# Quantum Chemistry/Example 9

Write a question and its solution that calculates the locations of the nodes of an electron in a 2s orbital.

## The Question

The question is to find the location of the radial node in a 2s electron for a hydrogen atom. To find the node one can start by analyzing, generally, how many nodes one should expect to see in a 2s electron system. There are two equations that give the number of nodes present in an orbital, the angular node equation and radial node equations:

${\displaystyle n-\ell -1=n_{r}}$

2. Angular Nodes:

${\displaystyle \ell =n_{a}}$

Therefore ℓ must be determined, and based on the table 1 data one can determine ℓ is equal to 0. And the n is equal to the ${\displaystyle n}$  which is 2, this comes from the number before the orbital type which tells you the principle quantum number.

Table 1: Orbitals and Quantum Number
Orbital Angular

Momentum

Quantum

Number (ℓ)

s 0
p 1
d 2
f 3

So how many nodes are there?

First analyze the number of angular nodes:

${\displaystyle \ell =n_{a}}$

${\displaystyle 0=n_{a}}$

Therefore, the number of angular nodes is 0.

${\displaystyle n_{r}=n-\ell -1}$

${\displaystyle n_{r}=2-0-1}$

${\displaystyle n_{r}=1}$

Therefore there is one radial node present in a 2s orbital, resulting in the question becoming where is the location of that radial node?

The Wavefunction

Now the next main step is to determine what wavefunction describes the wavefunction this scenario of the 2s electron. The wavefunction can be found online which is${\displaystyle :^{1}}$

${\displaystyle \psi _{2s}={\frac {1}{4(2\pi )^{\frac {1}{2}}}}\left({\frac {1}{a_{0}}}\right)^{3/2}\left(2-{\frac {r}{a_{0}}}\right){\text{e}}^{-r/(2a_{0})}}$

In the equation the ${\displaystyle \psi _{2s}}$  is the wavefunction for the 2s electron, 𝒓 is the radius, and ${\displaystyle a_{0}}$  is the Bohr radius.

Based on the equation we can then solve for the position of the electron. The best way to do this is to find where is the equation going to be equal to zero and what term that contains the position causes this. The first part ${\displaystyle {\frac {1}{4(2\pi )^{\frac {1}{2}}}}}$  is a constant thus won't change with the radius, the position, of the electron. So the only place that will change with the position of the electron are the 𝒓 terms. With ${\displaystyle {\text{e}}^{-r/(2a_{0})}}$  the 𝒓 term can be any number and the term won't be zero unless the 𝒓 is approaching infinity, while the ${\displaystyle \left(2-{\frac {r}{a_{0}}}\right)}$  can potentially be equal to zero since it has 2 subtracted by the position term. Therefore one can set this term equal to zero and solve for 𝒓.

${\displaystyle 0=2-{\frac {r}{a_{0}}}}$

${\displaystyle 2={\frac {r}{a_{0}}}}$

${\displaystyle 2a_{0}=r}$

Therefore, we get the solution to the position of the radial node which is 𝒓 ${\displaystyle =2a_{0}}$ , so when 𝒓 ${\displaystyle =2a_{0}}$  the probability of the electron being there is 0 all around the nucleus creating a node. The ${\displaystyle a_{0}}$  has a length of 52.9 pm which means the node is 105.8 pm in radius away from the nucleus.

In Conclusion

In conclusion the approach to the problem of finding nodes for an electron in an orbital boils down to first finding the number of theoretical nodes, then determining the wavefunction, analyzing the wavefunction's variables and solving for 0. After this is all done you will have the solution to the radial node locations. Further problems can be solved as well, because they are follow up questions that are made easier to solve after finding the nodes, like the position of the electron in it's most probable state. This problems follows the solution of arranging the P=${\displaystyle \psi _{2s}}$ ${\displaystyle \left(\psi _{2s}^{*}\right)}$ ${\displaystyle \left(r^{2}\right)}$  then finding the derivative of the wavefunction and then simplifying it. The final step is to find the zero points, where the 𝒓 is equal to zero which will give the most probable locations of the electron. The practical applications of finding the nodal locations can help with understanding how orbitals work which can help with making molecular orbital diagrams and SALCs that can be used to determine the way atoms and molecules bond. Other applications include understanding the energy levels of the bonds and orbitals to predict possible interactions between molecules and atoms, for research purposes and chemical engineering.

Reference

1) Branson, J. The Radial Wavefunction Solutions. https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html (accessed November 16, 2021).

By Dmitry Ivanov