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Hydrogen (H) electron density

I will not pay too much attention to the theoretical proof of these equations, concentrating more on the visualization of these electron-density clouds. This visualization will be provided for Atomistry - chemical elements website.

Hydrogen electron density

The radial electron density distribution for an electron in the ground state of a hydrogen atom is given by the probability density function \(\rho(r)\), which describes the likelihood of finding the electron in a thin spherical shell at a distance \(r\) from the nucleus. In atomic theory, the electron density is proportional to the square of the wave function, multiplied by \(r^2\) due to spherical symmetry in spherical coordinates.

For the hydrogen atom in its ground state \((n=1)\), the wave function \(\psi_{1s}(r)\) is: \[\psi_{1s}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}\] where \(a_0\)​ is the Bohr radius.

The electron density \(\rho(r)\), is given by:

\[\rho(r) = |\psi_{1s}(r)|^2 \cdot 4 \pi r^2\]

Substituting \(\psi_{1s}(r)\) into this expression, we get:

\[\rho(r) = \left( \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \right)^2 \cdot 4 \pi r^2 = \frac{4}{a_0^3} r^2 e^{-2r/a_0}\]

Therefore, the radial electron density distribution for the ground state of a hydrogen atom is:

\[\rho(r) = \frac{4}{a_0^3} r^2 e^{-2r/a_0}\]

The Bohr radius is defined by: \[a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2}\] where: \(\epsilon_0 \) is the permittivity of free space; \(\hbar\) is the reduced Planck's constant, \(m_e\) is the electron mass, and \(e\) is the elementary charge and it is equal to \(5.29177210544(82) \times 10^{−11} m \approx 0.0529 nm \approx 0.529 \mathring{A} \). This is not normalized integral and the integration from 0 to \(\infty \) will not be equal to 1.


Published: 2024-10-31 16:14:10
Updated: 2024-10-31 19:17:21

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