Potential energy of electron in excited hydrogen atom

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)

• 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 $n\hbar$. Aug 21, 2019 at 17:01

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.
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 = $$3.40\ \mathrm{eV}$$.
• 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. Aug 21, 2019 at 17:09