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The total energy of an electron in the first excited state of hydrogen atom is about $-3.4 \mathrm{eV}$. Its kinetic energy in this state is:
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Verified Answer
The correct answer is:
$3.4 \mathrm{eV}$
$$
\mathrm{K} . \mathrm{E} .=\left|\frac{1}{2} P . E\right|
$$
But P.E is negative
$$
\begin{aligned}
\therefore \text { Total energy } & =\left|\frac{1}{2} P \cdot E\right|-P \cdot E \\
& =-\frac{P \cdot E}{2}=-3.4 \mathrm{eV} \\
\mathrm{K} . \mathrm{E} & =+3.4 \mathrm{eV}
\end{aligned}
$$
\mathrm{K} . \mathrm{E} .=\left|\frac{1}{2} P . E\right|
$$
But P.E is negative
$$
\begin{aligned}
\therefore \text { Total energy } & =\left|\frac{1}{2} P \cdot E\right|-P \cdot E \\
& =-\frac{P \cdot E}{2}=-3.4 \mathrm{eV} \\
\mathrm{K} . \mathrm{E} & =+3.4 \mathrm{eV}
\end{aligned}
$$
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