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

Question

UnlockIAS · Accepted Answer

Correct — D, eV. Potential difference is defined as work per unit charge, V = W/q, so moving a charge q through a potential difference V does work W = qV on it. Here the charge is e coulomb and the potential difference is V volts, so the work done is eV joules. Because the electron starts from rest it has no initial kinetic energy, and by the work-energy theorem the whole of that work appears as kinetic energy: final energy = eV. That is the entire derivation, and the mass never enters it. Mass enters only if you go one step further and ask for the speed: setting the kinetic energy ½Mv² equal to eV and solving gives v = √(2eV/M). The setter has put that half-remembered formula in the option list on purpose. Dimensional analysis alone kills the other three without any physics. Coulomb × volt = joule, so only (d) has the dimensions of energy at all. A coulomb per volt is a farad, the unit of capacitance, so (a) e/V is a capacitance. (c) eV/M is joules per kilogram, which is energy per unit mass — the same dimensions as velocity squared, and in fact eV/M equals ½v², the kinetic energy carried by each kilogram rather than the energy the question asks for. (b) MeV multiplies energy by a mass and gives joule-kilograms. Note the typography: in the booklet (A) and (C) are printed as stacked fractions, e over V and eV over M, which our transcription flattens to e/V and eV/M — (C) eV/M and (D) eV are genuinely different options and not two renderings of the same thing. Because (d) is exactly right, that printing quirk does not damage the question.

Why the other options are wrong — (a) Printed in the booklet as e over V. Dividing charge by potential difference gives coulombs per volt, which is the farad — the unit of capacitance, not of energy. The option exists for the candidate who remembers that e and V belong together in the formula but not which way round; the definition V = W/q rearranges to W = qV, a product, never a quotient.  (b) A double trap. Taken as algebra it is M × e × V, mass times charge times potential, whose units are kilogram-joules — not an energy. Taken as a symbol, 'MeV' is the standard particle-physics abbreviation for megaelectronvolt, which genuinely is an energy, so a candidate who reads the three characters instead of the algebra can convince himself this is the right dimensional shape.  (c) Printed as eV over M, and the strongest wrong answer. It is the trap for anyone who reaches for the speed formula v = √(2eV/M) and stops halfway: eV/M is ½v², the energy per unit mass in joules per kilogram, a specific energy. The stem asks for the final energy in joules, not the energy per kilogram, and dividing by the mass is exactly the step the work-energy theorem does not require.

Concept: Electric potential is defined as potential energy per unit charge, and the volt is defined so that one joule of work is done in moving one coulomb through one volt. Everything on this card follows from that single definition. A charge q released from rest and allowed to fall through a potential difference V has work qV done on it by the electric field, and since the field is conservative that work is exactly the loss of electrostatic potential energy, which reappears as kinetic energy. It is the electrical version of a stone falling through a height h and arriving with kinetic energy mgh — the field does the work, the body picks up the speed, and the identity of the body only matters if you want the speed rather than the energy. This is also the origin of the electronvolt as a unit: one electronvolt is defined as the energy an electron gains falling through one volt. Since the 2019 redefinition of the SI fixed the elementary charge at exactly 1.602176634 × 10⁻¹⁹ coulomb, one electronvolt is exactly 1.602176634 × 10⁻¹⁹ joule. Physicists quote particle energies in eV, keV, MeV, GeV and TeV precisely because those numbers are the accelerating voltages, which is why an 'X-ray tube run at 100 kV' and 'X-rays of up to 100 keV' say the same thing twice.

State PCS 2025 Q90 — Science & Technology PYQ with Answer

State PCS 2025 Q90 — Science & Technology PYQ with Answer