When a hydrogen atom is raised from the ground state to an excited state, (a) both K.E. and P.E. increase (b) both K.E. and P.E. decrease (c) the P.E. decreases and K.E. increases (d) the P.E. increases and K.E. decreases.

Answer: (d). KE = (ke²)/(2r) and PE = -ke²/r. In an excited state r is larger: KE decreases, PE becomes less negative (increases). So the answer is (d) the P.E. increases and K.E. decreases.

When a hydrogen atom is raised from the ground state to an excited… | NEET 1995 (AIPMT)

Q59NEET 1995 · AIPMTEasy

Energy of the Electron & Energy Levels

When a hydrogen atom is raised from the ground state to an excited state,

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Answer: (d)

$KE = \frac{ke^2}{2r}$ and $PE = -\frac{ke^2}{r}$ .
In an excited state $r$ is larger: KE decreases, PE becomes less negative (increases).
So the answer is (d) the P.E. increases and K.E. decreases.

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