Question

Can someone help me understand how the IR-raman spectrum of the inversion of ammonia is direct evidence of quantum tunnelling? Does it arise solely from the mathematical aspect (i.e. wave functions)?

Answer 1

I attached an article about this

Answer 2

Without getting too mathematical, because I can't explain this using math or physics cuz it's probably way beyond my level, but i THINK this has to to do with the ground state vibrational level of nitrogen atom. It fluctuates between differenrent energy levels.
And I'm taking a guess here but I think the different peaks on the IR depend on the roational state of th emoleecule at different temperatures.

Answer 3

Isn't nuclear decay a form of tunnelling?

Answer 4

yeah i think it is. Is the vibrational level of the N atom or the vibrational level of the bond between N and H?

Answer 5

nitrogen atom.

Answer 6

Is this for quantum based p-chem? This reminds me of wavefunctions and eigenfunctions we did for hydrogen :P

Answer 7

it's for quantum mechanics (its an intro class, but this is for a presentation I'm working on, so it's beyond what were covering)

umm, so the 2 finite potential energy wells are dictated by the vibrational state of the N atom?

Answer 8

I believe so. The two wavefunctions for each well aren't independent according to the first page, since both of the wavefunctions for each well have different forms.

Answer 9

I think thats why nitrogen doesn't exist in one state.

Answer 10

I could be wrong though. This is pretty heavy stuff for me :/

Answer 11

thanks i appreciate your input, dude. but i've been thinking about this for too long for me to absorb what you're saying, my brain is scrambled. so i'm gonna take a break, and get back to you on this later.

Answer 12

HAHA. I feel you. That's exactly how I felt my final p-chem class! But good discussion so far i think haha

Answer 13

http://www.youtube.com/watch?v=gENVB6tjq_M
http://en.wikipedia.org/wiki/Virial_theorem

Just something to read on while I write. 2. link is more relevant then the first one.

Answer 14

Okay..

In classical mechanics the turning point \(\large x_{tp}\) of a harmonic oscillator occurs when its kinetic energy equals 0, which is exactly when its potential energy \(\large \frac{1}{2}kx^{2}\) is equal to the total energy, \(E\).
In other words what we are going to have a lot of fun with is: (particle in a box theory):
\[\Large x^{2}_{tp}=\frac{ 2E }{ k } \rightarrow x _{tp}=\pm \left( \frac{ 2E }{ k } \right)^{1/2}\]
Solve the Schrödinger equation and we that the total energy is:
\[\Large E _{v}=\left( v+\frac{ 1 }{ 2 } \right) \hbar \omega\]\[\Large \omega = \left( \frac{ k }{ m } \right)^{1/2}\]\[\Large v=0,1,2,4,...\]

We want to use the probability of finding the oscillator stretched beyond a displacement \(\large x_{tp}\) which most be the sum of the probabilities \(\Large \Psi^{2} dx\). We get that:
\[\Large P=\int\limits_{x_{tp}}^{\infty} \Psi _{v}^{2} dx\]
The variable of integration is best expressed in terms of \(\large y=x/ \alpha\) where
\[\Large \alpha=\left( \frac{ \hbar^2 }{ mk } \right)^{1/4}\]
We then get by evaluating the right side:
\[\Large y _{tp}=\frac{ x _{tp} }{ \alpha }=\left( \frac{ 2(v+\frac{ 1 }{ 2 })\hbar \omega }{ \alpha^2 k} \right)^{1/2}=(2v+1)^{1/2}\]
At the lowest state of energy \(\large y_{tp}\)=1
PART 1

Answer 15

The probability from above then becomes:
\[\Large P=\int\limits_{x _{tp}}^{\infty} \Psi _{0}^{2}dx=\alpha N _{0}^{2}\int\limits_{1}^{\infty}\exp \left( -y^2 \right) dy\]
The integral is a special case of the error function (so I was told), which is defined as:
\[\Large \sf erf \it z \sf = 1-\frac{ 2 }{\pi ^{1/2} }\int\limits_{z}^{\infty} \exp \left( -\it y^2 \right) \it dy\]
Using the relations we can calculate procentage of observations of any oscillator in any state \(v\) that will be found stretched to a classically forbidden extent

Source; Physical Chemistry, Atkins.

Answer 16

To give an example:
\[\Large P=\frac{ 1 }{ 2 }\left( 1-\sf erf(1) \right)=\frac{ 1 }{ 2 }\left( 1-0.843 \right)=0.079\]
We can from this say that 7,9 per cent of a large number of observations, any oscillator in the state \(v=0\) will be found stretched to a classically forbidden extent.

Answer 17

It can be seen that these tunneling probabilities are independent of the force constant and mass of the oscillator. The probability of being found in forbidden regions decreases with with increasing \(v\) and is in principle gone when \(v\) approach infinity (like for macromolecules). But for molecules like ammonia the vibrational ground state is very significant.

Answer 18

it was all good until you brought in the error function because I haven't seen it. :S but i have that atkins book, so i'll look it up. I'm a little confused on the outcome/interpretation of the result though, because i've read that mass is inversely proportional to the tunnelling probability because the de Broglie wavelength particle influences the probability of crossing the finite barrier. But you're saying that the math says it's independent.

Answer 19

@abb0t @Frostbite Thanks a lot guys!

Answer 20

Atkins is probably one of the best p-chem books to have!

Answer 21

True, I'm so happy I'm a owner of Quanta, Matter and Change: A Molecular Approach to Physical Chemistry.

Answer 22

haha i have it on my iPad !

Answer 23

I tried using other books, but Atkins by far is one of the better ones I've used. Kind of weak in thermo tho.