# Probability wave peaks and quantum number

Tags:
1. Apr 6, 2015

### Chemer

Hi,
Just wanted to ask, the principal quantum number represent the number of peaks of the probability wave and I think more the value of n, more the energy of the electron as the wave has more peaks so higher frequency,am I right? Then in azimuthal quantum number, the orbitals with same energy have different number of peaks, means different energy? But why? I'm referring to the point in this site:

Maybe I've misunderstood the point because I'm really a beginner, please guide.

2. Apr 6, 2015

### Simon Bridge

Depends... principle quantum number of what?
Usually, "n", in text books, indexes the descrete energy eigenstate.
This can correspond to the number of antinodes in the associated energy eigenfunction.
But it does not have to... check the context.

I did not find a reference to "azimuthal quantum number" in your link.
The context is "atomic wavefunctions in chemistry" and the site encourages you to think of n as indexing the average distance from the nucleus.
Since an atom is a 3d object, there are several ways to get peaks in the probability density function. You will have seen that n is the number of radial peaks.

Last edited: Apr 6, 2015
3. Apr 6, 2015

### Chemer

I was referring to this point on the site:

"Each wave pattern is identified by an integer number n, which in the case of the atom is known as the principal quantum number. The value of n tells how many peaks of amplitude (antinodes) exist in that particular standing wave pattern; the more peaks there are, the higher the energy of the state."

And that:

"When l = 0, the orbital is spherical in shape. If l = 1, the orbital is elongated into something resembling a figure-8 shape, and higher values of l correspond to still more complicated shapes— but note that the number of peaks in the radial probability distributions (below) decreases with increasing l."

Can you explain it in easy words? Aren't the peaks representing the frequency and energy of the probability wave?

File size:
39.8 KB
Views:
214
4. Apr 6, 2015

### Simon Bridge

Put simply, l and n states are different and have different rules.
The peaks do not represent frequency or energy of the wave.
The wavey shapes on the graph are not waves in the sense of water or sound waves.

5. Apr 6, 2015

### Chemer

The peaks of probability wave are the points where the electron is mostly likely to be found, and these peaks form the electronic cloud boundary with a radius in agreement to that predicted by Bohr's model. And each cloud has a specific wave pattern represented by specific number of peaks, the principal quantum number. So, didn't the orbits were quantized in Bohr model, and greater n greater number of peaks, so greater energy? Am I right? Please guide.
and then what the peaks represents in case of the sub shells?

6. Apr 7, 2015

### Simon Bridge

What is the radius for the n=1 peak?
What is the Bohr model radius for n=1?
Do these agree?

7. Apr 7, 2015

### Chemer

I realized now, from the Ist image the average radius of n= 3 agrees with Bohr's radius of n=1 that's 53 pm I think. But I don't understand why?
And what does these peaks represent?
Also when we say electron in 2s has lower energy than 2p, because of screening effect,I couldn't understand what is meant by p orbital being higher than s? They've same n, so same energy but the energy lowers because of the screening effect?

8. Apr 7, 2015

### Simon Bridge

The peaks represent the maxima of the probability density function.

The 2s and 2p electrons would have the same energy if the electrons did not have any effect on each other.

If you have a 5 proton nucleus (B) with two electrons in 1s, can you see that this looks from farther away like a Li nucleus?
That effect is called "screening".
Add two more electrons, what is the lowest state they can go to?

9. Apr 7, 2015

### Chemer

"The peaks represent the maxima of the probability density function."

So, as the increase in l, means increase in zero probability region thus the number of peaks decrease, am I right? And in case of n, the energy is nothing to do with probability density but with the electron's distance from the nucleus?

"The 2s and 2p electrons would have the same energy if the electrons did not have any effect on each other.

"If you have a 5 proton nucleus (B) with two electrons in 1s, can you see that this looks from farther away like a Li nucleus? That effect is called "screening".
Add two more electrons, what is the lowest state they can go to?"

Do you mean as Li has 3 protons, 2 are shielded by its 1s electrons and the outermost electron feels the effect of only one proton? So, in boron 2 protons are shielded by 1s electrons, then the 2s state has now lower energy and electrons will go to 2s? What about the remaining nuclear charge? How will it effect the electrons preference? And the n has changed as well, so does it have any effect on this phenomena?