SpectroscopyThe study and analysis of spectra.

Spectrum - the range and distribution of an effectIn science it particularly refers to the distribution of some

effect with respect to energy.

Examples: -the political spectrum-the microwave spectrum-the optical emission spectrum of the sun-the optical absorption spectrum of a liquid-the electron energy loss spectrum of a material-nuclear magnetic resonance spectroscopy

We want to measure how an emission, absorption, or other effectbehaves as a function of wavelength or energy.

So we need to be able to separate light/particles/EM waves bywavelength/energy.

For light, we can use dispersion, the dependence of the index of refraction on wavelength.

Resolution: The minimum difference between two discrete values that can be distinguished bya measuring device.

The Great Compromise: Resolution vs. Signal

Diffraction Grating

Atomic Spectra

Atoms exhibit sharp emission lines instead of continuous distributions.Absorption lines are in the same place as emission lines.

Johannes Balmer found that the visible and near UV emission lines from hydrogen were at

where ( )λn nmn

n=

−

364 6

4

2

2. n = 3 4 5, , ...

Balmer’s formula fit with the general expression of Rydberg and Ritz:

where 1 1 1

2 2λmn

Rm n

= −

n m>

R is called the Rydberg constant.

For Hydrogen, .R mH = × −1097 107 1.For the Balmer Series, m=2.

Thomson Model of an Atom

- Electrons in a fluid

- Fluid carries most of the atom’s mass

- Fluid is positively charged (+Ne) so that the atom is neutral.

Problems- No stable stationary arrangement exists (no static equilibrium)

- Electrons must move and therefore they must accelerate

- Electrons that accelerate must radiate electromagnetic waves

- Calculated frequencies for possible arrangements don’t match the observed frequencies of emitted light.

Rutherford ScatteringIdea: use energetic particles to probe the structure of the atom.

Uranium emits two kinds of radioactive particles, """" and $$$$. They showed that the alpha particle was really a helium nucleus, so, it had a charge of +2e. The alpha particles were all emitted with the same kinetic energy.

Alpha particles could be observed as a flash of light when they struck a zinc sulfide screen.

Allow a beam of alpha particles to collide with a thin metal foil.Observe the number of alpha particles scattered at each angle.

Expected result:

Actual Result: Most alpha particles did scatter at very small angles. But, manymore large angle scattering events occurred than would beexpected from the Thomson Model.

Explanation: Positive charge in the atom must be concentrated, not spread outuniformly. Alpha particles which miss the “nucleus go prettymuch straight through. Alpha particles that come close or strikethe “nucleus” will scatter through large angles.

Rutherford Scattering TheoryAssume the atom has a nucleus which is a point charge, +Q.The alpha particle will feel a repulsive force while it is near the nucleus: F=kq""""Q/r2. Assume the nucleus is fixed in position and will not be moving after the alpha passes through.So kinetic energy will be conserved and the alpha particle will not change speed.

Classical Mechanics can be used to show that the scattering angle, , depends on b, the impactθparameter, such that

.bkq Q

m v=

α

α

θ2 2

cot

All alpha particles striking with an impact parameter of b or less will scatter through an angle of or greater.θ

Since we don’t know the impact parameter of an individual event, how do we analyze thedata?

Calculate the fraction of alpha particles which will be scattered through an angle of orθgreater.

Assume that the foil is thin enough so that all nuclei are available for scattering (no blockingor channeling) and also assume there will be negligible multiple scattering events.

The area per atom (nucleus) will be ,A

N

A

nAt nt= =

1

where A is the area of the beam, N is the number of nuclei in the path of the beam, n is theatom density in the foil, and t is the thickness of the foil.

An alpha particle must strike within an area of to scatter through an angle of orπ b2 θlarger. The fraction of alpha particles which will strike this area will be

.

fb

nt

b nt=

=π

π2

2

1

We can combine this with the relationship between impact parameter and scattering angle,

,bkq Q

m v

k e Ze

E

kZe

Ek k

=

=

=

α

α

θ θ θ2

2

2

2

2 2 2cot

( )( )cot cot

to get

.f

ntk Z e

Ek

=

π θ2 2 4

2

2

2cot

How do we find the number of alpha particles at a particular angle?

The Intensity of alpha particles scattered at angles between and will be θ θ θ+ d

.

− =−

−

I dfI ntk Z e

Ed

k0

02 2 4

23

2

2

πθ

θ θcos

sin

Of these, only some of the alpha particles with the correct angle will strike our detector,

.

( )( ) ( )I I df

A

r rd

I A ntk Z e

r E

II A ntk Z e

r E

I A ntk Z e

sc sc

k

sc

k

sc

= − =

=

=

00

2 2 4

2 23

02 2 4

2 23

02 2 4

2 22

2

22

22 2 2

4

π θ θ

θ

θθ

θ

θ θ θ

sin

cos

sin sin

cos

sin cos sinr Ek

2 24

1

2sin

θ

If the nucleus is not a point charge, could we tell that from Rutherford scattering?

If the alpha particles are energetic enough, they could getcloser than R in a head-on collision. This would cause

deviations from the predictions of Rutherford scatteringtheory.

Closest approach would be the turning point, that is, thepoint in a head-on collision where the alpha particle has

zero kinetic energy.

E E

E V

E m vk e Ze

r

rkZe

m v

kZe

E

i closest

k closest

kclosest

closestk

==

= =

= =

1

2

2

4 2

2

2

2

2

α

α

( )( )

Bohr Model of the AtomElectrons are held in the atom by the Coulomb force.

Taking the inward radial components of givesr r

F ma=

.kZe

r=m

vr

2

2

2

The stable orbits which satisfy this equation are circles and ellipses. Forsimplicity, Bohr assumed circular orbits.The energy of a circular orbit of radius r can be calculated.

The electric potential energy is ,EkZe

rP

2

= −

The kinetic energy is ,E mvkZe

2rK12

22

= =

The total energy is . E E EkZe

2rkZe

rkZe

2rK P

2 2 2

= + = − = −

The frequency of the orbit is .fv

2 rkZe mr

2 r=

kZe 4 mr

2 2 2

3 2= =ππππ ππππππππ/ /

/

Classically, the electron should radiate electromagnetic waves at thisfrequency. When it does, the atom loses energy (E becomes morenegative) and the radius of the orbit must decrease.

As the radius decreases, the frequency of the orbit will increase and sowill the frequency of the radiated waves.

The orbit should decay very quickly with the electron spiraling into thenucleus. As this happens, the frequency of the emitted radiation shouldbe continuously changing. The observed emission spectrum from awhole bunch of atoms should be a continuous spectrum.

Obviously, this isn’t correct.

Bohr made two further postulates:

Only certain orbits (radii) are stable, called stationary states. The orbital radii are quantized, taking on only discrete values.

Radiation is emitted when an electron jumps to a lower energy orbit, .E E hf hci f− = = / λλλλ

To decide which orbits are the stationary states, he used thecorrespondence principle:

For large distances and energies, the quantum and classical results should be the same.

The result is that radii are quantized such that .r n rn2

1=Does this satisfy the correspondence principle?

.

( )

( )

hf=E E

hkZe 4 m

r

kZe2r

kZe2r

hkZe 4 m

n r

kZe

2n r

kZe

2 n 1 r

h

kZe mrn

n n 1

n n 1

h

kZe mr

n 2n 1

n 12

rh

4 kZe m kZe m

a

Z

i f

2 2

n3 2

2

n

2

n 1

2 2

313 2

2

21

2

21

2 21

32 2

2 2

2 21

2

1

2

2 2

2

20

−

=−

−−

= − +−

=− −

−

=−

−≈

= = =

−

/

/

( )

( )

( )

( )

/

/

ππππ

ππππ

ππππ

ππππ

ππππh

Summary of the Stationary States:

rn a

Zn

20= a

mke0 0529nm0

2

2= =h

.

EkZe2r

kZ e

2n a

Z

nEn

2

n

2 2

20

2

2 0= − = − = − Emk e

213 6eV0

2 4

2= =h

.

vkZemr

kZ e

mn a

Zkenn

2

n

2 2

20

2

= = =h

ββββ αααα= = =

= ≈

v

cZn

kec

Zn

Zn

1137

n2

h

L mv r mZke

n

n a

Zm

Zken

n

Zmkenn n n

2 20

2 2 2

2= = = =h h

hh

Spectra of hydrogen-like atoms The result is that radii are quantized such that .r n rn

21=

hf=E E

hf=Z E

n

Z E

n

f=Z E

h1

n1

n

1=

Z E

hc1

n1

n=Z R

1n

1n

i f

20

i2

20

f2

20

f2

i2

20

f2

i2

2

f2

i2

−

−−

−

−

−

−

λλλλ R

E

hc1 097 10 m0 7 -1= = ×.

Reduced massReduced massReduced massReduced massReduced mass is a concept that allows one to solve the two-body problem ofReduced mass is a concept that allows one to solve the two-body problem ofReduced mass is a concept that allows one to solve the two-body problem ofReduced mass is a concept that allows one to solve the two-body problem ofmechanics as if it were a one body problem. Given two bodies, one with massmechanics as if it were a one body problem. Given two bodies, one with massmechanics as if it were a one body problem. Given two bodies, one with massmechanics as if it were a one body problem. Given two bodies, one with mass mmmm1111and the other with mass and the other with mass and the other with mass and the other with mass mmmm2222, they will each orbit the center of mass of the system of, they will each orbit the center of mass of the system of, they will each orbit the center of mass of the system of, they will each orbit the center of mass of the system oftwo bodies. The equivalent one-body problem, with the position of one body withtwo bodies. The equivalent one-body problem, with the position of one body withtwo bodies. The equivalent one-body problem, with the position of one body withtwo bodies. The equivalent one-body problem, with the position of one body withrespect to the other as the unknown, is that of a single body of inertial mass respect to the other as the unknown, is that of a single body of inertial mass respect to the other as the unknown, is that of a single body of inertial mass respect to the other as the unknown, is that of a single body of inertial mass mmmmreducedreducedreducedreduced::::

m1

1m

1m

m m

m mmM

m Mreduced

1 2

1 2

1 2

= =+

=+

=+

µµµµ

where where where where m m m m is the electron mass and is the electron mass and is the electron mass and is the electron mass and M M M M is the nuclear mass.is the nuclear mass.is the nuclear mass.is the nuclear mass.

The reduced mass is always less than the mass of each body.The reduced mass is always less than the mass of each body.The reduced mass is always less than the mass of each body.The reduced mass is always less than the mass of each body.Using the reduced mass, the Rydberg constant becomesUsing the reduced mass, the Rydberg constant becomesUsing the reduced mass, the Rydberg constant becomesUsing the reduced mass, the Rydberg constant becomes

Rmk e4 c

k e4 c

mk e4 c

11 m M

R1

1 m M

2 4

3

2 4

3

2 4

3= → =+

=

+

∞π π πh h h

µµµµ/ /

where where where where m m m m is the electron mass and is the electron mass and is the electron mass and is the electron mass and M M M M is the nuclear mass. One consequence is thatis the nuclear mass. One consequence is thatis the nuclear mass. One consequence is thatis the nuclear mass. One consequence is thatthere are slight isotopic (same element, different number of neutrons) shifts inthere are slight isotopic (same element, different number of neutrons) shifts inthere are slight isotopic (same element, different number of neutrons) shifts inthere are slight isotopic (same element, different number of neutrons) shifts inatomic spectra. atomic spectra. atomic spectra. atomic spectra. The largest effects will be seen for the lightest nuclei. The largest effects will be seen for the lightest nuclei. The largest effects will be seen for the lightest nuclei. The largest effects will be seen for the lightest nuclei.

Reduced Mass Correction for HydrogenReduced Mass Correction for HydrogenReduced Mass Correction for HydrogenReduced Mass Correction for Hydrogen

( )R R1

1 m M1 09737 10 m

11 0 0005486u 1 00728u

1 09678 10 mH7 1 7 1=

+

= ×

+

= ×∞− −

/.

( . / . ).

( )R R1

1 m M1 09737 10 m

11 0 0005486u 2 01355u

1 09707 10 mD7 1 7 1=

+

= ×

+

= ×∞− −

/.

( . / . ).

The red line (longest wavelength) in the Balmer series will be seen atThe red line (longest wavelength) in the Balmer series will be seen atThe red line (longest wavelength) in the Balmer series will be seen atThe red line (longest wavelength) in the Balmer series will be seen at

( )

λλλλ23

2 2

1

R1

2

13

365R

=−

=

( )

λλλλ23H 7

365R

365(1.09678 10 m

656 467nm= =×

=)

.

λλλλ23D 7

365R

365(1.09707 10 m

656 294nm= =×

=)

.

∆λ∆λ∆λ∆λ23 656 467nm 656 294nm 0 173nm= − =. . .

Does the Bohr Model work when there is more than one electron?

It didn’t seem so since the spectraturned out to be very complicated.

One would expect the forcebetween electrons to have someeffect. Henry Moseley reasonedthat the innermost electrons wouldbe less affected by the outerelectrons (remember that radius ofthe orbit goes as n2).

So the x-ray spectrum was theplace to look. Moseley measuredthe x-ray emission spectrum frommany elements.

According to Bohr theory, the energylevels are proportional to Z2, and so thedifferences in energy levels and thefrequency of the emitted light should alsobe proportional to Z2.

Moseley measured the x-ray emissionspectrum from many elements then plottedthe square root of the frequency versus theatomic number. The plots were fit bystraight lines as the Bohr model predicted,expect with a non-zero intercept.

f A Z bn= −( )

For the K-series, a good fit is obtainedfor b=1. This is explained by the fact thatthere is one electron left in the n=1 statewhich provides partial shielding of thenuclear charge to higher orbits.

The L-series requires b=7.4 to get agood fit, so there are an average of 7.4electrons closer in than n=3 after one isejected.

Auger Effect and Auger Electron Spectroscopy

200 250 300 350 400 450 500 550

Au

ger

Ele

ctro

n S

ign

al

Energy (eV)

N

O

In

N

dN/dE

In

The Franck-Hertz ExperimentDemonstrated that a minimum energy was required beforeDemonstrated that a minimum energy was required beforeDemonstrated that a minimum energy was required beforeDemonstrated that a minimum energy was required beforeelectrons could cause excitations in a gas of mercury as theyelectrons could cause excitations in a gas of mercury as theyelectrons could cause excitations in a gas of mercury as theyelectrons could cause excitations in a gas of mercury as theytraveled through. Demonstrated that electrons in Atoms cantraveled through. Demonstrated that electrons in Atoms cantraveled through. Demonstrated that electrons in Atoms cantraveled through. Demonstrated that electrons in Atoms canbe excited to higher energy levels by non-optical means. be excited to higher energy levels by non-optical means. be excited to higher energy levels by non-optical means. be excited to higher energy levels by non-optical means.

This minimum energy was observed as a critical accelerating voltage at which theelectron current through the tube decreased dramatically because the electrons nowhad enough energy to excite the mercury, but in doing so they would no longer haveenough energy to reach the collecting plate.

Further increasing theaccelerating voltage wouldcause the electrons toagain gain kinetic energyafter the loss of energy tothe mercury. If they gainenough kinetic energy,they can excite a secondmercury atom. Thus,oscillations in the currentversus accelerating voltagewere observed.

The voltage spacingbetween peaks depends onthe energy spacingbetween the ground stateand first excited state of the gas atoms.