Question

A proton carrying $1{\text{MeV}}$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circular of the same radius in the same field?
(A) $1{\text{MeV}}$
(B) ${\text{0}}{\text{.5MeV}}$
(C) ${\text{4MeV}}$
(D) ${\text{2MeV}}$

Answer 1

To solve this question, we need to use the formula for the radius of the circular path described by a charged particle when it enters a magnetic field. From there we can find out its kinetic energy. Then substituting the values for the proton and the alpha particle, we will get the relation between their kinetic energies.
Formula used: The formula used for solving this question is given by
$r = \dfrac{{mv}}{{qB}}$, here $r$ is the radius of the circular path followed by a charged particle of mass $m$ and of charge $q$ when it enters in a magnetic field of $B$ with a velocity of $v$.

Complete step-by-step solution:
Let ${B_0}$ be the magnitude of the uniform magnetic field given in this question.
We know that the radius of the circular path followed by a charged particle when it enters in a magnetic field is given by
$r = \dfrac{{mv}}{{qB}}$
$\Rightarrow mv = qBr$
Taking square both sides, we have
${m^2}{v^2} = {q^2}{B^2}{r^2}$
Dividing both sides by $2m$
$\dfrac{{{m^2}{v^2}}}{{2m}} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$
$\dfrac{1}{2}m{v^2} = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$
We know that the kinetic energy is $K = \dfrac{1}{2}m{v^2}$. So we have
$K = \dfrac{{{q^2}{B^2}{r^2}}}{{2m}}$................. (1)
According to our assumption, $B = {B_0}$. Also, according to the question, when a proton enters the uniform magnetic field, it describes a circular path of radius $R$. Let ${K_p}$ be its kinetic energy. Also we know that for a proton, the charge is $e$. Therefore substituting $B = {B_0}$ $K = {K_p}$, $r = R$ and $q = e$ in (1) we get
${K_1} = \dfrac{{{e^2}{B_0}^2{R^2}}}{{2{m_p}}}$ ………….(2)
Here we have assumed the mass of a proton to be ${m_p}$.
Now, according to the question, an $\alpha$-particle describes a circular of the same radius in the same field. Let ${K_2}$ be the kinetic energy of the α-particle. That is we have $r = R$, and $B = {B_0}$ in this case. Also we know that the $\alpha$-particle is similar to the helium nucleus, whose charge is twice that of the proton, and mass is four times of the proton, that is, $q = 2e$ and $m = 4{m_p}$. Substituting these values in (1) we get the kinetic energy of the $\alpha$-particle as
${K_2} = \dfrac{{{{\left( {2e} \right)}^2}{B_0}^2{R^2}}}{{2\left( {4{m_p}} \right)}}$
On simplifying, we get
${K_2} = \dfrac{{{e^2}{B_0}^2{R^2}}}{{2{m_p}}}$ ………….(3)
From (2) and (3)
${K_2} = {K_1}$
According to the question, the kinetic energy of the proton is ${K_1} = 1{\text{MeV}}$. Substituting this above, we get
${K_2} = 1{\text{MeV}}$
Thus, the kinetic energy of the $\alpha$-particle is also equal to $1{\text{MeV}}$.

Hence, the correct answer is option A.

Note: The circular path followed by the charged particle is due to the fact that the magnetic force always acts perpendicular to the velocity of the charged particle. So this force will provide the required centripetal force for the charge to move in a circular path.