Rotational Transitions, Diatomic

The Planck Hypothesis

In order to explain the frequency distribution of radiation from a hot cavity (blackbody radiation) Planck proposed the ad hoc assumption that the radiant energy could exist only in discrete quanta which were proportional to the frequency. This would imply that higher modes would be less populated and avoid the ultraviolet catastrophe of the Rayleigh-Jeans Law.

The quantum idea was soon seized to explain the photoelectric effect, became part of the Bohr theory of discrete atomic spectra, and quickly became part of the foundation of modern quantum theory. http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3

The Photoelectric Effect

The remarkable aspects of the photoelectric effect when it was first observed were:

1. The electrons were emitted immediately -

no time lag! 2. Increasing the intensity of the light increased the number of photoelectrons, but

not their maximum kinetic energy! 3. Red light will not cause the ejection of

The details of the photoelectric effect were in electrons, no matter what the intensity! direct contradiction to the expectations of very 4. A weak violet light will eject only a few well developed classical physics. electrons, but their maximum kinetic

energies are greater than those for intense The explanation marked one of the major steps light of longer wavelengths! toward quantum theory. http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html#c2 http://hyperphysics.phy-astr.gsu.edu/hbase/grexp.html#c1

Properties of Molecules

The investigation of molecular structure parallels the study of atomic structure in that the methods of quantum mechanics are applied along with the information obtained from molecular spectra.

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Molecular Spectra The most commonly observed molecular spectra involve electronic, vibrational, or rotational transitions. For a diatomic molecule, the electronic states can be represented by plots of potential energy as a function of internuclear distance. Electronic transitions are vertical or almost vertical lines on such a plot since the electronic transition occurs so rapidly that the internuclear distance can't change much in the process. Vibrational transitions occur between different vibrational levels of the same electronic state. Rotational transitions occur mostly between rotational levels of the same vibrational state, although there are many examples of combination vibration-rotation transitions for light molecules. http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/molec.html#c1

Rotational Spectra

Incident electromagnetic waves can excite the rotational levels of molecules provided they have an electric dipole moment. The electromagnetic field exerts a torque on the molecule. The spectra for rotational transitions of molecules is typically in the microwaveregion of the electromagnetic spectrum. The rotational energies for rigid molecules can be found with the aid of the Shrodinger equation. The diatomic molecule can serve as an example of how the determined moments of inertia can be used to calculate bond lengths.

The illustration at left shows some perspective about the nature of rotational transitions. The diagram shows a portion of the potential diagram for a stable electronic state of a diatomic molecule. That electronic state will have several vibrational states associated with it, so that vibrational spectra can be observed. Most commonly, rotational transitions which are associated with the ground vibrational state are observed.

Rotational Energies

The classical energy of a freely rotating molecule can be expressed as rotational kinetic energy

where x, y, and z are the principal axes of rotation and Ix represents Index the moment of inertia about the x-axis, etc. In terms of the angular momenta about the principal axes, the expression becomes Molecular spectra concepts

The formation of the Hamiltonian for a freely rotating molecule is accomplished by simply replacing the angular momenta with the corresponding quantum mechanical operators.

Diatomic molecules

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Diatomic Molecules

For a diatomic molecule the rotational energy is obtained from the Schrodinger equation with the Hamiltonian expressed in terms of the angular momentum operator.

More detail

where J is the rotational angular momentum quantum number and I is the moment of inertia. Index Determining the rotational constant B Molecular spectra concepts

enables you to calculate the bond length R. The allowed transitions for the diatomic molecule are regularly spaced at interval 2B. The measurement and

Rotational transitions identification of one spectral line allows

one to calculate the moment of inertia and then the bond length.

Examples HCl CN CH

Table of diatomic molecule parameters

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Rotational Transitions, Diatomic

For a rigid rotor diatomic molecule, the selection rules for rotational transitions are ΔJ = +/-1, ΔMJ = 0 .

The rotational spectrum of a diatomic molecule consists of a series of equally spaced absorption lines, typically in themicrowave region of Index the electromagnetic spectrum. The energies of the spectral lines are Molecular

2(J+1)B for the transitions J -> J+1. spectra For real molecules like the example of concepts HCl , the successive transitions are a bit lower than predicted because centrifugal distortion lengthens the molecule, increasing its moment of inertia.

Moment of inertia for diatomic molecule

Vibration/rotation transitions in HCl

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Moment of Inertia, Diatomic

The moment of inertia about thecenter of mass is

From the center of mass definition

and

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Vibrational Spectra of Diatomic Molecules

The lowest vibrational transitions of diatomic molecules approximate the quantum harmonic oscillator and can be used to imply the bond force constants for small oscillations. The following is a sampling of transition frequencies from the n=0 to n=1 vibrational level for diatomic molecules and the calculated force constants.

Frequency Force constant Molecule x1013 Hz N/m HF 12.4* 970 HCl 8.66 480 HBr 7.68 410 HI 6.69 320 CO 6.42 1860 NO 5.63 1530 * From vibrational transition 4138.52 cm- 1 in Herzberg'stabulation. http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibspe.html#c1

Vibration-Rotation Spectrum of HCl

Add annotation to spectrum

Index

Molecular spectra concepts

Reference Tipler & Llewellyn Sec. 9-4

A classic among molecular spectra, the infrared absorption spectrum of HCl can be analyzed to gain information about both rotation and vibration of the molecule.

The absorption lines shown involve transitions from the ground to first excited vibrational state of HCl, but also involve changes in the rotational state. The rotational angular momentum changes by 1 during such transitions. If you had a transition from j=0 in the ground vibrational state to j=0 in the first excited state, it would produce a line at the vibrational transition energy. As observed, you get a closely spaced series of lines going upward and downward from that vibrational level difference. The splitting of the lines shows the difference in rotational inertia of the two chlorine isotopes Cl-35(75.5%) and Cl-37(24.5%).

From the spectrum above, you can examine details about the following:

Bond force constant Bond length

j for peak intensity Relative intensities

Pure rotational transitions in HCl

Vibration-rotation spectrum of HBr

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Bond Force Constant for HCl

By treating the vibrational transition in the HCl spectrum from its ground to first excited state as a quantum harmonic oscillator, the bond force constant can be calculated. This transition frequency is related to the molecular parameters by:

The desired transition frequency does not show up directly in the observed spectrum, because there is no j=0, v=0 to j=0, v=1 transition; the rotational quantum number must change by one unit. It can be approximated by the midpoint between the j=1,v=0->j=0,v=1 transition and the j=0,v=0- Index >j=1,v=1 transition. This assumes that the difference between the j=0 and j=1 levels is the same for the ground Molecula and first excited state, which amounts to assuming that the r spectra first excited vibrational state does not stretch the bond. This concepts "rigid-rotor" model can't be exactly correct, so it introduces some error. Referenc e For the HCl molecule, the needed reduced mass is Tipler & Llewelly n Sec. 9-4

Note that this is almost just the mass of the hydrogen. The chlorine is so massive that it moves very little while the hydrogen bounces back and forth like a ball on a rubber band!

Substituting the midpoint frequency into the expression containing the bond force constant gives:

Despite the approximations, this value is quite close to the value given in the table.

Pure rotational transitions in HCl

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Bond Length of HCl A bond length for the HCl molecule can be calculated from the HCl spectrum by assuming that it is a rigid rotor and solving the Schrodinger equation for that rotor. For a free diatomic molecule the Hamiltonian can be anticipated from the classical rotational kinetic energy

and the energy eigenvalues can be anticipated from the nature of angular momentum.

Assuming that the bond length is the same for the ground and first excited states, the difference between the j=1,v=0- >j=0,v=1 transition and the j=0,v=0->j=1,v=1 transition frequencies can be used to estimate the bond length. The separation between the two illustrated vibration-rotation transitions is assumed to be twice the rotational energy change from j=0 to j=1.

Substitution of numerical values leads to an estimate of the bond length r:

This compares reasonably with the value r=.127 nm obtained from pure rotational spectra.

Other Heteronuclear Diatomic Molecules Index Equilibrium Separation (nm) Molecule Dissociation Energy(eV) (Bond length) Tables BN 4.0 0.128 Data CO 11.2 0.113 References HBr ... 0.141 Krane Ch 9 HCl 4.4 0.127

HF 5.8 0.092 Beiser, Concepts... NO 7.0 0.115 Sec 13.5 PbO 4.1 0.192

PbS 3.3 0.239

Ionic Diatomic Molecules Homonuclear Diatomic Molecules

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Properties of Ionic Diatomic Molecules

Equilibrium Separation (nm) Molecule Dissociation Energy(eV) (Bond length) NaCl 4.26 0.236

NaF 4.99 0.193

NaBr 3.8 0.250

NaI 3.1 0.271

NaH 2.08 0.189

LiCl 4.86 0.202

LiH 2.47 0.239

LiI 3.67 0.238

KCl 4.43 0.267

KBr 3.97 0.282

RbF 5.12 0.227

RbCl 4.64 0.279

CsI 3.57 0.337 http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diatomic.html#c2

Homonuclear Diatomic Molecules

Equilibrium Separation (nm) Molecule Dissociation Energy(eV) (Bond length) H-H 4.5 0.075

N-N 9.8 0.11

O-O 5.2 0.12

F-F 1.6 0.14

Cl-Cl 2.5 0.20

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Raman Scattering

When light encounters molecules in the air, the predominant mode of scattering is elastic scattering, called Rayleigh scattering. This scattering is responsible for the blue color of the sky; it increases with the fourth power of the frequency and is more effective at short wavelengths. It is also possible for the incident photons to interact with the molecules in such a way that energy is either gained or lost so that the scattered photons are shifted in frequency. Such inelastic scattering is called Raman scattering.

Like Rayleigh scattering, the Raman scattering depends upon the polarizability of the molecules. For polarizable molecules, the incident photon energy can excite vibrational modes of the molecules, yielding Index scattered photons which are diminished in energy by the amount of the vibrational transition energies. A spectral analysis of the scattered light Scattering under these circumstances will reveal spectral satellite lines below the concepts Rayleigh scattering peak at the incident frequency. Such lines are called

"Stokes lines". If there is significant excitation of vibrational excited states Atmospheric of the scattering molecules, then it is also possible to observe scattering at optics frequencies above the incident frequency as the vibrational energy is concepts added to the incident photon energy. These lines, generally weaker, are called anti-Stokes lines. Molecular spectra Although finding some application in vibrational spectroscopy of concepts molecules, the use of direct infrared sources for such spectroscopy is usually much easier. Raman spectroscopy has found some application in Reference remote monitoring for pollutants. For example, the scattering produced by Thornton a laser beam directed on the plume from an industrial smokestack can be and Rex used to monitor the effluent for levels of molecules which will produce Sec 11.1 recognizable Raman lines.

Raman scattering can also involve rotational transitions of the molecules from which the scattering occurs. Thornton and Rex picture a photon of energy slightly than the energy separation of two levels being scattered, with the excess energy released in the form of a photon of lower energy. Since this is a two-photon process, the selection rule is J = +/-2 for rotational Raman transitions. The sketch below is an idealized depiction of a Raman line produced by interaction of a photon with a diatomic molecule for which the rotational energy levels depend upon one moment of inertia. The upper electronic state of such a molecule can have different levels of rotational and vibrational energy. In this case the upper state is shown as being in rotational state J with scattering associated with an incoming photon at energy matching the J+2 state.

Since the Raman effect depends upon the polarizability of the molecule, it can be observed for molecules which have no net dipole moment and therefore produce no pure rotational spectrum. This process can yield information about the moment of inertia and hence the structure of the molecule.

In Raman scattering, an intense monochromatic light source (laser) can give scattered light which includes one or more "sidebands" that are offset by rotational and/or vibrational energy differences. This is potentially very useful for remote sensing, since the sideband frequencies contain information about the scattering medium which could be useful for identification. Current projects envision Raman scattering as a tool for identification of mineral forms on Mars. Such remote sensing could become a major tool in planetary exploration.

Raman scattering from lunar soil

Some history

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C. V. Raman

C. V. Raman discovered the inelastic scattering phenomenon which bears his name in 1928 and for it he was awarded the Nobel Prize for Physics in 1930.Raman scattering produces scattered photons which differ in frequency from the radiation source which causes it, and the difference is related to vibrational and/or rotational properties of the molecules from which the scattering occurs. It has become more prominent in the years since powerful monochromatic laser sources could provide the scattering power.

Raman scattering from lunar soil

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Circular Orbit Index Gravity supplies the necessary centripetal force to hold a satellite in orbit about the earth. The circular orbit is a special case since orbits are generally Gravity ellipses, or hyperbolas in the case of objects which are merely deflected by concepts the planet's gravity but not captured. Setting the gravity force from the univeral law of gravity equal to the required centripetal force yields the Orbit description of the orbit. The orbit can be expressed in terms of concepts the acceleration of gravity at the orbit.

The force of gravity in keeping an object in circular motion is an example ofcentripetal force. Since it acts always perpendicular to the motion, gravity does not do work on the orbiting object if it is in a circular orbit.

Calculation

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Earth's Gravity

The weight of an object is given by W=mg, the force of gravity, which comes from the law of gravity at the surface of the Earth in the inverse square law form:

Index

At standard sea level, the acceleration of gravity has the value g = 9.8 m/s2, Gravity but that value diminishes according to the inverse square law at greater concepts distances from the earth. The value of g at any given height, say the height of an orbit, can be calculated from the above expression. Orbit concepts Above the earth's surface at a height of h = m = x 106 m, which corresponds to a radius r = x earth radius, the acceleration of gravity is g = m/s2 = x g on the earth's surface.

Please note that the above calculation gives the correct value for the acceleration of gravity only for positive values of h, i.e., for points outside the Earth. If you drilled a hole through the center of the Earth, the acceleration of gravity would decrease with the radius on the way to the center of the Earth. If the Earth were of uniform density (which it is not!), the acceleration of gravity would decrease linearly to half the surface value of g at half the radius of the Earth and approach zero as you approached the center of the Earth.

Hole through center of Earth

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Binary Circular Orbit

From the gravity force and the necessary centripetal force:

If you are riding on one of the masses, the relative motion equation has the same form if you substitute the reduced mass

Index

Gravity which gives the orbit equation: concepts

Orbit concepts

This leads to Kepler's 3rd law (the Law of Periods) which is useful for the analysis of the orbits of moons and binary stars.

Since the period T of the orbit is given by

then the motion equation can be written

which reduces to

If we use the convenient astronomical units

 r = a in AU (astronomical units)  G = 4π2

(solar masses)

m in

 T in years then this takes the form

This applies to circular orbits where a is the radius, and to elliptical orbits where a is the semi-major axis.

and from just the period and orbit radius you can obtain the sum of the masses m1+ m2. If you can obtain the individual orbit radii r1 and r2 then you can use thecenter of mass condition

with the measured mass sum to obtain the individual masses

These relationships are important in the study of visual binaries.

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Reduced Mass

The relative motion of two objects that are acted upon by a central force can be described by Newton's 2nd Law as if they were a single mass with a value called the "reduced mass".

From Newton's 3rd Law :

The relative acceleration of the two masses is

Since a2 is negative, this can be rewritten in terms of the magnitudes of the quantities:

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Inde

Force from spherical shell of mass M

Gravitational lens

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Gravity

Gravity is the weakest of the four fundamental forces, yet it is the dominant force in the universe for shaping the large scale structure of galaxies, stars, etc. The gravitational force between two masses m1 and m2 is given by the relationship:

Index This is often called the "universal law of gravitation" and G the universal gravitation constant. It is an example of an inverse square law force. The force is always attractive Gravity and acts along the line joining the centers of mass of the two masses. The forces on the Concept two masses are equal in size but opposite in direction, obeying Newton's third law. Viewed as an exchange force, the massless exchange particle is called the graviton. s

The gravity force has the same form as Coulomb's law for the forces between electric charges, i.e., it is an inverse square law force which depends upon the product of the two interacting sources. This led Einstein to start with the electromagnetic force and gravity as the first attempt to demonstrate the unification of the fundamental forces. It turns out that this was the wrong place to start, and that gravity will be the last of the forces to unify with the other three forces. Electroweak unification (unification of the electromagnetic and weak forces) was demonstrated in 1983, a result which could not be anticipated in the time of Einstein's search. It now appears that the common form of the gravity and electromagnetic forces arises from the fact that each of them involves an exchange particle of zero mass, not because of an inherent symmetry which would make them easy to unify.

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Examples of Trajectories

Common misconceptions about guns: A dropped bullet will hit the ground before one which is fired from a gun. As shown in the illustration of a horizontal launch, gravity acts the same way on Index both bullets, giving them the same downward acceleration and making them strike the ground at the same time if the bullet is fired horizontally over level ground. Trajectory concepts Bullets fired from high-powered rifles drop only a few inches in hundreds of yards. Fired at twice the speed of sound, a bullet will drop over 3 inches in 100 yards, and at 300 yards downrange will have dropped about 30 inches. Plug in numbers into the bullet drop calculation to see for yourself. Ammunition manufacturers contribute to this misconception by stating the drop of their projectiles as just the extra drop caused by frictional drag compared to an ideal frictionless projectile.

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Drop of a Bullet

If air friction is neglected, then the drop of a bullet fired horizontally can be treated as an ordinary horizontal trajectory. The air friction is significant, so this is an idealization.

If the muzzle velocity is v = m/s = ft/s = mi/hr = km/hr Index and the distance downrange is Trajectory R = m = ft = yards, concepts Then the amount of drop of the bullet below the horizontal would be d = m = cm = inches

If the gun is fired on level ground at a height of m = ft, then the bullet will hit the ground in seconds, having traveled a distance of meters = feet.

To hold the drop to cm = inches at the downrange distance R above would require a muzzle velocity of m/s = ft/s.

Index

Motion concept s

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Freefall

In the absence of frictional drag, an object near the surface of the earth will fall with the constant acceleration of gravity g. Position and speed at any time can be calculated from themotion equations.

Illustrated here is the situation where an object is released from rest. It's position and speed can be predicted for any time after that. Since all the quantities are directed downward, that direction is chosen as the positive direction in this case.

Index

At time t = s after being dropped, Trajectory concepts the speed is vy = m/s = ft/s ,

The distance from the starting point will be

y = m= ft Enter data in any box and click outside the box.

Note that you can enter a distance (height) and click outside the box to calculate the freefall time and impact velocity in the absence of air friction. But the calculation assumes that the gravity acceleration is the surface value g = 9.8 m/s2, so the height is great enough for gravity to have changed significantly the results will be incorrect.

Free fall with air friction

Free fall from great height

Free fall experiment with spark timer

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Peak at m at Vertical Trajectory t= Vertical motion under the influence of gravity can be described by the basic motion equations. Given the constant acceleration of gravity g, the position and speed at any time can be calculated from the motion equations:

Index You may enter values for launch velocity and time in the boxes below and click outside the box to perform the calculation. Trajectory concepts

For launch speed v0y = m/s = ft/s and time t = s ,

The values below are output values; those boxes will not accept input for calculation. The velocity will be

vy = m/s = ft/s s and the height will be y = m = ft

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Index

Trajectory concepts

Calculation

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Horizontal Launch

All the parameters of a horizontal launch can be calculated with the motion equations, assuming a downward acceleration of gravity of 9.8 m/s2.

Time of flight

t = s Vertical impact velocity

vy = m/s Launch velocity Index

v = m/s 0 Trajectory Height of launch concepts h = m Horizontal range

R = m

Calculation is initiated by clicking on the formula in the illustration for the quantity you wish to

calculate.

Include demonstration apparatus

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General Ballistic Trajectory

The motion of an object under the influence of gravity is determined completely by the acceleration of gravity, its launch speed, and launch angle provided air friction is negligible. The horizontal and vertical motions may be separated and described by the general motion equations for constant acceleration. The initial vector components of the velocity are used in the equations. The diagram shows trajectories with the same launch speed but different launch angles. Note that the 60 and 30 degree trajectories have the same range, as do any pair of launches at complementary angles. The launch at 45 degrees gives the maximum range. Index

Trajectory concepts

Calculation

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At time t = sec:

Horizontal velocity

vx = m/s.

Horizontal distance

x = m. Index

Trajectory concepts

Vertical velocity

vy = m/s.

For launch velocity v0 = m/s, launch angle θ = degrees: Vertical position

y = m.

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Index

Trajectory concepts

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Index

Trajectory concepts

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Index

Trajectory concepts

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For launch velocity

v0 = m/s, launch angle

θ = degrees, The horizontal range is Index

R = m. Trajectory The total time of flight is concepts t = s. The peak height is

h = m.

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Will it clear the fence?

The basic motion equations can be solved simultaneously to express y in terms of x.

For launch velocity

v0 = m/s = ft/s, launch angle Index θ = degrees, and horizontal range Trajectory x = m = ft, concepts

the calculated height is

y = m = ft. The time of flight is

t = s.

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Where will it land?

The basic motion equations give the position components x and y in terms of the time. Solving for the horizontal distance in terms of the height y is useful for calculating ranges in situations where the launch point is not at the same level as the landing point.

Index

Trajectory concepts

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Where will it land?

Launch velocity

v0 = m/s = ft/s, launch angle

θ = degrees, Index and trajectory height y = m = ft, Trajectory concepts

The two calculated times are t1 = s and Note that the value y in the illustration is downward t = s. 2 and it is presumed that upward is positive. To The corresponding ranges are reproduce the scenario in the diagram, the input value x1 = m = ft of y should be negative.

and x2 = m = ft.

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Launch Velocity

The launch velocity of a projectile can be calculated from the range if the angle of launch is known. It can also be calculated if the maximum height and range are known, because the angle can be determined.

From the range relationship, the launch velocity can be calculated. For range Index

R = m = ft, Trajectory and launch angle concepts θ = degrees,

the launch velocity is

v0 = m/s = ft/s.

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Launch Velocity

Index

Trajectory concepts

For range

R = m = ft, and peak height h = m = ft, the launch velocity is v0 m/s = ft/s.

The required launch angle is

θ = degrees.

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Angle of Launch

Variation of the launch angle of a projectile will change the range. If the launch velocity is known, the required angle of launch for a desired range can be calculated from the motion equations.

From the range relationship, the angle of launch can be determined. For range

R = m = ft, Index and launch velocity Trajectory v0 = m/s = concepts ft/s. there are two solutions for the launch angle.

θ1 = degrees,

θ2 = degrees,

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