An electron of mass M kg and charge e coulomb travels from rest through a potential difference of V volts. The final energy in joules would be
eV
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.
| Option | Units | Verdict |
|---|---|---|
| eV | coulomb × volt = joule | ✓ Correct |
| eV/M | joule per kilogram — energy per unit mass | Not an energy |
| e/V | coulomb per volt = farad, a capacitance | Not an energy |
| MeV | kg × joule | Not an energy |
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.
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