Question:A proton (q = +e) and an alpha particle (q = +2e) are accelerated by the same voltage V. Which gains the greater kinetic energy, and by what factor?

The electric potential energy per unit charge at any point in space is termed as the electric potential at that point.

The electric potential is given by the following:

\(V = \frac{{PE}}{q}\)

Here, q is the charge.

When a charge q moves through a potential difference V, the change in potential energy is given by the following:

\(\Delta PE = qV\) … (i)

The charge on the proton,\({q_{\rm{p}}} = + e\).

The charge on the alpha particle,\({q_\alpha } = + 2e\).

Accelerating voltage is V.

When the charged particle is accelerated by a potential difference V, the potential energy gets converted into kinetic energy.

From the conservation of energy,

\(\begin{array}{c}\Delta KE + \Delta PE = 0\\\Delta KE = - qV\end{array}\)

As the potential difference is the same, the change in kinetic energy depends on the charge only.

\(\begin{array}{c}\frac{{\Delta K{E_\alpha }}}{{\Delta K{E_p}}} = \frac{{ - {q_\alpha }V}}{{ - {q_{\rm{p}}}V}}\\\frac{{\Delta K{E_\alpha }}}{{\Delta K{E_p}}} = \frac{{2eV}}{{eV}}\\\frac{{\Delta K{E_\alpha }}}{{\Delta K{E_p}}} = 2\end{array}\)

Thus, alpha particles gain higher kinetic energy compared to the proton by a factor of 2.