Griffiths' Quantum Mechanics Problem 4.27 : Diatomic particles

[hmparticle9]

I don't understand. What is wrong with my post #29?
$$ a = \sqrt{\frac{\hbar}{4} \frac{m_1+m_2}{m_1m_2}}$$

 

[kuruman]

I don't understand. What is wrong with my post #29?
$$ a = \sqrt{\frac{\hbar}{4} \frac{m_1+m_2}{m_1m_2}}$$

It's incomplete. You didn't do what the last sentence in part (d) is asking you to do. You need to come up with a number. That's the fun part.

 

[hmparticle9]

I get $2.8 \times 10^{-10} \text{m}$

 

[kuruman]

I get $2.8 \times 10^{-10} \text{m}$

It's the correct order of magnitude but not what I got. Please show explicitly the numbers that you substituted for the symbols in the equation below including units in the SI system, e.g.
$\hbar = 1.05 \times 10^{-34}~\text{J}\cdot \text{s}$, etc. etc.
$$a=\sqrt{\frac{\hbar ^2}{\Delta E}\frac{m_1+m_2}{m_1 m_2}}$$ I hope you are doing this on a spreadsheet. It facilitates troubleshooting.

Note the $\hbar^2$ under the radical. That's not what you have. The correct order of magnitude in your calculation is a fluke because number "4" in your equation is not a pure number; it is an energy difference and has units of energy.

 

[hmparticle9]

Look at post #9. my expression for $a$ is correct

 

[hmparticle9]

$\hbar$ is the same as you. Mass of carbon $2 \times 10^{-26}$ and mass of oxygen $2.656 \times 10^{-26}$

To make the dimensions work out I said:
$$\Delta \nu \approx 4 \text{ cm}^{-1} = 12 \times 10^{10} \text{ s}^{-1}$$

 

[kuruman]

Look at post #9. my expression for $a$ is correct

I looked. $\Delta E$ in that post the energy difference between two adjacent energy levels $$\Delta E_n=E_{n+1}-E_{n}=\frac{\hbar^2}{2I}[n(n+1)-(n-1)n]=\frac{\hbar^2}{I}n$$ and I wholeheartedly agree. That's part (c), it is correct and is behind us.

Now we are in part (d). Note that the energy difference in part (c) depends on the value of $n$ which means that it is not constant. Adjacent energy levels are not equally spaced, therefore their photon energies are not equal. Thus, you cannot invoke the equation in post #9, arbitrarily declare $\Delta E$ in that expression equal to 4 cm^-1 because that is the only energy around and say that you have solved the problem. First you need to understand what this 4 cm^-1 energy represents physically, then find a mathematical expression for it which better be independent of $n$ and then put in that expression the 4 cm^-1 energy.

In short, do what I suggested in post #16

For example, the red dashed line occurs at about 34 cm-1. First explain with words, not equations, what energy that is. Then write a mathematical expression for it. Do the same for the orange dashed line and the energy at 38 cm-1. The distance, in energy, between the orange and red dashed lines is 4 cm-1. Write another mathematical expression for it.

The first mathematical expression mentioned above is $\Delta E_n$ and the answer to part (c). What is the second mathematical expression mentioned in the last sentence?

 

[kuruman]

Mass of carbon $2 \times 10^{-26}$ and mass of oxygen $2.656 \times 10^{-26}$

These numbers are meaningless without units.

 

[hmparticle9]

Are we interested in:
$$\Delta \nu = \frac{\hbar^2}{I}?$$

 

[kuruman]

Are we interested in:
$$\Delta \nu = \frac{\hbar^2}{I}?$$

We are interested in extracting $a$ from the data. You need to have a plan for that. I have already given you an outline in post #16

For example, the red dashed line occurs at about 34 cm^-1. First explain with words, not equations, what energy that is. Then write a mathematical expression for it. Do the same for the orange dashed line and the energy at 38 cm^-1. The distance, in energy, between the orange and red dashed lines is 4 cm^-1. Write another mathematical expression for it.

 

[hmparticle9]

So I said, in words,

"The red line shows a dip in the intensity of detected radiation. This means that the sample of CO has absorbed energy."

And you said that was correct. Okay. So we want a mathematical expression for this energy. Surely this dip in energy has to equal the energy of the photon that has been omitted?

 

[kuruman]

Yes, and there are many such dips at different energies corresponding to transitions between different n values.

 

[kuruman]

OK, just to keep you focused. Shown on the right is the energy level diagram (first 5 energies) for the rotational states. It is drawn to scale. The numbers on the vertical axis are $n(n+1).$ Multiply any number by $\frac{\hbar^2}{2I}$ and you have the energy of that level.

Can you use this diagram to make a list of the energies of the missing photons that appear as dips? How many are there if you limit the transitions to only between shown levels?

 

[hmparticle9]

"Can you use this diagram to make a list of the energies of the missing photons that appear as dips?"

$$0, 2\frac{\hbar^2}{2I}, 6\frac{\hbar^2}{2I}, 12\frac{\hbar^2}{2I}$$

There are 4 transitions.

 

[renormalize]

There are 4 transitions.

Remember that transitions can occur between any pair of levels, so there are more than 4 possible.

 

[kuruman]

Please specify the $n$ numbers between which these transitions occur.
Also

1. Note that a photon with zero energy cannot exist much like an object with zero mass does not exist.
2. Justify what criteria you used to include some transitions but not others.

 

[hmparticle9]

The transitions occur between 5 energy levels, from $n=0$ to $n=4$.
I only included transitions between neighbouring energy levels because #26. I should really include all possible transitions.

Well if we limit transitions between shown energy levels from post #43, then from the 5 shown we need to choose 2. Hence $5 \choose 2$