A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an deutron to describe a circle of same radius in the same field?

Question

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an deutron to describe a circle of same radius in the same field? (A) 0.5 MeV (B) 4 MeV (C) 2 MeV (D) 1 MeV

Tardigrade Admin · Accepted Answer

R =( mu / qB ) ⇒ KE =( q 2 B 2 R 2/2 m ) propto ( q 2/ m ) Since charge on both is same but mass of deutron is double of proton. ( KE ) text deutron =(( KE ) text proton /2)=0.5 MeV

Q. A proton carrying $1 M e V$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an deutron to describe a circle of same radius in the same field?

0.5 MeV

4 MeV

2 MeV

1 MeV

$R = \frac{m u}{qB}$
$\Rightarrow K E = \frac{q ^{2} B ^{2} R ^{2}}{2 m} \propto \frac{q ^{2}}{m}$
Since charge on both is same but mass of deutron is double of proton.
$(K E)_{deutron} = \frac{( K E ) _{proton}}{2} = 0.5 M e V$

Q. A proton carrying 1MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an deutron to describe a circle of same radius in the same field?