[KZ99]

The bond length of CO is 1.128 Å. At what wavenumbers do the first three rotational transitions appear

The equations you nee are
μ= (m1m2/m1+m2)
I = μr0^2
B= h/(8∏^2)IbC
ΔE = 2B (J+1)

For the first equation I got

(12 x 16) / 28 = [(6.857 /1000)/ avagadro number] = 1.138 x 10^-26

For the second equation I got

(1.138 x 10^-26) x (1.128 x 10^-10)^2 = 1.44 x 10^-46 kg/m^2

For the third equation I got

planck's constant / [(8∏^2) x (1.44 x 10^-46 ) x (speed of light)] = 1.93cm^-1

I can't go any further, I'm stuck from here on!!!   thank you

[Corribus]

I didn't check your numbers, but assuming your answers for the first three equations are correct, now you just need to solve for the transition energies of the lowest three transitions.

So what are the first three transitions?  (hint: the only thing you haven't used yet is J).

[KZ99]

I didn't check your numbers, but assuming your answers for the first three equations are correct, now you just need to solve for the transition energies of the lowest three transitions.

So what are the first three transitions?  (hint: the only thing you haven't used yet is J).

I think something just hit my mind when I you told me that i haven't used J yet, thank you very much for that

Okay
I have already found B

E=  BJ (J+1)    [J would be 1 for the first energy level as they are quantised]

My problem is there is another equation that is
E= B (2J+2) ( should I use this for the second level, or the previous one by just substituting 2 for J)

[Corribus]

Careful!  You are confusing ΔE for E.

The energy levels of rigid rotor are given by E_J = BJ(J + 1) where J = 0, 1, 2, 3, 4...

So, the lowest energy level (J = 0) has energy 0.  The next lowest energy level (J = 1) has energy 2B.  The next, 6B, and so forth.

The energy difference between energy levels, ΔE_{J --> J+1} = E_J+1 - E_J.

I encourage you to do the algebra to generate the formula you posted in your opening post.

Now, having that - what are the first three transitions, and what are the energy gaps, ΔE_{J --> J+1}?

(Edit - Borek, if you're listening, the reaction arrows don't work when placed in subscripts!)

[KZ99]

Thanks for the reply
I have a very good idea now and I did spot the E and ΔE error I have done after you told me. I did the algebra you told but didnt post here as it is very long

okay
Assume I have calculated the energies I need to use the formula below to get the wave numbers

E= (plank's constant x speed of light x wavenumber)

My value for B= 1.93cm^-1

Now for the first energy level (J=0) as you have told, E= 0 joules

How to calculate the wavenumber, or is it just 1..93cm^-1

Second energy level J=1

ΔE_{j j+1} = 2B (J+1)

ΔE = 2 x 1.93 x (2)
    = 7.72J

Third energy level J=2

ΔE_{j+1 j+2} = 4B (J+1)

ΔE = 4 x 1.93 x (3)
    = 23.16J
 
Have i worked everything out correctly

[Corribus]

I don't have a calculator with me so I cannot easily check your math.  You can use wavenumbers (cm^-1) in place of Joules. 

E(J = 0) = 0.  Joules or wavenumbers, makes no difference.  (Which should make sense - classically, if there is no energy, there is no rotation!)

E(J = 1) = B1(1+1) = 2B.

Therefore the first transition ΔE_{J = 0 --> J = 1} = E₁ - E₀ = 2B - 0 = 2B.  Which according to your values should be just shy of 4 wavenumbers.

Note that this will be the same value you get from using the other formula ΔE = 2B(J+1).  In this equation, J is taken from the "origin" or "initial" state in the transition (J = 0).  That can be a little confusing, I know.  Probably better to write it as something like ΔE_{initial-->final} = 2B(J_initial+1).

Anyway, you'll see you get 2B when you put J = 0 into that equation.

Ok.  For rigid rotor the only allowed transitions are ΔJ = ±1. So the second transition will originate from J = 1 and the third transition will originate from J = 2. 

Now try one more time.

[KZ99]

Thanks for the reply

I don't have a calculator with me so I cannot easily check your math. 

Sorry I wanted to ask you if my method was correct

Therefore the first transition ΔEJ = 0 --> J = 1 = E1 - E0 = 2B - 0 = 2B.  Which according to your values should be just shy of 4 wavenumbers.

Yes 1.93 x 2 = 3.86 Joules

.  For rigid rotor the only allowed transitions are ΔJ = ±1. So the second transition will originate from J = 1 and the third transition will originate from J = 2.

Okay here what you are telling me is that I should calculate the energy difference, so the first transition is 2B. My 2nd and 3rd transistional energy levels are correct in my previous post right?

You then use those energies and use
E= (plank's constant x speed of light x wavenumber)

to get the wavenumbers right?! I might be a bit

[Corribus]

My 2nd and 3rd transistional energy levels are correct in my previous post right?

The second is correct, third is not.  Use your formula.

You then use those energies and use
E= (plank's constant x speed of light x wavenumber)

to get the wavenumbers right?!

You can express the energy in terms of wavenumber because the relationship is linear (unlike wavelength), but if you want to express in terms of Joules, then yes you would have to do the conversion like this.

[KZ99]

Thank you for the reply

Okay I am a bit stuck again

To go from J  J+1     it is

[(J+1) (J+1+1)] - [J(J+1)]}

Simplify this you get

2J + 2

Now to go from J₂ to J₃  is it

{[(J+1)(J+2)(J+3)] - [(J+1) (J+1+1)]}

Just want to know if what I have done just above is correct?

[Corribus]

You're doing some strange substitutions and you're dropping off your B's, so let's look at it again.

For a transition energy between state J and state J+1, is equal to the energy of state J+1 minus the energy of state J.

The energy of state J is E_J = BJ(J+1)

Therefore the transition energy for any transition is ΔE_J = 2B(J+1), where in this case J refers to the lower energy state.

So the first transition (from J = 0 to J = 1) is 2B.  The second transition (from J = 1 to J = 2) is 4B.  The third transition (from J = 2 to J = 3) is 6B. 

Notice a pattern?

For a rigid rotor in a linear molecule, the transitions are evenly spaced, and the spacing is equal to 2B.  You can see this by taking the derivative of the ΔE formula with respect to J.  (dΔE/dJ = 2B).

So now, I realize I've pretty much given you the answer to your question because all you have to do now is put in your value for B, but I encourage you to test yourself.  What is the transition energy of the TENTH lowest energy transition?

[KZ99]

You're doing some strange substitutions and you're dropping off your B's, so let's look at it again.

For a transition energy between state J and state J+1, is equal to the energy of state J+1 minus the energy of state J.

The energy of state J is E_J = BJ(J+1)

Therefore the transition energy for any transition is ΔE_J = 2B(J+1), where in this case J refers to the lower energy state.

So the first transition (from J = 0 to J = 1) is 2B.  The second transition (from J = 1 to J = 2) is 4B.  The third transition (from J = 2 to J = 3) is 6B. 

Notice a pattern?

For a rigid rotor in a linear molecule, the transitions are evenly spaced, and the spacing is equal to 2B.  You can see this by taking the derivative of the ΔE formula with respect to J.  (dΔE/dJ = 2B).

So now, I realize I've pretty much given you the answer to your question because all you have to do now is put in your value for B, but I encourage you to test yourself.  What is the transition energy of the TENTH lowest energy transition?

Thanks for the reply

i will put the full working below for 0 to 1

B[(J+1)(J+1+1)]- B[J(J+1)]
B(J² + 3J + 2 – J² – J)
B(2J+2)

1 to 2
B[(J+1+1)(J+1+1+1)]- B[(J+1)(J+1+1)]
B[(J+2)(J+3)] -B[(J+1)(J+2)]
B[(J²+5J+6)]- B[(J² + 3J + 2)]
B[(2J + 4)    I am not getting the expected answer 

And transition energy of the tenth lowest energy transition  is
22B (J+1)  please tell me this is correct

[Corribus]

And transition energy of the tenth lowest energy transition  is
22B (J+1)  please tell me this is correct

Once you put 10 in for J, there's no longer a J in the expression.  The answer is 22B.

Basically, if ΔE = 2B(J+1) for the transition from the J^th level to the (J+1)^th level, just put the value of J in that expression.  Whatever the J of the original state, add 1, multiply by 2B, and that's the energy of the transition. 

So if J = 420, the energy of the transition is 842B.

[KZ99]

Basically, if ΔE = 2B(J+1) for the transition from the J^th level to the (J+1)^th level, just put the value of J in that expression.  Whatever the J of the original state, add 1, multiply by 2B, and that's the energy of the transition. 

Got it thank you ever so much for all of this. Too many equations in physical chemistry and it is killing me