Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to:

We begin by recalling the energy principle for a charged particle moving through an electric potential difference. The work done by the electric field is converted into kinetic energy. Mathematically we state the relation

$$\text{Work done} \;=\; qV \;=\; \tfrac12 m v^{2},$$

where $$q$$ is the charge on the particle, $$V$$ is the potential difference, $$m$$ is the mass of the particle and $$v$$ is the final speed acquired from rest.

We now apply this relation separately to the hydrogen ion (proton) and to the singly ionised helium atom.

For the hydrogen ion (symbolically $$\text{H}^{+}$$):

Its charge is that of one proton, so $$q_{\text{H}} = +e$$. We denote its mass by $$m_{\text{H}}$$. Substituting these into the energy equation we have

$$q_{\text{H}} V = \tfrac12 m_{\text{H}} v_{\text{H}}^{2}.$$

Explicitly, that reads

$$e\,V = \tfrac12\,m_{\text{H}}\,v_{\text{H}}^{2}.$$

Solving for the speed $$v_{\text{H}}$$ gives

$$v_{\text{H}}^{2} = \frac{2eV}{m_{\text{H}}}, \qquad\text{so}\qquad v_{\text{H}} = \sqrt{\frac{2eV}{m_{\text{H}}}}.$$

For the singly ionised helium atom (symbolically $$\text{He}^{+}$$):

Although the helium nucleus contains two protons, the phrase “singly ionised” means it has lost only one electron, so its net charge is still just one elementary charge. Hence

$$q_{\text{He}} = +e.$$

The mass of a helium nucleus is essentially four times the proton mass, so we write

$$m_{\text{He}} = 4\,m_{\text{H}}.$$

Inserting these values into the same energy equation gives

$$q_{\text{He}} V = \tfrac12 m_{\text{He}} v_{\text{He}}^{2}$$

$$e\,V = \tfrac12 \left(4m_{\text{H}}\right) v_{\text{He}}^{2}.$$

Rearranging to isolate $$v_{\text{He}}^{2}$$:

$$v_{\text{He}}^{2} = \frac{2eV}{4m_{\text{H}}} = \frac{1}{4}\,\frac{2eV}{m_{\text{H}}},$$

and therefore

$$v_{\text{He}} = \sqrt{\frac{1}{4}\,\frac{2eV}{m_{\text{H}}}} = \frac{1}{2}\,\sqrt{\frac{2eV}{m_{\text{H}}}}.$$

Computing the ratio of the two speeds:

$$\frac{v_{\text{H}}}{v_{\text{He}}} = \frac{\sqrt{\dfrac{2eV}{m_{\text{H}}}}} {\dfrac{1}{2}\sqrt{\dfrac{2eV}{m_{\text{H}}}}} = \frac{\sqrt{\dfrac{2eV}{m_{\text{H}}}}}{\sqrt{\dfrac{2eV}{m_{\text{H}}}}/2} = 2.$$

Thus,

$$v_{\text{H}} : v_{\text{He}} = 2 : 1.$$

Hence, the correct answer is Option C.