Question

A helium ion and a hydrogen ion are accelerated from rest through a potential difference of V to velocities ${v_{He}}$ and ${v_H}$ ​ respectively. If helium has lost one electron, the ratio of $\dfrac{{{v_{He}}}}{{{v_H}}}$ is:
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{2}$
(C) $1$
(D) $\sqrt 2$

Answer 1

Hint : Here we can approach the question by using the concept of ionization of atoms. The acceleration of the ions in this case causes it to get excited from one energy level to the other. At each state, the electrons contain a value of energy which is the energy level. We should compare the energy gained by both the ions after its acceleration from the state of rest.

Complete Step By Step Answer:
In our question, we are given a helium and a hydrogen ion. Both these ions are accelerated from rest through a potential difference of V to velocities ${v_{He}}$ and ${v_H}$ ​ respectively. It is also said that the helium ions lose one electron further.
In such a situation, the charge of the H and the He ions will be the same. Thus they gain the same value of kinetic energy after traversing through a field that is equal to a value of qV.
Equating the kinetic energies, we obtain
$\dfrac{1}{2}{m_{He}}{v^2}_{He} = \dfrac{1}{2}{m_H}{v^2}_H$
$\Rightarrow \dfrac{{v_{He}^2}}{{v_H^2}} = \dfrac{{{m_H}}}{{{m_{He}}}}$
Now, upon taking the ratio of the masses of Hydrogen and Helium, we obtain
$\dfrac{{{m_H}}}{{{m_{He}}}} = \dfrac{1}{4}$
$\Rightarrow \dfrac{{v_{He}^2}}{{v_H^2}} = \dfrac{1}{4}$
$\therefore \dfrac{{{v_{He}}}}{{{v_H}}} = \dfrac{1}{2}$
Thus we can conclude that option B is the correct answer among the given options..

Note :
The kinetic energies that are equated in the question, is equal to the field qV as it is given that the potential difference is equal. We know that in an equipotential surface, extra work will not be required to move the bodies. The value of q can be substituted by the charge of an electron since electrons are involved in this process. Hence the value of kinetic energy is equal to eV.