Question

Ionized hydrogen atoms and $\alpha -$particles with the same momenta enter perpendicular to a constant magnetic field, $B$. The ratio of their radii of their paths ${r_H}:{r_\alpha }$ will be:
(A) $2:1$
(B) $1:2$
(C) $4:1$
(D) $1:4$

Answer 1

To solve this question, we need to use the formula for the radius of the circular path followed by a charged particle when it enters in a magnetic field. Then, equating the momenta for both the particles we can get the final answer.
Formula used: The formula used for solving this question are given by
$r = \dfrac{{mv\sin {{\theta }}}}{{qB}}$, 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 $v$, an angle of ${{\theta }}$ with it.

Complete step-by-step solution:
We know that the radius of a charged particle inside a magnetic field is given by the question
$\Rightarrow$ $r = \dfrac{{mv\sin {{\theta }}}}{{qB}}$
Now, according to the question both the particles enter perpendicular to the magnetic field. This means that ${{\theta }} = {90^ \circ }$. Substituting this above, we get
$\Rightarrow$ $r = \dfrac{{mv}}{{qB}}$ ………... (1)
Now, we know that the momentum is given by
$\Rightarrow$ $p = mv$.........................(2)
Putting (2) in (1) we get
$\Rightarrow$ $r = \dfrac{p}{{qB}}$
For the $\alpha -$particle, we get the radius as
$\Rightarrow$ ${r_\alpha } = \dfrac{{{p_\alpha }}}{{{q_\alpha }B}}$.............(3)
Also, for the ionized hydrogen atom, we get the radius as
${r_H} = \dfrac{{{p_H}}}{{{q_H}B}}$..............(4)
Dividing (3) by (4) we get
$\Rightarrow$ $\dfrac{{{r_\alpha }}}{{{r_H}}} = \dfrac{{{p_\alpha }}}{{{p_H}}} \times \dfrac{{{q_H}}}{{{q_\alpha }}}$
According to the question, the momentum of the $\alpha -$particle and the ionized hydrogen atom are equal, that is, ${p_\alpha } = {p_H}$. Substituting this above, we get
$\dfrac{{{r_\alpha }}}{{{r_H}}} = \dfrac{{{p_H}}}{{{p_H}}} \times \dfrac{{{q_H}}}{{{q_\alpha }}}$
$\Rightarrow \dfrac{{{r_\alpha }}}{{{r_H}}} = \dfrac{{{q_H}}}{{{q_\alpha }}}$.............. (5)
Now, since an ionized hydrogen atom is nothing but a proton. So the charge on the hydrogen atom is
$\Rightarrow$ ${q_H} = e$..............(6)
Also, we know that an alpha particle is similar to a helium nucleus whose charge is equal to two times that of the proton. So the charge on the alpha particle is
$\Rightarrow$ ${q_\alpha } = 2e$..............(7)
Putting (6) and (7) in (5) we get
$\dfrac{{{r_\alpha }}}{{{r_H}}} = \dfrac{e}{{2e}}$
$\Rightarrow \dfrac{{{r_\alpha }}}{{{r_H}}} = \dfrac{1}{2}$
Finally taking the reciprocal we get
$\Rightarrow \dfrac{{{r_H}}}{{{r_\alpha }}} = 2$
Thus, the ratio of their radii of their paths ${r_H}:{r_\alpha }$ is equal to $2:1$.

Hence, the correct answer is option A.

Note: The phenomenon of the circular path of a charged particle moving perpendicular to the magnetic field, which is discussed in this question, is the working principle of a cyclotron. A cyclotron is used for producing a beam of charged particles by accelerating them to a high speed.