NDA (II) 2018 General Ability Test · Question 72 of 146

Correct — B, 1·6 × 10⁻¹⁶ J. The work done in moving a charge q through a potential difference V is W = qV, and for a free electron that work appears as kinetic energy. The magnitude of the electronic charge is 1·6 × 10⁻¹⁹ C and the potential difference is 1 kV, which is 10³ V. Multiplying, the energy gained is 1·6 × 10⁻¹⁹ × 10³ = 1·6 × 10⁻¹⁶ J. In the units physicists actually use for this, the same answer reads as 1 keV, because one electronvolt is defined as the energy an electron gains across one volt, namely 1·6 × 10⁻¹⁹ J, and a kilovolt gives it a thousand of them.

• (a)1·6 × 10⁻¹⁹ J — This is the energy for a potential difference of one volt — that is, one electronvolt. The factor of 10³ from the kilo in '1 kV' has been dropped, which is the single most common slip in this calculation.
• (c)1 × 10⁻¹⁹ J — Loses both the 1·6 factor from the electronic charge and the factor of a thousand from the kilovolt. The charge on an electron is 1·6 × 10⁻¹⁹ C, not 1 × 10⁻¹⁹ C.
• (d)1 × 10⁻¹⁶ J — The right power of ten, so the kilovolt has been handled correctly, but the 1·6 from the electronic charge has been rounded away. The paper offers this deliberately to catch a student who tracks exponents but not coefficients.

Potential difference is defined as work done per unit charge, so a charge q taken through a potential difference V has work qV done on it. For a free charge with nothing to resist its motion, that work goes entirely into kinetic energy — an electron accelerated from rest through V volts arrives with kinetic energy eV. Because the joule is an awkwardly large unit at this scale, the energy is usually quoted in electronvolts, and the numerical value in electronvolts is simply the accelerating voltage.

Two habits secure the mark. First, convert the prefix before multiplying — write 1 kV as 10³ V on the page, because leaving it as 'k' is how the factor of a thousand gets lost. Second, keep the coefficient and the exponent as separate pieces of the answer and check both against the options, since the paper has planted one distractor that gets the exponent right and the coefficient wrong and another that does the reverse. It is also worth knowing the shortcut the working reveals: an electron through V volts carries V electronvolts, so 1 kV gives 1 keV, 100 kV gives 100 keV, and so on — the numbers used to describe electron microscopes and X-ray tubes.

• The work done on a charge moved through a potential difference is W = qV.
• The magnitude of the electronic charge is 1·6 × 10⁻¹⁹ C.
• One electronvolt is 1·6 × 10⁻¹⁹ J, the energy an electron gains across a potential difference of one volt.
• 1 keV = 10³ eV = 1·6 × 10⁻¹⁶ J; 1 MeV = 10⁶ eV.
• An electron accelerated through V volts acquires a kinetic energy of V electronvolts, which is the scale used for electron microscopes and X-ray tubes.

• Forgetting the kilo and answering 1·6 × 10⁻¹⁹ J.
• Rounding the electronic charge to 1 × 10⁻¹⁹ C and losing the 1·6 factor.
• Treating the electronvolt as a unit of potential rather than of energy.

NDA sets this as a one-line substitution, and the options are always built so that a slip in either the coefficient or the exponent lands on a wrong answer that looks plausible.

• practice — not a real PYQ

  An electron accelerated from rest through a potential difference of 500 V acquires a kinetic energy of

  • (a)500 J
  • (b)500 eV
  • (c)1·6 × 10⁻¹⁹ eV
  • (d)500 keV

  Answer(b) 500 eV — an electron taken through V volts gains V electronvolts of kinetic energy, by definition of the unit.

• practice — not a real PYQ

  One electronvolt is equal to

  • (a)1·6 × 10⁻¹⁹ J
  • (b)1·6 × 10⁻¹⁶ J
  • (c)1·6 × 10⁻¹² J
  • (d)1 J

  Answer(a) 1·6 × 10⁻¹⁹ J — the energy an electron gains across a potential difference of one volt, numerically equal to the electronic charge.