CHEM 305

Electromagnetic radiation and matter

http://www.colorado.edu/physics/2000/waves_particles/

What is an electromagnetic wave?

Electromagnetic waves are produced by the motion of electricallycharged particles. These waves are called electromagnetic radiationbecause they radiate from electrically charged particles. They cantravel through vacuum, air, as well as other substances.

λ

λ

λν = c

ν = 1/λ

E = hν

c= 3 x 108 m.s-1

~

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Can one use radiowaves to look at proteins?How about x-rays?

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Radiation Process Information gained

X-rays Details of electronic structure

Ultraviolet & Details of electronic structurevisible and bond energies in molecules

Infrared Internuclear distances and forceconstants

transitionsof innerelectrons

transitionsof outerelectrons

changes invibrational-rotational state

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Radiation Process Information gained

Far infrared & Internuclear distancesmicrowave

Radio frequency magnetic environment of nucleus

changes inrotational state

change of nuclear spinorientation

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An electromagnetic wave depends on both the electric and magnetic fieldstrength, i.e. as described by Maxwell’s equations

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+ -

If we model the electron above as a mass on a spring and pull the mass downwards,it will feel a restorative force given by Hooke’s law, namely

F = -kx (1.0)

where k is the spring constant. The total force will be

F = m d2x + kx =0 (1.1)dt2

and the frequency of oscillation is given by

ν = 1 k or ω = k (1.2)2π m m

½ ½

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Equation 1.1 can be solved to give,

x = ½ x0 (eiω(t-t0) + e-iω(t-t0))x = x0 cos[ω(t-t0)]. (1.3)

From this we can calculate the velocity,

v = dx/dt = -ωx0 sin[ω(t-t0)] (1.4)

The total energy of the system is the kinetic energy plus the potential energy,

E = ½ mv2 + ½ kx2

= ½ m (ω2x02 sin2 [ω(t-t0)]) + ½ kx0

2 cos2[ω (t-t0)]= ½ kx0

2 {sin2[ω (t-t0)] + cos2[ω(t-t0)]}= ½ kx0

2 (1.5)

We can also write the energy in terms of momentum p = mv,

E = p2 + kx2 (1.6)2m 2

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For a long time, this classical description was sufficient to describe all observedphenomena. A problem arose when it came time to describe blackbodyradiation. In this case, equation 1.6 has to be written in terms of Hamiltonianoperators, i.e.

E ψ = - h2 d2ψ + kx2 ψ = Kψ + Pψ2m dx2 2

where ψ is a wavefunction and h is Planck’s constant. This equation is the Schrödinger equation.

Erwin Schrödinger

Nobel Laureate1933

^ ^kinetic energy operator

potential energy operator

As an aside….

What is blackbody radiation? Why can’t it be described classically?

see http://www.phys.virginia.edu/classes/252/black_body_radiation.html for moredetails.

Heated “bodies” radiate – i.e. emit energye.g. ring on your stove.

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Harmonic oscillator

In the previous three slides, we described the generation of electromagneticwaves as an oscillation of the charges. We then modelled this oscillation as aweight on a spring.

F = m d2x + kx =0dt2

with the frequency of oscillation given by

ν = 1 k or ω = k2π m m

This is a harmonic oscillator.ma

kx

½ ½

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For an example, go to http://www3.adnc.com/~topquark/fun/JAVA/dho/dho.htmland run the applet with a damping of zero (and launch). What you see is:

Position cosine function (equation 1.3)

Velocity dx/dt => sine function (equation 1.4)

Energy constant (equation 1.5)

Conclusion:

- The harmonic oscillator canbe used to describe EM waves.

Can it describes all types of EM waves?Also those from diatomics, i.e. which arisefrom the vibration of molecules?

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The harmonic oscillator as a model for a diatomic molecule

m1 m2

x1 x2

We can write two equations of motion:

m1 d2x1 = k (x2 – x1 – l0) (1.8)dt2

m2 d2x2 = - k (x2 – x1 – l0) (1.9)dt2

where l0 is the equilibrium length of the spring. These equations indicate that if the spring is stretched, i.e. z2 – z1 > l0, then the force acting on m1 is towards

the right and that on m2 is towards the left (negative sign).

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If we add equations 1.8 and 1.9, we get

m1 d2x1 + m2 d2x2 = 0dt2 dt2

or

d2 (m1 x1 + m2 x2) = 0dt2

We can define a center of mass coordinate:

X = m1x1 + m2x2 = m1x1 + m2x2m1 + m2 M

This allows us to write

M d2X = 0dt2

which implies that the center ofmass moves uniformly in time.

m1 m2

x1 x2X

m1

m2

x1 x2X

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But we are interested in the motion on the two masses, i.e. the relative motion, therefore we need to define a relative coordinate

ξ = x2 – x1

To write the equation of motion in terms of this relative coordinate, we devideequation 1.8 by m1 and equation 1.9 by m2 and subtract the new equations from each other, i.e.

d2x2 – d2x1 = - k (x2 – x1 – l0) – k (x2 – x1 – l0)dt2 dt2 m2 m1

d2x = - kx 1 + 1 = - kx m1 + m2 = -kxdt2 m1 m2 m1m2 µ

where x = ξ – l0 and µ is the reduced mass. In other words, we have that

µ d2x + kx = 0dt2

m1 m2

x1 x2µ

Reduce a 2 body problem into a 1 body problem

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As was done for the harmonic oscillator, we can now determine the potentialenergy of the system. Recall that previously, we found the potential energy tobe U = ½ kx2. We can write a mathematical expansion of the energy for ourdiatomic molecule by using a Taylor expansion of the potential energy U(ξ) about the equilibrium position l0.

The expansion is:

U(ξ) = U(l0) + dU (ξ – l0) + 1 d2U (ξ – l0)2 + 1 d3U (ξ – l0)3 +… (2.0)dξ 2! dξ2 3! Dξ3

where each derivative is evaluated at ξ = l0.

To simplify equation 2.0, we can look atthe potential energy curve and make afew definitions:

• ∆U(ξ) = U(ξ) – U(l0)• dU/dξ @ l0

l0

U(ξ

)ξ

U(l0)

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As a result, we have

∆U(ξ) = 1 k (ξ – l0)2 + 1 γ (ξ – l0)3 +… (2.1)2 6

= ½ kx2 + (1/6) γx3 + ….

If we keep to small displacements of x, we can simplify the equation aboveto

∆U(x) = ½ kx2

which is a quadratic equation.

As you can see from the graph,this equation is a goodestimate for small x.

l0 ξ

U(ξ

)

At this point: We can describe EM waves – but in order to describee.g. scattering – we need to introduce two more concepts

1- DAMPED

2- DRIVEN

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We can now take our weight on a spring and submerge the entire system ina viscous medium:

The force will therefore contain an additional termdue to friction,

F = m d2x + f dx + kx =0dt2 dt

This is a damped harmonic oscillator.

As before, it can be shown that a solution to this equation is:

x = exp{-f t/2m} cos(ω’(t-t0)) with ω’ = k – f 2 ½ (1.7) m 2m

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Proof:

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For an example, go to http://www3.adnc.com/~topquark/fun/JAVA/dho/dho.htmland run the applet with a damping of 0.06, 0.12, 0.24, 0.32, 0.5, and 1.8 (usinglaunch after each increase in damping). What you see is:

Position exponential decay * cosine function (equation 1.7)

Velocity ???

Energy ???

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Driven harmonic oscillators

Up to this point, we have considered both undamped and damped harmonicoscillators. In both cases, no driving (additional) force was applied, i.e.

F = ma + kx = 0 harmonicF = ma + f v + kx = 0 damped harmonic

In the following section, we will see what effect applying a driving force has onthe system. We are interested in doing this because the interaction of radiationwith matter is described by a driven and damped harmonic oscillator.

Let us consider a driving force of F0 cos(ωt). What will the equation of force be?

F = ma + f v + kx = F0 cos(ωt)md2x + f dx + kx = F0 cos(ωt)

dt2 dtExample:

http://webphysics.ph.msstate.edu/javamirror/explrsci/dswmedia/drivosc.htm

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A solution to this equation is:

x = x’ cos(ωt) + x’’ sin(ωt)

with

x’ = F0 m (ω02 – ω2)

m2 (ω02 – ω2)2 + f 2ω2

and

x’’ = F0 f ωm2 (ω0

2 – ω2)2 + f 2ω2

with ω0 = k ½ is the frequency of the mass, and ω is the frequency of the drivingm

force (example – the sinusoidal box on web site).

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0

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Two conditions are important:

Scattering limit:m2(ω0

2 – ω2)2 >> f 2 ω2Resonance limit:ω0 ~ ω

x’ = F0 m (ω02 – ω2)

m2 (ω02 – ω2)2 + f 2ω2

x’’ = F0 f ωm2 (ω0

2 – ω2)2 + f 2ω2

x’ = F0m (ω0

2 – ω2)

x’’ = 0

x’ = F0 (ω0 – ω) τ2

2mω0 [1 + (ω0 – ω)2 τ2]

x’’ = F0 τ2mω0 [1+ (ω0 – ω)2 τ2]

where τ = 2m/f