An electron makes a transition from an excite | vedclass

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(C) For a hydrogen-like atom,the energy levels are given by:$T.E. = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$$P.E. = 2 	imes T.E. = -27.2 \frac{Z^2}{n^2} 	e

Vedclass · Accepted Answer

$K$.$E$. increases but $P$.$E$. and $T$.$E$. decrease.

Vedclass · Answer

(C) For a hydrogen-like atom,the energy levels are given by:$T.E. = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$$P.E. = 2 	imes T.E. = -27.2 \frac{Z^2}{n^2} 	ext{ eV}$$K.E. = -T.E. = 13.6 \frac{Z^2}{n^2} 	ext{ eV}$When an electron transitions from an excited state to the ground state,the principal quantum number $n$ decreases.As $n$ decreases:$1$. $K.E. = 13.6 \frac{Z^2}{n^2}$ increases.$2$. $T.E. = -13.6 \frac{Z^2}{n^2}$ becomes more negative,so it decreases.$3$. $P.E. = -27.2 \frac{Z^2}{n^2}$ becomes more negative,so 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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