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Question #cb6b8

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Answer 1 · 2016-02-07T04:56:39

Via Quantum Mechanics with lots and lots of math.

Explanation:

In Quantum Mechanics, there is the Schrodinger Equation . The equation is capable determining many properties of a particle in a system based on time and position of the particle.

In a steady state Schrodinger Equation (we are not interested in the time variable);
The equation is $E = - \frac{{¯ h}^{2}}{2 m} \nabla^{2} ψ + U ψ$ $E=-barh^2/(2m)grad^2psi+Upsi$ .
$ψ$ $psi$ is the wave function of a particle; a function that describes the nature of your particle.

$\nabla^{2} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}$ $grad^2=del^2/(delx^2)+del^2/(dely^2)+del^2/(delz^2)$ in Cartesian coordinate system.
$\nabla^{2} = \frac{1}{r} \frac{\partial}{\partial r} (r^{2} \partial \frac{ψ}{\partial r}) + sin (θ) \frac{\partial}{\partial θ} (sin (θ) \partial \frac{ψ}{\partial θ}) + \partial^{2} \frac{ψ}{\partial ϕ^{2}}$ $grad^2=1/rdel/(delr)(r^2delpsi/(delr))+Sin(theta)del/(deltheta)(Sin(theta)delpsi/(deltheta))+del^2psi/(delphi^2)$ in Polar Coordinate system.

Lets see the hydrogen atom, the easiest one.

$E = \frac{E_{1}}{n^{2}}$ $E=E_1/n^2$ . $n$ $n$ is the principle quantum number which corresponds to the orbital shell (also energy state).

What we want to do is to solve the Schrodinger equation to find $ψ$ $psi$ .

Solving the equation is very tedious and requires arbitrary constants $l$ $l$ and $m_{l}$ $m_l$ . Refer to Separation of Variables method .

Solving this implies that $ψ$ $psi$ has non-zero value if the values of $l$ $l$ has integer values and does not exceed $n - 1$ $n-1$ .

!! $l$ $l$ will determine your orbitals. $n$ $n$ determines the values of $l$ $l$ .!!

In other words, if $n = 3$ $n=3$ you must have $l = 0, 1 and 2$ $l=0,1 and 2$ . Energy level 3 has 3 orbitals.
$s, p and d$ $s, p and d$ orbitals. Other values of $l$ $l$ in this case will cause $ψ$ $psi$ to breakdown.

The electron will cease to exist.

Therefore, if $n = 3$ $n=3$ , you will have 3 solutions for $ψ$ $psi$ based on $l = 0, 1, 2$ $l=0,1, 2$ .

Sorry if I can't express this any simpler. The history of the atomic model from Neils Bohr onwards is very theoretical.

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