A deuteron of kinetic energy 50 KeV describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be

Question

A deuteron of kinetic energy 50 KeV describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be (A) 25 KeV (B) 50 KeV (C) 100 KeV (D) 200 KeV

Tardigrade Admin · Accepted Answer

K E=(B2 q2 r2/2 m) K E2=K E1 × (m1/m2) =50 × (2 m/m)=100 keV

Q. A deuteron of kinetic energy $50 Ke V$ describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be

25 KeV

50 KeV

100 KeV

200 KeV

$K E = \frac{B ^{2} q ^{2} r ^{2}}{2 m}$
$K E_{2} = K E_{1} \times \frac{m _{1}}{m _{2}}$
$= 50 \times \frac{2 m}{m} = 100 k e V$

Q. A deuteron of kinetic energy 50KeV describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be