Question

A proton and $\alpha $— particle (with their masses in the ratio of 1: 4 and charges in the ratio of 1: 2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii ${r_p}:{r_\alpha }$of the circular path described by them will be:
A.)$1:\sqrt 2 $
B.)1: 2
C.)1: 3
D.)$1:\sqrt 3 $

Answer 1

Hint: In these types of questions you need to remember the basic concepts of proton and alpha and also show that $r \propto \sqrt {\dfrac{m}{q}} $to find the correct option.

Complete step-by-step answer:
According to the question ratio of the mass of alpha particle and proton is given as $\dfrac{{{m_p}}}{{{m_\alpha }}} = \dfrac{1}{4}$ and the ratio of the charge of proton and alpha particle is given as $\dfrac{{{q_p}}}{{{q_\alpha }}} = \dfrac{1}{2}$

Since the potential difference is generated in them, therefore, energy in the particle is given by

$E = q\vartriangle V$Here $\vartriangle V$is the potential difference

We know that this energy will be converted into kinetic energy .i.e. $E = q\vartriangle V = \dfrac{1}{2}m{v^2}$ (equation 1) where v is the velocity of the particle

According to the given information uniform magnetic field is set up perpendicular to the velocities of the proton and alpha particle and since we know that whenever the velocity and the magnetic field are perpendicular to each other particle goes under a circular motion whose radii is represented as;

$r = \dfrac{{mv}}{{qB}}$ (Equation 2)

Here v is the velocity and B is the magnetic field and q is the charge in the particle
Finding the value of v form equation 1 and substituting in equation 2

$\dfrac{{2q\vartriangle V}}{m} = {v^2}$

\[v = \sqrt {\dfrac{{2q\vartriangle V}}{m}} \]

Substituting the value of V in the formula of radius of a circular path due to uniform magnetic field acting perpendicular to the velocities of particles.

$r = \dfrac{{mv}}{{qB}}$$ \Rightarrow $$r = \dfrac{m}{{qB}}\sqrt {\dfrac{{2q\vartriangle V}}{m}} $
$ \Rightarrow $$r = \dfrac{1}{B}\sqrt {\dfrac{{2m\vartriangle V}}{q}} $ (Equation 3)
Let use the equation 3 to find the relation of radii

For proton

${r_p} = \dfrac{1}{B}\sqrt {\dfrac{{2{m_p}\vartriangle V}}{{{q_p}}}} $

For alpha particle

${r_\alpha } = \dfrac{1}{B}\sqrt {\dfrac{{2{m_\alpha }\vartriangle V}}{{{q_\alpha }}}} $
By the equation 3, we can say that $r \propto \sqrt {\dfrac{m}{q}} $

Therefore $\dfrac{{{r_p}}}{{{r_\alpha }}} = \sqrt {\dfrac{{{m_p}}}{{{q_p}}}} \times \sqrt {\dfrac{{{q_\alpha }}}{{{m_\alpha }}}} $ (equation 4)
Substituting the given values in equation 4
$ \Rightarrow $ $\dfrac{{{r_p}}}{{{r_\alpha }}} = \sqrt {\dfrac{1}{4}} \times \sqrt {\dfrac{2}{1}} $

Hence, $\dfrac{{{r_p}}}{{{r_\alpha }}} = \dfrac{1}{{\sqrt 2 }}$

Therefore option A is the correct answer.

Note: In this question magnetic field was a key to solving the issue so we should know what the magnetic field is; it is a vector field that describes the magnetic influence of electrical charges in relative motion and magnetized materials. A charge which moves parallel to a current of other charges is experiencing a force perpendicular to its direction.