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 :
- (a) 2 E=3 U
- (b) E=U
- (c) 2 E=U
- (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.
- 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.
- 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
- 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.