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The total energy of an electron in the first excited state of hydrogen is about $-3.4$ eV . Its kinetic energy in this state is
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The correct answer is:
$3.4 \mathrm{eV}$
The kinetic energy is equal to the negative of total energy.
Kinetic energy of electron $K=\frac{Z e^2}{8 \pi \varepsilon_0 r}$
Potential energy of electron $U=-\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r}$
:. Total energy
$E=K+U=\frac{Z e^2}{8 \pi \varepsilon_0 r}-\frac{Z e^2}{4 \pi \varepsilon_0 r}$
$\begin{aligned} & \text { or } E=-\frac{Z e^2}{8 \pi \varepsilon_0 r} \\ & \text { or } E=-K \\ & \text { or } K=-E=-(-3.4)=3.4 \mathrm{eV} \\ & \end{aligned}$
Kinetic energy of electron $K=\frac{Z e^2}{8 \pi \varepsilon_0 r}$
Potential energy of electron $U=-\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r}$
:. Total energy
$E=K+U=\frac{Z e^2}{8 \pi \varepsilon_0 r}-\frac{Z e^2}{4 \pi \varepsilon_0 r}$
$\begin{aligned} & \text { or } E=-\frac{Z e^2}{8 \pi \varepsilon_0 r} \\ & \text { or } E=-K \\ & \text { or } K=-E=-(-3.4)=3.4 \mathrm{eV} \\ & \end{aligned}$
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