# Question #cb6b8

Feb 7, 2016

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 $E = - {\overline{h}}^{2} / \left(2 m\right) {\nabla}^{2} \psi + U \psi$.
$\psi$ is the wave function of a particle; a function that describes the nature of your particle.

${\nabla}^{2} = {\partial}^{2} / \left(\partial {x}^{2}\right) + {\partial}^{2} / \left(\partial {y}^{2}\right) + {\partial}^{2} / \left(\partial {z}^{2}\right)$ in Cartesian coordinate system.
${\nabla}^{2} = \frac{1}{r} \frac{\partial}{\partial r} \left({r}^{2} \partial \frac{\psi}{\partial r}\right) + S \in \left(\theta\right) \frac{\partial}{\partial \theta} \left(S \in \left(\theta\right) \partial \frac{\psi}{\partial \theta}\right) + {\partial}^{2} \frac{\psi}{\partial {\phi}^{2}}$ in Polar Coordinate system.

Lets see the hydrogen atom, the easiest one.

$E = {E}_{1} / {n}^{2}$. $n$ 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 $\psi$.

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

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

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

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

The electron will cease to exist.

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

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