# Bohr radius for Yukawa potential

The Bohr radius is defined as $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2m_{e}},$$ which can be regarded as an rough estimate of the ground state atomic orbital 'radius'.

If we go back to fundamentals and consider the wavefunction for the 1s orbital, we can find it has a radius
$$\psi_{1s} \sim e^{-\frac{r}{a_0}}$$ which justifies the definition of the Bohr radius.

Now, my question is about the Hydrogen atom with a Yukawa potential, $$V(r) \sim \frac{e^{-\frac{r}{r_0}}}{r}.$$ Compared with Coulomb's potential, this potential is more localized. Ideally, we can solve Schrödinger's equation and may find a similar 1s wavefunction form, $$\psi_{1s, Yukawa} \sim e^{-\frac{r}{\alpha}}$$ where $$\alpha$$ is the 'Bohr radius analog' for this Yukawa Hydrogen atom.

Does anyone know what the form of $$\alpha$$ should be? I tried to solve the equation by myself, but was trapped in the math.

• It is solved numerically; i.e. probably the solution cannot be expressed in closed form. link.springer.com/article/10.1007/s11433-004-0020-5 May 15, 2023 at 15:27
• Yes, of course. Thank you! May 24, 2023 at 15:35

• Since we are talking about the (s) ground state, integrate out angular dependence, and your radial hamiltonian, having absorbed a factor of $$m/\hbar^2$$ in its units, reduces to just $$H= -\frac{1}{2 r^2 }\partial_r r^2 \partial_r - \frac{e^{-r/r_0}}{a_0 ~r}.$$ I assume you checked your Ansatz wavefunction $$e^{-r/\alpha}$$ is not an exact eigenfunction of this hamiltonian, but that's not the point. Instead, you are meant to check whether variation of the free parameter α yields a stationary $$\langle H\rangle$$ on it!
• So, plug in and do the simple integrals, $$\langle H\rangle= {\int_0^\infty \!dr ~r^2 ~ \left [e^{-r/\alpha}\left ( -\frac{1}{2 r^2 }\partial_r r^2 \partial_r \right )e^{-r/\alpha} - {e^{-(r/\alpha)(2+\alpha/r_0)} \over a_0 ~r }\right ] \over \int_0^\infty \!\!dr ~r^2 ~ e^{-2r/\alpha} }$$ to find $$= \frac{1}{2 \alpha^2 } -{1\over a_0 \alpha (1+ \alpha/(2r_0))^2}.$$
• Finally, apply the minimum condition $$\partial_\alpha \langle H \rangle =0$$ to obtain $$a_0= \alpha\frac{1+\frac{3\alpha}{2r_0}}{ (1+\frac{\alpha}{2r_0})^3}.$$
It is evident that the Yukawa potential goes to the Coulomb potential as $$r_0\gg a_0\sim \alpha$$, so if you expand in $$\alpha/r_0$$ you find $$\alpha\sim a_0 \left (1 +\tfrac{3}{4}(\alpha/r_0)^2+...\right )\sim a_0 \left (1 +\tfrac{3}{4}( a_0/r_0)^2+O (a_0^3)\right ),$$ a lengthening w.r.t. the Coulombic Bohr radius!
This is, naturally, not unexpected. For infinite $$r_0$$, the Coulomb potential, one has an infinity of bound states, but for vanishing $$r_0$$ the entire potential vanishes and there can be no bound states. In fact, according to this ref, Table 3, at the critical $$r_0\sim a_0/1.2$$, even the ground state is lost.