JEE Main 2024 · 5 April · morning · Single correct · +4 / −1

An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is :

  1. (a) 2 E=3 U
  2. (b) E=U
  3. (c) 2 E=U
  4. (d) E=2 U
Answer and solution

Answer (c) 2 E=U

Solution

Physics sir · 3 steps

Let the centripetal force fix the kinetic energy, and you will find it is exactly half the size of U.

  1. Force balance. The Coulomb pull of the nucleus supplies the centripetal force: mv²/r = kZe²/r², where k = 1/4πε₀. Multiply both sides by r/2 to get 1/2mv² = kZe²/2r.
  2. Compare K and U. The potential energy of the electron is U = -kZe²/r. So the kinetic energy is K = kZe²/2r = -U/2.

    E = K + U = -U/2 + U = U/2

  3. So. E = U/2, which means 2E = U. That is option (c). Check the signs: U = -kZe²/r and E = -kZe²/2r, both negative, with U twice as large in size.

The idea to keep. For any orbit held by an inverse-square force, K = -E and U = 2E. Remember the set K : U : E = 1 : -2 : -1 and these questions take seconds.

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.