Modern I: Finding energy involving a rotator HCl molecule.

[baltimorebest]

Homework Statement

The energy of a rotator ( a dumbbell-shaped object) is given by L^2/2I, where L is the angular momentum and I is the rotational inertia (moment of inertia) of the dumbbell. Consider an HCl molecule, in which the atomic weights of H and Cl are 1 and 35, respectively, and their internuclear separation is 0.127 nm. Calculate the energy required to excite the molecule from the ground state to the first rotational excited state.

Homework Equations

The Attempt at a Solution

All I know is that the equation E₁-E₀= (h_bar)²/mR² is the proper equation for diatomic molecules. However I am not sure how to apply that idea to a case where the molecule is like the one in this problem.

 

[fzero]

Did you try to compute the moment of intertia, I? How about the values of L² for the ground and first excited states?

 

[baltimorebest]

Well that was one of my problems. When I compute those values, what do I use for the mass? In the book, they have strictly diatomic molecules, so the masses are whatever the mass of the atom is. In this case, they are different atoms, so I am unsure. Is it reduced mass?

 

[fzero]

The problem specifies the atomic weights for you.

 

[baltimorebest]

Right, but do I sum them? The equation is I=mr^2. We are given .127 nm as the separation (so I assume you divide that by 2 to get the radius), but for M, I still don't understand what value to use.

 

[fzero]

Right, but do I sum them? The equation is I=mr^2. We are given .127 nm as the separation (so I assume you divide that by 2 to get the radius), but for M, I still don't understand what value to use.

Go back to the definition of moment of inertia. You need to find the center of mass and the moment of inertia around an axis through the c.o.m. It's actually not true that mr² is the moment of inertia of a diatomic molecule formed from two of the same atom. There's a factor that you missed.

 

[baltimorebest]

I believe the factor I missed was a 1/2. So would the center of mass be (m1*m2)/(m1+m2)

 

[fzero]

Yes.

 

[baltimorebest]

Ok that mass value is 1.6E-27 kg. Now I multiplied that by (.127nm)^2 to get 2.62E-47. I also got no units... That doesn't seem right.

 

[fzero]

The moment of inertia has units of mass (length)². L²/(2I) has units of energy.

 

[baltimorebest]

Ohh ok right. Sorry. Ok so I have 2.62E-47 m^2 for I. Now for L^2/2I, I know it's equal to 0 for n=0. When n=1 isn't it going to be (h_bar)^2/(2*2.62E-47 m^2)?

 

[baltimorebest]

When I do that, I am getting units that are incorrect. I think my thinking is correct, but my final answer can't be.

 

[fzero]

What are the eigenvalues of L²? What is the value of L² for the first excited state?

Also, you're missing the factor of mass unit in I.

 

[baltimorebest]

I fixed the unit issue for I. I have that fine now.

We haven't discussed eigenvalues at all. However, I believe the value of L^2 for the first excited state to be h_bar^2 since n=1.

 

[fzero]

You should check your notes or text, since even the formula you quote in your original post is missing a factor of 2. The L² eigenvalues are $$\ell(\ell+1)\hbar^2$$. This yields an extra factor of 2 for the energy of the first excited state compared to your results.

 

[baltimorebest]

I'm sorry but that last post confused me more. Nowhere in class or in the book does it say anything about eigenvalues. However, as far as the factor of 2, I think you are right. I think he left it out of the denominator of the original equation I gave.

 

[baltimorebest]

Unfortunately I have to get offline in a few minutes. I have to turn this in early tomorrow morning. If there is any way you can describe the solution before then, I can read it before I turn it in, and follow up with you tomorrow. I just need something to turn in tomorrow, and I can discuss it with you tomorrow to make sure I understand it. If not, that's ok too. Thanks.

I also believe the answer is supposed to be 1.33 meV, but I am not entirely sure.

 

[fzero]

The energy of the first excited state is

$$E_1 = \frac{2\hbar^2}{2I}.$$

If you have something different in your notes for the case you covered in class, then try to find the derivation in your text. I think you have a correct result for I, so you just have to justify the rotational spectrum that I gave you as correct.

 

[baltimorebest]

Yeah I will check my notes again in the morning before class. So the answer should be E=h_bar^2/I which is equal to (6.58E-16)^2/2.62E-47. This seems to give a strange result for the units though since the units for h_bar are ev*s and the units for I are kg*m^2

 

[fzero]

Yeah I will check my notes again in the morning before class. So the answer should be E=h_bar^2/I which is equal to (6.58E-16)^2/2.62E-47. This seems to give a strange result for the units though since the units for h_bar are ev*s and the units for I are kg*m^2

You can convert eV to MKS units.

 

[baltimorebest]

So I finally set h_bar equal to 6.63E-34 J*s. When I plugged this into the equation you gave me, and used my value of I, the result was 1.7E-20 Joules. Hope this is right.