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(A) For a hydrogen atom,the kinetic energy $(K.E.)$ and potential energy $(P.E.)$ of an electron in an orbit of radius $r$ are given by:$K.E. = \frac{

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$P.E.$ increases and $K.E.$ decreases

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(A) For a hydrogen atom,the kinetic energy $(K.E.)$ and potential energy $(P.E.)$ of an electron in an orbit of radius $r$ are given by:$K.E. = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{2r}$$P.E. = -\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r}$As the hydrogen atom is raised from the ground state to an excited state,the principal quantum number $n$ increases,which causes the orbital radius $r$ to increase $(r \propto n^2)$.Since $K.E. \propto \frac{1}{r}$,as $r$ increases,$K.E.$ decreases.Since $P.E. \propto -\frac{1}{r}$,as $r$ increases,the magnitude of $P.E.$ decreases,but because it is negative,the value of $P.E.$ increases (becomes less negative).Therefore,$P.E.$ increases and $K.E.$ decreases.

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

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