NDA (II) 2017 General Ability Test · Question 79 of 150

Correct — B, 13·6 eV. On the Bohr model the energy of the hydrogen electron in the nth level is minus 13·6 divided by n squared, in electronvolts. In the ground state n equals one, so the electron sits at minus 13·6 eV. Ionisation means lifting it to n equal to infinity, where the energy is zero, so the energy that must be supplied is 13·6 eV. All four options carry the same figure, and the question is really about the unit. An electronvolt is the energy an electron gains crossing one volt, about 1·6 times ten to the minus nineteen joule, which is the natural size for atomic processes.

• (a)13·6 MeV — A mega-electronvolt is a million times too large. Energies of that size belong to nuclear physics — binding energies per nucleon, alpha and beta decay, fission fragments — not to pulling one electron off an atom. Confusing the two scales by a factor of a million is the whole point of this option.
• (c)13·6 Joule — The joule is an everyday unit, far too coarse for a single atom. In joules the answer is about 2·18 times ten to the minus eighteen; 13·6 joule would be roughly the energy needed to ionise ten to the nineteen hydrogen atoms at once.
• (d)Zero — Zero would mean the electron is already free, which is the energy of the n equal to infinity level, not of the ground state. The hydrogen electron is bound, and energy must be spent to remove it — that is why the bound-state energy is written as a negative number.

Ionisation energy is the energy needed to take the most loosely held electron completely away from a neutral, gaseous atom in its ground state. For hydrogen the Bohr model gives the level energies as minus 13·6 divided by n squared electronvolts, so the ground state lies at minus 13·6 eV and the free electron at zero. Because the levels are discrete, the atom can only absorb the exact amounts that carry it from one level to another, which is why hydrogen has a line spectrum rather than a continuous one.

When every option repeats the same number, the question is testing whether you know the right order of magnitude. Atomic and chemical processes are measured in electronvolts — a few eV for a chemical bond, roughly 4 to 25 eV for the first ionisation energies of the elements. Nuclear processes are measured in millions of electronvolts. Hydrogen's 13·6 eV sits comfortably inside the atomic band, so the mega-electronvolt option can be dismissed on scale alone.

• The Bohr energy levels of hydrogen are given by minus 13·6 divided by n squared electronvolts.
• One electronvolt is about 1·602 times ten to the minus nineteen joule, so 13·6 eV is about 2·18 times ten to the minus eighteen joule.
• The ionisation energy is the same as the binding energy of the ground-state electron, taken with the opposite sign.
• The energy needed to lift the hydrogen electron from n equal to one to n equal to two is 10·2 eV, the first line of the Lyman series.
• Nuclear binding energies run to several mega-electronvolts per nucleon, about a million times the atomic scale.

• Reading the number without checking the unit when every option repeats the same figure.
• Mixing the atomic scale in electronvolts with the nuclear scale in mega-electronvolts.
• Calling the ground-state energy 13·6 eV instead of minus 13·6 eV — the bound state is negative and the ionisation energy is the positive amount you must supply.

NDA asks for this value directly, and also as excitation energies between levels or as a spectral-series question built on the same formula.

• practice — not a real PYQ

  The energy of the electron in the second orbit of a hydrogen atom, on the Bohr model, is

  • (a)-13·6 eV
  • (b)-6·8 eV
  • (c)-3·4 eV
  • (d)-1·51 eV

  Answer(c) -3·4 eV — using E = -13·6/n² with n = 2 gives -13·6/4 = -3·4 eV.

• practice — not a real PYQ

  One electronvolt is approximately equal to

  • (a)1·6 × 10⁻¹⁹ joule
  • (b)1·6 × 10⁻¹² joule
  • (c)9·1 × 10⁻³¹ joule
  • (d)6·6 × 10⁻³⁴ joule

  Answer(a) 1·6 × 10⁻¹⁹ joule — it is the charge on an electron multiplied by one volt.