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

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

U=14πϵ0q1q2r,\begin{align*} U &= - \frac{1}{4 \pi \epsilon_0} \frac{ q_1 q_2 }{r}, \tag{9} \end{align*}

where UU is potential energy of the system electron-proton, q1,2q_{1,2} are charges of electron/proton (they have the same magnitude ee) and rr 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 UU of the system electron-proton is

U=14π8.851012F/m(1.61019C)22.121010mU=1.091018J1eV1.61019JU=6.80eV.\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*}

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