JEE Main 2024 · 30 January · morning · Single correct · +4 / −1

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5ᵗʰ excited state of a hydrogen atom is :

  1. (a) 4
  2. (b) 1
  3. (c) 1/2
  4. (d) 1/4
Answer and solution

Answer (c) 1/2

Solution

Physics sir · 3 steps

In any Bohr orbit, potential energy is minus twice the kinetic energy, so the level number does not matter.

  1. Force balance. The Coulomb pull supplies the centripetal force: mv²/r = ke²/r². Multiply both sides by r and halve to get the kinetic energy.

    K = 1/2mv² = ke²/2r, U = -ke²/r = -2K

  2. Take magnitudes. |K|/|U| = (ke²/2r)/(ke²/r) = 1/2. The radius cancels, so this ratio is the same in every orbit.
  3. So. The 5th excited state is n = 6, but n never enters the ratio. The answer is 1/2, option (c).

The idea to keep. For any electron held by a 1/r² force: U = -2K and E = -K. Ratios between K, U and E are fixed and do not depend on n, so do not waste time finding the orbit.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2024.