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The angular momentum of electron in an excited H atom is $\frac{h}{\pi}$. The potential energy (PE) of electron is?

Let $\frac{h}{\pi}=\frac{nh}{2\pi}$, therefore $n=2$.

So, $$E=\frac{-13.6}{4}=\pu{-3.4eV}$$

The answer is $\pu{-6.8eV}$. What’s wrong with my solution? (I think the issue is with the formula, yet I would like to confirm it)

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    $\begingroup$ The Bohr model is fundamentally flawed and incorrect. The (orbital) angular momentum of an electron in an orbital with quantum number $l$ is given by $\sqrt{l(l+1)}\hbar$, not $n\hbar$. $\endgroup$ Aug 21, 2019 at 17:01

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This answer owes a big debt to @orthocresol, who got me back on track after my over-simplified answer to the OP’s question.

In the long obsolete Bohr theory of the hydrogen atom, the total energy of the $n$-th energy level, $E_n$, is $-2.179\times10^{-18}/n^2\ \mathrm J$, which is approximately $-2.18\times10^{-18}/n^2\ \mathrm J$. This is depicted in the figure below, with original figure reference therein.

Fig. 6.8 in Kotz et al.

For the $n$-th energy level, $E_n$ equals the sum of the kinetic energy, $T_n$, and the potential energy, $V_n$. Using the virial theorem, as per this link, the average kinetic energy, $T_\mathrm{ave}$, is minus one half of the average potential energy, $V_\mathrm{ave}$. Thus $$-V_\mathrm{ave}=2 T_\mathrm{ave}$$

Let $T_n$ be estimated by $T_\mathrm{ave}$ and $V_n$ be estimated by $V_\mathrm{ave}$. For integer $n\ge1$, $E_n=-2.18\times10^{-18}/n^2\ \mathrm J$. With $n = 2$, $E_2=-5.45\times10^{-19}\ \mathrm J$. Since $1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J$, $E_2 = -3.40\ \mathrm{eV}$. Therefore, $-3.40\ \mathrm{eV}=V_\mathrm{ave}+T_\mathrm{ave}=V_\mathrm{ave}-V_\mathrm{ave}/2=V_\mathrm{ave}/2$, so $V_\mathrm{ave}=-6.80\ \mathrm{eV}$. Then $T_\mathrm{ave}=3.40\ \mathrm{eV}$. In summary: average potential energy = $-6.80\ \mathrm{eV}$, average kinetic energy = $3.40\ \mathrm{eV}$, and total energy = $3.40\ \mathrm{eV}$.

This link is recommended for further information.

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    $\begingroup$ Actually, the -6.8 makes complete sense in the context of quantum mechanics: it is a known result (the virial theorem) that for a hydrogen atom, $\langle V\rangle = -2\langle T\rangle$, where $V$ and $T$ are potential and kinetic energies respectively. Now, because -3.4 is the total energy $E = T + V$ (the formula for $E$ is derived from a Hamiltonian which includes both kinetic & potential energies), this leads to the result that $T = +3.4$ and $V = -6.8$. What is baffling is the simultaneous and incoherent use of quantum mechanics with the old Bohr model for angular momentum. $\endgroup$ Aug 21, 2019 at 17:09

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