## Related questions with answers

A hydrogen atom is in its first excited state (n = 2). Calculate the potential energy of the system.

Solutions

VerifiedOnly interaction between electron and nucleus (proton), that we are considering, is electrostatic interaction, which can be described as

$\begin{align*} U &= - \frac{1}{4 \pi \epsilon_0} \frac{ q_1 q_2 }{r}, \tag{9} \end{align*}$

where $U$ is potential energy of the system electron-proton, $q_{1,2}$ are charges of electron/proton (they have the same magnitude $e$) and $r$ is distance between electron and nucleus, which is radius of orbit (result (2)). Minus sing in (9) denoted that this system is bound (we would have apply energy to this system in order to break it apart). Therefore, potential energy $U$ of the system electron-proton is

$\begin{align*} U &= - \frac{ 1 }{ 4 \pi \cdot 8.85 \cdot 10^{-12} \: \mathrm{F/m} } \cdot \frac{ \left( 1.6 \cdot 10^{-19} \: \mathrm{C} \right)^2 }{2.12 \cdot 10^{10} \: \mathrm{m} } \\ U &= - 1.09 \cdot 10^{-18} \: \mathrm{J} \cdot \frac{1 \: \mathrm{eV}}{ 1.6 \cdot 10^{-19} \: \mathrm{J} } \tag{10} \\ U &= \boxed{ - 6.80 \: \mathrm{eV} }. \tag{11} \end{align*}$

**Given information**

State of the hydrogen atom: $n=2$,

We need to calculate the potential energy of the system $U$.

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