Feb 12, 2023 · The bohr radius is derived by setting the centripetal force of electron to equal the coulomb force. $$\frac{mv^2}{r}=\frac{kq_1q_2}{r^2}.$$ The electron is not a particle orbiting the electron according to quantum mechanics. Its position is defined by its wave function. So, does the bohr radius make any sense quantum mechanically?

For the Bohr radius rb = h^2 ε0 /π m e^2; take the centrifugal force equals the static electric attractive force in the atom. The ratio of any two of the above numbers is the fine structure constant α=e^2/ε0 hc; and what is notable is that this number is non dimensional, and the three electron radii are obtained in independent processes ...

Jun 19, 2020 · Even though the Bohr model might be replaced by more sophisticated models, the Bohr radius still lives on as one of the 4 fundamental constants in what is known as atomic units: It is very popular in software for atomic and molecular scale simulations.

Feb 23, 2015 · So the Bohr radius for a muonium atom is $\frac{1}{0.995} = 1.005$ times larger than a that of a hydrogen atom. But this, again, is the distance between the two particles rather than the size of the atom. So we multiply by $\frac{\mu}{m_e}$ again to get the distance of the electron to the barycenter of the system (as we did for positronium).

May 30, 2022 · According to Wikipedia, the Bohr radius (already described here) is:. a physical constant, approximately equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.

May 15, 2023 · 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.

Jan 15, 2022 · $\begingroup$ @watertruck Bohr radius is a natural unit made of the constants in the problem. It is ...

Aug 6, 2020 · $\begingroup$ Since we know the Bohr model is false and there is no simple "radius" in real atoms, what do you mean by either of the relation being "correct" here? $\endgroup$ – ACuriousMind ♦ Commented Aug 6, 2020 at 14:52

May 31, 2020 · $^\dagger$ The Bohr radius provides the characteristic length scale for the problem. If you wish, you could recall that ground state wavefunction is of the form $\psi(r,\theta,\phi)=c e^{-r/a}$ , and evaluate the expected value of $\frac{1}{r}$ ; if you do so, you will find $\langle \frac{1}{r}\rangle = \frac{1}{a}$ , justifying the inversion ...

Aug 25, 2016 · From the Bohr/De Broglie postulate we have n λ = 2πr where λ is the De-Broglie wavelength , r is the radius corresponding to n and n is the quantum number. An electron in the state n=2 has more energy than that at n=1; That implies that the De- Broglie wavelength associated with the electron should also decrease ?