qp=e,qd=e,qa=2e,mp=m,md=2m,ma=4m$

$⟹rp:rd=\sqrt{2}:2:\sqrt{2}=1:\sqrt{2}:1$