Question

A proton, a deutron and an $\alpha$ particle accelerated through the same potential difference enter a region of uniform magnetic field, moving at right angles to $B$. What is the ratio of their radii.?

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

Answer 1

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

Proton Deutron $\alpha$particle

mass=m mass=2m mass=4m

charge=q (one proton + one neutron) (two proton+two neutron)

charge=q charge=2q

as radius of path in magnetic field is given by

$r=\frac{m\vartheta }{qB}=\frac{\sqrt{2m\left(KE\right)}}{qB}$

and KE gained by particle

$KE=qV$

$(KE{)}_{proton}=qV$ $\therefore r_{proton}=\frac{\sqrt{2mqV}}{qB}$

$(KE{)}_{deuton}=qV$ $\therefore r_{deuton}=\frac{\sqrt{4mqV}}{qB}$

$(KE{)}_{\alpha }=2qV$ $\therefore r_{\alpha }=\frac{\sqrt{8m\ast 2qV}}{2qB}$

$\therefore r_{proton}:r_{deutron}:r_{\alpha }=\sqrt{2}:\sqrt{4}:\frac{\sqrt{16}}{2}$

$=\sqrt{2}:2:2$

$=1:\sqrt{2}:\sqrt{2}$