# Schrodinger Wave Equation

Schrodinger wave equation is a mathematical wave equation developed by Schrödinger to describe the wave motion of an electron in a hydrogen atom. Here, the nucleus is surrounded by a vibrating electron wave which may be mathematically compared with stationary wave formed by a vibrating string fixed between two points.

The electron in the hydrogen atom is under constraint imposed by the attraction of the nucleus. An electron motion can be described by an equation analogous to that used to describe a stationary wave system whose solution in one dimension is given by the sine wave equation, $\Psi=A\sin\frac{2πx}{\lambda}\;\;\text{__(1)}$

where, $\Psi=\text{amplitude of the wave}$ $\lambda=\text{wavelength of the wave}$ $x=\text{distance from the origin}$ $A=\text{a constant}$

Differentiating with respect to $x$, $\frac{d\Psi}{dx}=A\frac{2π}{x}\cos\frac{2πx}{\lambda}$ $\text{and,}\;\;\frac{d^2\Psi}{dx^2}=A\frac{2π}{\lambda}\frac{2π}{\lambda}\left(-\sin\frac{2πx}{\lambda}\right)$ $=-\frac{4π^2}{\lambda^2}A\sin\frac{2πx}{\lambda}=-\frac{4π^2}{\lambda^2}\Psi$ $\therefore\frac{d^2\Psi}{dx^2}+\frac{4π^2}{\lambda^2}\Psi=0\;\;\text{__(2)}$

This is the equation for one dimensional standing wave which can be extended to describe motion of an electron in three dimensions by a second order partial differential equation.

$\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}+\frac{4π^2}{\lambda^2}\Psi=0\;\;\text{__(3)}$

where, $\Psi$ is a function of Cartesian coordinates $x$, $y$, $z$. This equation may also be written as, $\nabla^2\Psi+\frac{4π^2}{\lambda^2}\Psi=0\;\;\text{__(4)}$ where, $\nabla^2=\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}$

From de Broglie equation, $\lambda=\frac{h}{mv}\;\;\text{__(5)}$ From equations $(4)$ and $(5)$, $\nabla^2\Psi+\frac{4π^2m^2v^2}{h^2}\Psi=0\;\;\text{__(6)}$

If we consider the electron as a particle, the total energy $(E)$ of the system is the sum of kinetic energy $\left(\frac{1}{2}mv^2\right)$ and potential energy $(V)$. $\therefore E=\frac{1}{2}mv^2+V$$v^2=\frac{2(E-V)}{m}\;\;\text{__(7)}$

In order to calculate values for the energy states, the value of $v^2$ is substituted into equation $(6)$ to obtain a fundamental form of Schrodinger wave equation for a single particle in three dimensions. $\nabla^2\Psi+\frac{8π^2m}{h^2}(E-V)\Psi=0\;\;\text{__(8)}$ $\text{or,}\;\;\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}+\frac{8π^2m}{h^2}(E-V)\Psi=0\;\;\text{__(9)}$

The potential energy $(V)$ is given by, $V=\frac{-Ze^2}{r}\;\;\text{__(10)}$ where, $+Ze=\text{charge of the nucleus}$ $Z=\text{atomic number}$ $-e=\text{charge on the nucleus}$ $r=\text{distance between the electron and the nucleus}$

Hence, the equation $(9)$ for one electron atomic system in three dimensions may be written as $\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}+\frac{8π^2m}{h^2}\left(E+\frac{Ze^2}{r}\right)\Psi=0\;\;\text{__(11)}$

## Application of Schrodinger Wave Equation to Hydrogen Atom

Schrödinger wave equation for one electron atomic system is, $\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}+\frac{8π^2m}{h^2}\left(E+\frac{Ze^2}{r}\right)\Psi=0$ For hydrogen atom, $Z=1$, $\therefore\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}+\frac{8π^2m}{h^2}\left(E+\frac{e^2}{r}\right)\Psi=0\;\;\text{__(12)}$

This equation $(12)$ results from the assumptions that;

1. the behaviour of an electron in an atom is analogous to a system of standing wave.
2. de Broglie equation $\left(\lambda=\frac{h}{mv}\right)$ describes the wavelength of an electron.

Justification for the above equation for hydrogen atom is found in the fact that the solution of equation $(12)$ gives values of energy which agree well with those obtained experimentally from atomic spectra or calculated from the equation given by Bohr’s atomic model. $E=-\frac{2π^2me^4Z^2}{h^2n^2}\;\;\text{__(13)}$

That is, Schrodinger wave equation leads to exactly the same allowed energies as those deduced from the Bohr’s model. Beside that, the most probable radius for $1s$ electron in H atom from wave mechanics comes out to be the same as determined by Bohr $0.529\;Å$ from his atomic model. $r=\frac{h^2n^2}{4π^2me^2Z^2}\;\;\text{__(14)}$

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