The energy gained is eV joules — option 1.

The derivation. Potential difference is defined as work done per unit charge :

\(V=\dfrac{W}{q}\qquad\Rightarrow\qquad W=qV\)

For an electron the charge is e, so the work done on it by the field is eV. Since it starts from rest, all of that work becomes kinetic energy :

\(\tfrac{1}{2}Mv^{2}=eV\qquad\Rightarrow\qquad v=\sqrt{\dfrac{2eV}{M}}\)

Note that the energy does not involve the mass at all; the mass enters only when one asks for the speed.

A dimensional check disposes of the rest.

The unit that comes from this. The electron volt is defined as exactly this quantity for a potential difference of one volt :

\(1\ \text{eV}=1.6\times10^{-19}\ \text{J}\)

It is the natural unit of energy in atomic and nuclear physics because the energies involved are so small in joules. Note the trap in option 4: MeV is read by a physicist as “mega electron volt”, a million electron volts — a real unit, but here it stands for mass times charge times voltage, which is not.

Where this is used. Electron guns in cathode-ray tubes and electron microscopes, X-ray tubes, and particle accelerators all work by exactly this principle: apply a potential difference and the charged particle emerges with energy qV.

Hence, the answer is eV.