Here we have introduced the dipole moment μ of ρ as an integral over the charge distribution. Setting V(0) as the zero energy, the interaction becomes Where we took the origin 0 somewhere within ρ. That is, V is given by a two-term Taylor expansion, If the electric field is of macroscopic origin and the charge distribution is microscopic, it is reasonable to assume that the electric field is uniform over the charge distribution. If this charge distribution is non-polarizable its interaction energy with an external electrostatic potential V(r) is Before turning to quantum mechanics we describe the interaction classically and consider a continuous charge distribution ρ(r). The Stark effect originates from the interaction between a charge distribution (atom or molecule) and an external electric field. Therefore, the formerly degenerate energy levels will split into slightly lower and slightly higher energy levels. However, in an electric field, there will be hybrid orbital (also called quantum superposition) of the 2s and 2p states where the electron tends to be to the left, which will acquire a lower energy, and other hybrid orbital where the electron tends to be to the right, which will acquire a higher energy. For example, in the Bohr model, an electron has the same energy whether it is in the 2s state or any of the 2p states. The Stark effect can lead to splitting of degenerate energy levels. Other things equal, the effect of the electric field is greater for outer electron shells, because the electron is more distant from the nucleus, so it travels farther left and farther right. In another way of viewing it, if an electronic state has its electron disproportionately to the left, its energy is lowered, while if it has the electron disproportionately to the right, its energy is raised. An electric field pointing from left to right, for example, tends to pull nuclei to the right and electrons to the left.
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