Question

A proton and alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by the proton is $1\,{\text{MeV}}$, the energy acquired by the alpha particle will be:
A. $1\,{\text{MeV}}$
B. $4\,{\text{MeV}}$
C. $0.5\,{\text{MeV}}$
D. $1.5\,{\text{MeV}}$

Answer 1

Use the equation for the magnetic force and centripetal force acting on a charge particle moving in the magnetic field. Also use the equation for the kinetic energy of a particle. These equations give the relation between the kinetic energy, charge and mass of the charged particle.

Formulae used:
The force ${F_B}$ on a charged particle moving in the uniform magnetic field $B$ is
${F_B} = qvB$ …… (1)
Here, $q$ is the charge on the particle and $v$ is the speed of the particle.
The centripetal force ${F_C}$ on a particle in circular motion is
${F_C} = \dfrac{{m{v^2}}}{R}$ …… (2)
Here, $m$ is the mass of the particle, $v$ is the speed of the particle and $R$ is the radius of the circular path.
The kinetic energy $K$ of a particle is given by
$K = \dfrac{1}{2}m{v^2}$ …… (3)
Here, $m$ is the mass of the particle and $v$ is the velocity of the particle.

Complete step by step answer:
The proton and the alpha particle are moving in a uniform magnetic field $B$ at right angles to the magnetic field $B$ in the circles of the same radius.
The kinetic energy ${K_P}$ acquired by the proton is $1\,{\text{MeV}}$.
${K_P} = 1\,{\text{MeV}}$
Calculate the kinetic energy ${K_\alpha }$ acquired by the alpha particle.
The force ${F_B}$ of the moving charged particle in the magnetic field is balanced by the centripetal force ${F_C}$ acting on the charged particle.
${F_B} = {F_C}$
Substitute $qvB$ for ${F_B}$ and $\dfrac{{m{v^2}}}{R}$ for ${F_C}$ in the above equation.
$qvB = \dfrac{{m{v^2}}}{R}$
$\Rightarrow qB = \dfrac{{mv}}{R}$
Rearrange the above equation for the speed $v$ of the charged particle.
$v = \dfrac{{qBR}}{m}$

Calculate the kinetic energy of a charged particle.
Substitute $\dfrac{{qBR}}{m}$ for $v$ in equation (3).
$K = \dfrac{1}{2}m{\left( {\dfrac{{qBR}}{m}} \right)^2}$
$\Rightarrow K = \dfrac{{{q^2}{B^2}{R^2}}}{{2m}}$ …… (4)
The magnetic field $B$ and the radius $R$ of the circular path is the same for both proton and alpha particles.
Rewrite the equation (4) for the kinetic energy of the proton and the alpha particle.
${K_P} = \dfrac{{q_P^2{B^2}{R^2}}}{{2{m_P}}}$ …… (5)
Here, ${m_P}$ is the mass of the proton and ${q_P}$ is the charge on the proton.
${K_\alpha } = \dfrac{{q_\alpha ^2{B^2}{R^2}}}{{2{m_\alpha }}}$ …… (6)
Here, ${m_\alpha }$ is the mass of the alpha particle and ${q_\alpha }$ is the charge on the alpha particle.
The mass of the alpha particle is four times the mass of the proton and the charge on the alpha particle is twice the charge on a proton.
${m_\alpha } = 4{m_P}$
${q_\alpha } = 2{q_P}$
Divide equation (6) by equation (5).
$\dfrac{{{K_\alpha }}}{{{K_P}}} = \dfrac{{\dfrac{{q_\alpha ^2{B^2}{R^2}}}{{2{m_\alpha }}}}}{{\dfrac{{q_P^2{B^2}{R^2}}}{{2{m_P}}}}}$

$\Rightarrow \dfrac{{{K_\alpha }}}{{{K_P}}} = \dfrac{{q_\alpha ^2{m_P}}}{{{m_\alpha }q_P^2}}$

Substitute $4{m_P}$ for ${m_\alpha }$, $2{q_P}$ for ${q_\alpha }$ and $1\,{\text{MeV}}$ for ${K_P}$ in the above equation and rearrange it for ${K_\alpha }$.
$\dfrac{{{K_\alpha }}}{{1\,{\text{MeV}}}} = \dfrac{{{{\left( {2{q_P}} \right)}^2}{m_P}}}{{\left( {4{m_P}} \right)q_P^2}}$
$\Rightarrow {K_\alpha } = \dfrac{4}{4}\left( {1\,{\text{MeV}}} \right)$

$\Rightarrow {K_\alpha } = 1\,{\text{MeV}}$
Therefore, the kinetic energy acquired by the alpha particle is $1\,{\text{MeV}}$.

**Hence, the correct option is A.

Note: **
There is no need to convert the unit of kinetic energy of proton in the SI system of units as the ultimate result is to be measured in MeV and all the remaining units get cancelled.