A proton, a deuteron and an α-particle enter a region of perpendicular magnetic field (to their velocities) with same kinetic energy. If rp, rd and rα are the radii of circular paths of these particles, then

Question

A proton, a deuteron and an α-particle enter a region of perpendicular magnetic field (to their velocities) with same kinetic energy. If rp, rd and rα are the radii of circular paths of these particles, then (A) rα=rd < rp (B) rα=rd=rp (C) rα < rd < rp (D) rα>rd>rp

Tardigrade Admin · Accepted Answer

As we know that radius of circle, r=(m v/B q)=(1/2)((2 m v2/B q v)) ⇒ r=(2 E/B q v) Energy is same. So, r propto 1 / q but in case of deuteron and α-particle q is same and q for proton is least ∴ rp >rα=rd

Q. A proton, a deuteron and an $α$ -particle enter a region of perpendicular magnetic field (to their velocities) with same kinetic energy. If $r_{p}, r_{d}$ and $r_{α}$ are the radii of circular paths of these particles, then

$r_{α} = r_{d} < r_{p}$

$r_{α} = r_{d} = r_{p}$

$r_{α} < r_{d} < r_{p}$

$r_{α} > r_{d} > r_{p}$

As we know that radius of circle,
$r = \frac{m v}{Bq} = \frac{1}{2} (\frac{2 m v ^{2}}{Bq v})$
$\Rightarrow r = \frac{2 E}{Bq v}$
Energy is same. So, $r \propto 1/ q$ but in case of deuteron and $α$ -particle $q$ is same and $q$ for proton is least
$∴ r_{p} > r_{α} = r_{d}$

Q. A proton, a deuteron and an α-particle enter a region of perpendicular magnetic field (to their velocities) with same kinetic energy. If rp​,rd​ and rα​ are the radii of circular paths of these particles, then

rα​=rd​<rp​

rα​=rd​=rp​

rα​<rd​<rp​

rα​>rd​>rp​