Question

${H^ + },H{e^ + }$ and ${O^{2 + }}$ all have the same kinetic energy passes through the region in which there is uniform magnetic field perpendicular to their velocity. The masses of ${H^ + },H{e^ + }$ and ${O^{2 + }}$ are $1$, $14$ and $16amu$ respectively. Then,
A) ${H^ + }$ will be deflected most
B) ${O^{2 + }}$ will be deflected most
C) $H{e^ + }$ and ${O^{2 + }}$ will be deflected equally
D) All will be deflected equally

Answer 1

Hint:- Using the expression of radius of curvature of the particle moving in the magnetic field –
$qvB = \dfrac{{m{v^2}}}{r}$
Find the relation between radius, mass and charge.
Then, put the values of charge and mass in that relation.

Complete step by step answer:
From the question, we know that the particle is moving in the uniform magnetic field perpendicular to their velocity. So, when the charged particles enter a magnetic field, then a force acts on the particle which will act as centripetal force.
$\therefore qvB = \dfrac{{m{v^2}}}{r} \\\ \Rightarrow r = \dfrac{{mv}}{{qB}} \cdots \left( 1 \right) \\\$
It is also given in the question, ${H^ + },H{e^ + }$ and ${O^{2 + }}$ all have the same kinetic energy. So, let the kinetic energy be $K$.
We know that –
$K = \dfrac{1}{2}m{v^2} \\\ \Rightarrow {v^2} = \dfrac{{2K}}{m} \\\ \therefore v = \dfrac{{\sqrt {2K} }}{m} \cdots \left( 2 \right) \\\$
Putting the value of $v$ from equation $\left( 2 \right)$ in equation $\left( 1 \right)$, we get –
$\Rightarrow r = \dfrac{m}{{qB}}\sqrt {\dfrac{{2K}}{m}} \\\ \Rightarrow r = \dfrac{m}{{qB\sqrt m }}\sqrt {2K} \\\ \Rightarrow r = \dfrac{{\sqrt {2Km} }}{{qB}} \\\ \Rightarrow r = \dfrac{{\sqrt {2K} }}{B} \times \dfrac{{\sqrt m }}{q} \\\$
In the above expression, we can see that, $\dfrac{{\sqrt {2K} }}{B}$ is the constant value as the kinetic energy is also the same.
$\therefore r\alpha \dfrac{{\sqrt m }}{q} \cdots \left( 3 \right)$
Now, according to the question, we know that masses of ${H^ + },H{e^ + }$ and ${O^{2 + }}$ are $1$, $14$ and $16amu$ respectively. So, putting values for ${H^ + },H{e^ + }$ and ${O^{2 + }}$ in equation $\left( 3 \right)$, we get –
$\Rightarrow {r_{{H^ + }}}\alpha \dfrac{{\sqrt 1 }}{1} \\\ \therefore {r_{{H^ + }}}\alpha 1 \\\$
$\Rightarrow {r_{H{e^ + }}}\alpha \dfrac{{\sqrt 4 }}{1} \\\ \therefore {r_{H{e^ + }}}\alpha 2 \\\$, and
$\Rightarrow {r_{{O^{2 + }}}}\alpha \dfrac{{\sqrt {16} }}{2} \\\ \therefore {r_{{O^{2 + }}}}\alpha 2 \\\$
Therefore, from the above we can conclude that $H{e^ + }$ and ${O^{2 + }}$ will get deflected equally and since the radius of ${H^ + }$ is smallest it will deflect most.

Hence, the correct option is (A) and (C).

Note: - Remember that it is given in the question that kinetic energy is the same for all and particles pass through the region of uniform magnetic field. Since, the magnetic field is uniform it is also constant and kinetic energy is also constant for this. So, $\dfrac{{\sqrt {2K} }}{B}$ is made as constant.