An electron makes a transition from an excite | vedclass

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Question

(C) For a hydrogen-like atom,the energies are given by:Potential Energy $(P.E.)$ = $-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$Kinetic Energy

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Kinetic energy increases,but potential energy and total energy decrease.

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(C) For a hydrogen-like atom,the energies are given by:Potential Energy $(P.E.)$ = $-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$Kinetic Energy $(K.E.)$ = $-\frac{1}{2} (P.E.) = \frac{1}{8 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$Total Energy $(T.E.)$ = $\frac{1}{2} (P.E.) = -\frac{1}{8 \pi \varepsilon_0} \cdot \frac{Ze^2}{r}$As the electron transitions from an excited state to the ground state,the radius $(r)$ of the orbit decreases.Since $K.E. = \frac{k}{r}$,as $r$ decreases,$K.E.$ increases.Since $P.E. = -\frac{k'}{r}$,as $r$ decreases,$P.E.$ becomes more negative,meaning it decreases.Since $T.E. = -\frac{k''}{r}$,as $r$ decreases,$T.E.$ becomes more negative,meaning it decreases.Therefore,$K.E.$ increases,while $P.E.$ and $T.E.$ decrease.

An electron makes a transition from an excited state to the ground state of a hydrogen-like atom. Out of the following statements,which one is correct?

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