# How do we find the angular, radial nodes and the quantum numbers when the radial probability distribution curve's equation is given?

I am having serious conceptual issues with the Schrödinger equations. Please note that I only have basic school level (11th grade) knowledge of these equations and we are expected to know very little about it. But we are given these type of questions so I guess I will be able to understand things within the boundary of this question.

$$\psi_r = k_1 * e^{-r/k_2} * (r^2 - 5k_3r + 6k_3^2)$$

Solving the equation we get $r = 3 k_3 , 2 k_3$

It is also given that $k_3 =1$ . So $r = 3, 2$

1. Now, since there are 2 nodes therefore, $n = 3$ . Does the equation $\psi_r$ represents only radial nodes or all nodes?
2. What is the azimuthal quantum number?
3. How many angular nodes are present?
4. What is the angular momentum of the given orbital?
5. Is this equation of an atom or an orbital?

This equation was given as an MCQ question and we were asked to find the quantum states of the given orbital on the basis of the equation. But I guess we must have conceptual insight too to answer the question so asked it here.

• We can't really tell much about angular nodes. The angular part is simply not there. We have no idea what it might be. – Ivan Neretin Apr 8 '16 at 8:01

For a given orbital, there are two types of modes: radial nodes and angular nodes.

• The total number of nodes is $n-1$
• Number of angular nodes is $l$
• Number of radial nodes is $n-l-1$

In this question: If the given function represents the radial part of the wave function of an atomic orbital, then we get the number of radial nodes: $n-l-1=2$. i.e. $n-l=3$:

We have several possibilities:

1. $n=3$, $l=0$: then we have the orbital $\ce{3s}$
2. $n=4$, $l=1$: then we have the orbital $\ce{4p}$
3. $n=5$, $l=2$: then we have the orbital $\ce{5d}$

and so on.

Your equation is a wavefunction. Simply put, a wavefunction is a description of the state of a particle(s). In most cases these are electrons (at least in chemistry). Wavefunctions describe particles not atoms. There are many-particle particle wavefunctions. In other words, wavefunctions can describe the state of more than one particle. An orbital, on the other hand, can only describe the state of one particle (but some people say two at most). We obtain the state by solving the Schrodinger equation. For electrons in atoms, there are these things called quantum numbers which specify the state of the electron. These are the n,l,m,s that is talked about in introductory chemistry classes: https://en.wikipedia.org/wiki/Quantum_number The angular momentum as well as other properties can be calculated by looking at the quantum numbers (in wikipedia, you'll find that they list which quantum number is related to the magnitude of the angular momentum).

More specifically, the wavefunction gives information on the probability of finding a particle in a certain state. Nodes, are places where particles can't be located. The probability of finding particle in a node is 0. Thus, wavefunction describing an electron with a principal quantum number 3 (the "radial" part) would be "aware" of the nodes. Otherwise, it wouldn't be a valid description for the probability of finding an electron. Look, at your equation, when will the wave function be zero? I am not sure how many angular nodes are present, since I am not familiar with wavefunction that you are providing. I think it may be helpful if you gave us the context in which you found the equation. However, I think that your wave function has no angular part.

• Actually this was given as an mcq question and we were asked to find the quantum states , I don't know whether this wavefunction is real or hypothetical :/ – brainst Apr 8 '16 at 8:55
• @brainst then it only tells you information on the r part unless k1 is a function of the angular part – CoffeeIsLife Apr 8 '16 at 8:59
• So in short if $\psi_r$ is given then it would provide only the information of radial part and if $\psi (R,\theta)$ is given then it would provide the information about the angular part as well ? – brainst Apr 8 '16 at 9:02
• @brainst Also try to show some effort. Show what you attempted, what you understand.Otherwise people may not be too positive about your questions. – CoffeeIsLife Apr 8 '16 at 9:02
• Actually I thought that knowing that the probability goes zero at node and solving the quadratic was the atmost that I was able to do and I did shown that in the answer , other all is conceptual which I had doubt about (I wasn't sure whether the two nodes that I found was all or there were angular nodes too) , and thats what the question is about ... Is it not explained well? – brainst Apr 8 '16 at 9:05