For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### The Rigid Rotator III

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- The Rigid Rotator III
- When a Rotator is Subjected to Electromagnetic Radiation
- Selection Rule
- Frequencies at Which Absorption Transitions Occur
- Energy Absorption & Transition
- Energy of the Individual Levels Overview
- Energy of the Individual Levels: Diagram
- Frequency Required to Go from J to J + 1
- Using Separation Between Lines on the Spectrum to Calculate Bond Length
- Example I: Calculating Rotational Inertia & Bond Length

- Intro 0:00
- The Rigid Rotator III 0:11
- When a Rotator is Subjected to Electromagnetic Radiation
- Selection Rule
- Frequencies at Which Absorption Transitions Occur
- Energy Absorption & Transition
- Energy of the Individual Levels Overview
- Energy of the Individual Levels: Diagram
- Frequency Required to Go from J to J + 1
- Using Separation Between Lines on the Spectrum to Calculate Bond Length
- Example I: Calculating Rotational Inertia & Bond Length 29:18
- Example I: Calculating Rotational Inertia
- Example I: Calculating Bond Length

### Physical Chemistry Online Course

### Transcription: The Rigid Rotator III

*Hello and welcome back to www.educator.com and welcome to Physical Chemistry.*0000

*Today, we are going to close out our discussion of the rigid rotator which is a model for a rotating diatomic molecule.*0004

*Let us jump right on in.*0011

*I think I'm going to work with blue today.*0014

*I hope you guys can hear me.*0017

*In the last lesson, we saw that the allowed energies…*0019

*… of the quantum mechanical rigid rotator R A sub J*0046

*is equal to H ̅² / 2I × J × J + 1, where J is a quantum number that runs from 0, 1, 2, and so on.*0056

*And I here is equal to the reduced mass × the radius of the rotation, the distance between the two masses.*0073

*When a molecule rotator is subjected to electromagnetic radiation, it undergoes transitions from 1 energy state to the next.*0083

*For the rigid rotator, we are saying it is going to go from 1 to 2, 2 to 3, 3 to 4.*0118

*You may be asking yourself can I make the jump from, if I hit it with enough energy can I make the jump from 1 to 3, level 5 to 8?*0141

*In the case of the rigid rotator, just like in the case of a harmonic oscillator, the answer is no.*0148

*Any transition has to be made is going to be made between successive levels, 1 to 2, 2 to 3, 3 to 4, and so on.*0153

*4 to 3, 3 to 2, 2 to 1 and so on.*0159

*For the rigid rotator, the selection rule tells us about how the quantum number can change.*0162

*What levels we can jump to?*0173

*The rigid rotator, the selection rule is that δ J is equal to + or – 1.*0176

*Recall the selection rule for the harmonic oscillator, we call that R.*0192

*Δ R = + or -1, it is the same thing.*0208

*It can only make transitions between adjacent energy levels.*0211

*In addition to this δ J equaling + or -1, in order for a rotating molecule to actually absorb energy and*0216

*spin faster, move to the next energy level, it has to have a permanent dipole.*0231

*We will talk about this one later but it is something that you should now.*0247

*The molecule must possess a permanent dipole.*0250

*Basically, the idea is this.*0259

*You were hitting something with some electromagnetic energy.*0262

*The electric field of the energy that we actually hit with it, the electromagnetic energy*0267

*that we hit with it needs to be able to twist the molecule so they can make it spinning.*0273

*But it needs to make it spin faster.*0283

*Imagine I grab it and I just spin it faster, the electromagnetic energy needs to do that.*0285

*It needs to grab the molecule and actually spin it faster.*0291

*It is the force acting on the molecule.*0295

*If there is no dipole moment, if there is no positive and negative, as if it is just neutral, the electric field has nothing to grab onto.*0299

*The molecule does not experience the force of the electromagnetic field so we cannot actually spin it.*0306

*It cannot grab it and turn it faster.*0313

*It has to have a permanent dipole moment, a positive and a negative.*0316

*In order for that positive and negative end to feel the force of the electric field.*0319

*That is all that is going on.*0324

*That is why it have to have that.*0326

*In addition, the molecules must posses a permanent dipole in order to transition.*0328

*Something like the molecule it has a rotational spectrum because I know it has a permanent dipole.*0344

*Whereas something like nitrogen does not because there is no positive or negative end.*0366

*Therefore, there is nothing for the electric field to grab onto.*0374

*It just does not experience any force in the electric field.*0377

*Let us talk about some difference in energies.*0383

*So δ E is just going to be the E of J + 1 - E of the level right below it.*0386

*It is going to equal H ̅² all / 2I × J + 1 × J + 1 + 1 - J × J + 1.*0399

*Should I go through the multiplication?*0420

*That is fine, I will go ahead and go through it.*0422

*It is not a problem.*0425

*= H ̅² / 2I ×, we have J² + 3 J + J + 1.*0426

*This is going to end up being J + 2 - J × J + 1.*0440

*When you multiply these out and start canceling terms, you are going to get the J² + 3 J + 2 - J² – J = H ̅ / 2I × 2 × J + 1.*0443

*The 2 cancel leaving you with the δ E = H ̅² / I × J + 1.*0474

*In terms of the actual plank constant, H² / 4 π² I × J + 1.*0493

*This gives us the actual energies and again J is equal to 0, 1, 2, and so on.*0505

*These expressions right here, this and this, one in terms of the H ̅, one in terms of H,*0519

*they give us the actual change in energy from level to level.*0525

*So δ E is actually also equal to H ν.*0532

*H ν is equal to H² / 4 π² I × J + 1.*0541

*This H cancels this H, therefore, we have ν is equal to H / 4 π² I × J + 1.*0551

*Remember, I = MR² and K.*0563

*These are the frequencies, this equation right here.*0570

*These are the frequencies at which absorption transitions occur.*0580

*There are going to be very specific frequencies at which absorption transitions occur.*0598

*It is going to go from 0 to 1, 1 to 2, 2 to 3, and so on.*0609

*Rotational transitions they happen with frequencies in the microwave region or sometimes in the far infrared region.*0617

*It is a microwave region but also happened in the far infrared also.*0646

*In practice, let me go back to blue here.*0655

*In practice, we write the frequency is equal to 2B × J + 1 or B is equal to H / 8 π² I.*0664

*This B right here it is called the rotational constant.*0689

*The energies, the frequencies are actually expressed in terms of this rotational constant.*0701

*Also instead of frequency, we also use wave number which is actually more common.*0711

*We also use that ν tilde which is called wave number.*0728

*The wave number is anything with a tilde on it is just equal to the thing ÷ C, the speed of light.*0737

*We have ν tilde = ν/ C = 2B/ C × J + 1 and this 2B/ C that is the wave number, this B ̃.*0744

*B ̃ is just B ÷ the speed of light.*0767

*In this particular case, the B ̃ is going to equal to H ÷ 8 π² CI.*0775

*That is it, we have just taken B and we divide it by the speed of light.*0786

*Again, it is going to be expressed in terms of the wave number which is going to be*0789

*in units of inverse cm, that is the wave number or frequency which is going to be Hz.*0793

*It is just different ways of expressing the same thing.*0799

*Let us stop and think about this for a second.*0802

*What this thing gives us the J=0, 1, 2, and so on, what this gives is the energy needed.*0811

*Units of wave numbers, this inverse cm is actually a unit of energy.*0833

*What this gives is the energy needed.*0838

*Let us put it this way.*0845

*It is the energy that needs to be absorbed, in order to actually make the transition to the next energy level.*0849

*In order to make the transition from level J to the next level up J + 1.*0880

*When I put a J2, what this in effect?*0898

*This is going to be 2 + 1 is going to be 3, 2 × 3 is 6, 6 × this B ̃ that is the amount of energy*0902

*that needs to be absorbed in order to make the transition from 2 to 3.*0912

*Once again, what this ν gives us, what this equation gives us?*0918

*I’m sorry I forgot the J + 1.*0926

*What this equation actually gives us, it gives us the energy that needs to be absorbed in order to make the transition from J to J + 1.*0932

*This J is a level at which it is, in order make the transition from that to the next one up.*0940

*Let us go ahead and calculate a few.*0947

*You so ν sub 0 = 2B ̃ × 0 + 1 = 2B ̃.*0953

*At a wave number, 2B ̃, the energy transitions from level 0 to level 1.*0976

*Now ν sub 1 is equal to 2B ̃ × 1 + 1 that is equal to.*0999

*1 + 1 is 2, this is 4B.*1008

*This is the amount of energy for B, whatever this is energy.*1012

*We will worry about values in just a minute but this is the amount that is necessary*1017

*to be absorbed by something at level 1 in order to go to level 2 or B.*1021

*4B must be absorbed to go from level 1 to level 2.*1038

*Now, we are at level 2, let us calculate this one.*1055

*This is going to be 2B ̃ × 2 + 1 that is going to be 6B.*1058

*This is the amount of energy when you are at level 2.*1067

*When you are at level 2, this is the amount of energy that you need to absorb in order to make it to level 3.*1070

*6B must be absorbed to go from level 2 to level 3, that is what is happening here.*1079

*Notice the pattern 2B, 4B, 6B and it goes on.*1098

*In order to make the transition from level 3 to level 4, you need to absorb 8B ̃.*1104

*In order to make the transition from level 4 to level 5, you need to absorb 10B ̃.*1109

*In each case you need to absorb more energy to go to the next energy level up.*1114

*It is going to be important here and it tends to be a little bit confusing here for kids.*1119

*Now, when we actually see the spectrum, the spectrum itself, the lines are evenly spaced and the lines 2B apart.*1124

*You get a spectrum that looks like this.*1153

*Line, line, line, line, and this distance right here is always going to be 2B because you are going 2B, 4B, 6B, 8B.*1156

*This is very important.*1166

*I will do this in red.*1167

*Let me move over here.*1169

*I will write it down here.*1178

*On the spectrum, the lines are evenly spaced.*1183

*This does not mean that the energy levels are evenly spaced, they are not.*1186

*Each transition from 0 to 1 requires energy of 2B ̃, from 1 to 2 requires 4B ̃, from 2 to 3 requires 6B ̃,*1217

*that is how much energy that needs to be absorbed.*1233

*The spectrum itself, the lines are evenly spaced.*1236

*This is 2B, 4B, 6B, 8B.*1240

*This difference between them is the 2B.*1242

*However, that does not mean that the energy levels themselves are the same and we will demonstrate that in just a minute.*1245

*This does not mean that the energy levels are evenly spaced.*1249

*They are only evenly space on the spectrum itself.*1252

*The energies are as follows.*1256

*The levels of the individual energy levels are E sub J is equal to H ̅² / 2I × J × J + 1.*1271

*The energy level of 0 is equal to, when I put 0 in for this, I end up with 0.*1291

*Energy level of 1, I put 1 in for here, 1 + 1 is 2, 2 × 1 is 2.*1300

*2 H ̅²/ 2I = 2 H ̅²/ 2I.*1307

*The 2 cancels and I’m left with H ̅² / I.*1314

*My energy level for 2, I'm going to end up with 6 H ̅² / 2I which is just going to be 3 H ̅² / I.*1320

*When I calculate the third energy level, I'm going to end up with 12 H ̅² / 2I which is going to be 6 H ̅² / I.*1332

*Let me just go ahead and do one more.*1344

*Energy of a4th level is going to be 20 H ̅² / 2I which is going to equal 10 H ̅² / I.*1346

*Notice that the spacing between the energy levels is not constant.*1359

*From here to here, I have 2 H ̅² / I.*1363

*From here to here, I have 3 H ̅² / I.*1367

*From here to here, I have 4 H ̅² / I.*1370

*The energy levels themselves are different.*1374

*It is the spectrum spacing that is the same.*1377

*Once again, the spacing between the lines of the spectrum are evenly spaced.*1386

*The space between them is always going to be 2B ̃.*1406

*Let us go ahead and draw what this looks like.*1410

*Let us see if I should go ahead and do it on this page.*1415

*I’m going to go ahead and do it on the next page.*1419

*Let me go ahead and start down here.*1423

*Let me start with the 0 level and I will go to 1 level and I will go to the 2 level and I will go to the 3 level.*1426

*In case I got 0, 1, 2, 3.*1445

*I’m going from the 0 level to the 1 level, I'm going to have to absorb 2B amount of energy.*1457

*In order to go from 1 to 2, I'm going to have to absorb a total of 4B energy.*1464

*In order to go from 2 to 3, I’m going to have to absorb the total of 6B.*1473

*The amount of energy in order to go from 3 to 4, sure enough I'm going to have to absorb 8B.*1479

*The spectrum is just going to tell me how much energy I’m actually absorbing.*1486

*Here is going to be to 0 level, I'm going to get a line at that.*1496

*I’m going to get a line here and here.*1502

*This is 2B, this tells me 4B, this is 6B, and this is 8B.*1510

*The spacing in between is constant on the spectrum.*1521

*This is the spectrum, the amount of energy that takes to go from 1 level to the next*1526

*is not constant because the levels, the energies themselves are not constant.*1532

*This definitely distinguish between the energies and the amount of energy that we need to absorb in order to make the transition.*1538

*That is what is going on here, it is the spectrum spacing.*1545

*This 2B that is constant on the spectrum not here.*1548

*Let us see what we have got.*1556

*We have frequency = 2B × J + 1 or wave number = 2B ̃ × J + 1.*1561

*In either case, we have J goes 0, 1, 2…*1575

*We have B equal 2H / 8 π² I.*1583

*We also have the B ̃ = H / 8 π I² C × I.*1590

*These expressions give the frequency and or wave number required to transition, required to go from level J to level J + 1.*1600

*Of course, the energy of each level is given by H ̅² / 2I × J × J + 1.*1633

*These are the actual energies of the levels starting with J = 0 point energy.*1643

*Since the rotational inertia is equal to the reduced mass × the radius², notice the I shows up in those expressions.*1685

*We can use the separation between lines on the spectrum which we know is 2B to find R, the bond length of the molecule.*1697

*Profound, this is how we find bond lengths for most molecules, always the ones that have permanent dipole moment.*1735

*Let us go ahead and do an example.*1751

*Transitions between the rotational energy levels occur in a microwave for infrared region of the electromagnetic spectrum as we said.*1760

*One such spectrum for hydrogen bromide HBR79 shows a separation between adjacent lines of 16.7 inverse cm.*1772

*Use this data to calculate the rotational inertia and the bond length R for the molecule.*1785

*We know the spacing is 2B.*1792

*2B ̃ is equal to this 16.70 inverse cm.*1805

*B ̃ is equal to H / 8 π² CI.*1815

*2B which is 2 H / 8 π² C × I is equal to 16.70.*1829

*Let us go ahead and find I.*1842

*We want calculate I and we want to calculate R.*1844

*The first thing that we need to do is we need to find I.*1846

*We rearrange and we get I is equal to 2H ÷ 8 π² × 16.7.*1849

*We have 2 × 6 point this is planks constant, 626 × 10⁻³⁴ J/ s ÷ 8 π² × C*1859

*which is going to be 3.0 × 10¹⁰ cm/ s and then × our 16.7 inverse cm.*1895

*And then when we end up actually solving for this, calculating it, which is not my favorite part.*1910

*I still do not like all these numbers floating around, they make crazy.*1917

*But in any case, 5.594 × 10⁻⁴⁶ and we end up with a unit of kg m².*1920

*We went ahead and we found I, that is the rotational inertia of this particular system which is the HBR molecule.*1932

*Let me see, do I have another page?*1943

*We know that I is equal to the reduced mass × R².*1954

*We have I, we have the reduced mass, we can find that.*1960

*We need to find R, the bond length.*1965

*The first thing we need to do now is go ahead and find the reduced mass.*1967

*Let us go ahead and do that on the next page.*1972

*The reduced mass is equal to mass 1 × mass 2 ÷ mass 1 + mass 2.*1978

*The mass 1, the bromine is going to bromide, it is going to be 79 atomic mass units × hydrogen*1990

*which is 1 atomic mass unit ÷ the 79 + 1, which is the 80 atomic mass units.*2002

*We are going to need to change that to kg.*2011

*It is 1.661 × 10⁻²⁷.*2013

*There are that many kg in 1 atomic mass unit and we end up with a reduced mass equal to 1.64 × 10⁻²⁷ kg.*2020

*We are ready to solve.*2037

*I is equal to reduced mass R² which means that R is going to equal the rotational inertia ÷ the reduced mass the square root of that*2039

*And that is going to equal 5.594 × 10⁻⁴⁶ kg m² ÷ the reduced mass which is 1.64 × 10⁻²⁷.*2051

*It is going to be kg, kg cancels kg.*2073

*This is the square root and the square m² gives us a bond length of 5.8 × 10⁻¹⁰ m or 5.8 angstroms.*2077

*There we go, we use our microwave spectra data and we know the distance between the absorption bonds is going to be 2B ̃.*2096

*We use that information to find the rotational inertia and we use the rotational inertia to find the bond length.*2108

*That is all.*2115

*Thank you so much for joining us here at www.educator.com.*2116

*We will see you next time, bye.*2118

1 answer

Last reply by: Professor Hovasapian

Sat Apr 23, 2016 7:28 PM

Post by Tram T on April 21, 2016

Dear Prof. Hovasapian,

-I have question on the part when you mentioned having permanent dipole moment is a must so the electromagnetic radiation could 'grab' on the bond and 'rotate' the rotator faster:

-Is the same thing applied for vibrational transition where the dipole moment of the molecule must change either from already having a dipole moment to having another dipole or from not having a dipole moment to have one during vibration of the molecule?

-Is it why having a change in dipole moment also a must in vibrational transition in order for the electromagnetic radiation to 'grab' and 'pull' on to the 2 bonding atoms to make the 2 bonding atoms oscillate at higher frequency?

-If then why vibrational selection rule does not require permanent dipole moment but 'a change' in dipole moment instead?

-What does 'a change' in dipole moment during vibration really mean?

Thank you!!