Cross section of ionization of atom by electrons per Lotz-Drevin ================================================================ In this law, we are talking about the interaction of an atom and an electron, which ionizes an atom. Equivalent electrons on the outer shell of the ionized atom are electrons with the same principal and orbital quantum numbers. In this case, the Lotz-Drewin approximation for the ionization cross section is considered. **Notation:** #. :math:`a_0` (:code:`a_0`) is :attr:`~symplyphysics.quantities.bohr_radius`. #. :math:`\mathrm{IE}_\text{H}` (:code:`IE_h`) is :attr:`~symplyphysics.quantities.hydrogen_ionization_energy`. .. TODO: find link TODO: fix file name TODO: move to `ionization` folder? .. py:currentmodule:: symplyphysics.laws.chemistry.cross_section_of_ionization_of_atom_by_electron_by_lotz_drevin .. py:data:: cross_section :attr:`~symplyphysics.symbols.chemistry.cross_section` of ionization. Symbol: :code:`sigma` Latex: :math:`\sigma` Dimension: :code:`area` .. py:data:: ionization_energy :attr:`~symplyphysics.symbols.basic.energy` of ionization of atoms. Symbol: :code:`E_i` Latex: :math:`E_\text{i}` Dimension: :code:`energy` .. py:data:: electron_energy :attr:`~symplyphysics.symbols.basic.energy` of ionizing electrons. Symbol: :code:`E` Latex: :math:`E` Dimension: :code:`energy` .. py:data:: first_coefficient A coefficient used in the calculation. Symbol: :code:`A` Latex: :math:`A` Dimension: :code:`dimensionless` .. py:data:: second_coefficient A coefficient used in the calculation. Symbol: :code:`B` Latex: :math:`B` Dimension: :code:`dimensionless` .. py:data:: electron_count A :attr:`~symplyphysics.symbols.basic.nonnegative_number` of equivalent electrons on the outer shell of the ionized atom. Symbol: :code:`N` Latex: :math:`N` Dimension: :code:`dimensionless` .. py:data:: law :code:`sigma = 2.66 * pi * a_0^2 * N * IE_h^2 / E_i^2 * A * (E / E_i - 1) / (E / E_i)^2 * log(1.25 * B * E / E_i)` Latex: .. math:: \sigma = \frac{2.66 \pi a_0^{2} N \mathrm{IE}_\text{H}^{2}}{E_\text{i}^{2}} \frac{A \left(\frac{E}{E_\text{i}} - 1\right)}{\left(\frac{E}{E_\text{i}}\right)^{2}} \log \left( 1.25 B \frac{E}{E_\text{i}} \right)