# How many quantum numbers are needed to define the probability of finding the electron in a given region of space in the hydrogen atom?

Feb 14, 2016

Three: $n$, $l$, and ${m}_{l}$.

We need the quantum level of the orbital, the shape/type ($s , p , d$, etc) of the orbital, and which of those orbitals it is in (e.g. ${p}_{x}$, ${p}_{y}$, or ${p}_{z}$ for $p$ orbitals).

PROBABILITY DENSITY

The probability density of an electron in an atomic orbital is defined as the statistical distribution of where an electron often appears in that orbital.

Probability density is defined as:

${\int}_{\text{allspace" psi^"*}} \psi d \tau$

where:

• $\psi$ is the wave function that describes the state of a quantum mechanical system such as an atom or molecule.
• ${\psi}^{\text{*}}$ is its complex conjugate; since Real Chemists™ use real numbers, ${\psi}^{\text{*}} = \psi$.
• "allspace" means the inclusion of all possible (finite) locations in space.

PROBABILITY DENSITY FOR THE HYDROGEN ATOM

For the hydrogen atom, which uses spherical coordinates ($\vec{r} , \theta , \phi$), the wave function $\psi$ can be split into two components: the radial and angular components, ${R}_{n l} \left(\vec{r}\right)$ and ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$, respectively.

$\setminus m a t h b f \left(\psi \left(\vec{r} , \theta , \phi\right) = {R}_{n l} \left(\vec{r}\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)\right)$

From this, the probability density is written as:

int_"allspace" psi^"*"(vecr,theta,phi)psi(vecr,theta,phi) d tau = stackrel("Radial Component")overbrace(int_(0)^(oo) R_(nl)^2(vecr)r^2dr) stackrel("Angular Component")overbrace(int_(0)^(pi) (Y_(l)^(m_l)(theta))^2sinthetad theta int_(0)^(2pi) (Y_(l)^(m_l)(phi))^2dphi)

• The radial component contains the quantum numbers $n$ (quantum level, such as $n = \textcolor{b l u e}{3}$ for the $\textcolor{b l u e}{3} s$ atomic orbital, etc) and $l$ (the angular momentum, which for instance is $l = \textcolor{b l u e}{2}$ for a $\textcolor{b l u e}{d}$ orbital).
• $\vec{r}$ is the radial distance (outwards in all directions).
• The angular component essentially sweeps out the changes away from spherical uniformity. For instance, an $s$ orbital has a constant $\theta$ and $\phi$, so it is a sphere. A $p$ orbital is not a sphere, so it has $\theta$ and $\phi$ dependence.
• $\theta$ is the angle from the $\setminus m a t h b f \left(z\right)$ axis (vertical) going clockwise towards the $x$ axis (horizontal), and $\phi$ is the angle from the $\setminus m a t h b f \left(y\right)$ axis (towards you) going counterclockwise (on the $x y$-plane).

WHAT DO WE PLOT?

For simplicity, we often don't plot ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$. We choose to plot the radial density distribution by using only the radial component.

Therefore, it is a graph of ${\vec{r}}^{2} {R}_{n l}^{2} \left(\vec{r}\right)$ vs. $\vec{r}$ with $x$-axis units of ${a}_{0} = 5.29177 \times {10}^{- 11}$ $\text{m}$, as shown below for a $2 s$ orbital:

TAKE-HOME MESSAGE

Since ${R}_{n l} \left(\vec{r}\right)$ uses the quantum numbers $n$ and $l$, and ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ uses the quantum numbers $l$ and ${m}_{l}$, we need:

1. $n$ to show the quantum level of the orbital that the electron might be found in.
2. $l$ to show the shape of the orbital that the electron might be found in.
3. ${m}_{l}$ to determine which of the $2 l + 1$ orbitals (${m}_{l} = 0 , \pm 1 , . . . , \pm l$) the electron might be found in.