Spectroscopy

It will be of interest to observe the interaction of radiation with matter for the purposes ofextracting atomic and molecular information. Such observation will give structural information. E-M radiation, being composed of electric and magnetic waves, requires a dipole or pointcharge in order to interact. Such a requirement gives rise to "selection rules" which governs whatspectral peaks will be seen. In general, we require that an atom or a molecule possessing a dipole moment be promotedfrom one quantum state to another. Let us examine the quantum mechanical allowances for sucha transition.

Selection Rules:

Derivation of the selection rules requires solution of the time-dependent Schrödingerequation. This is because the molecule is now subject to a time-varying potential fielddescribed by...

V t( ) μ E= μ Eo sin 2 π υ t( )= where is the dipole moment on the molecule and Eis the electromagnetic magnitude of the radiation offrequency, .

Since we will not be able to solve the time dependent problem explicitly, we will use anapproximate method known as the time dependent perturbation theory method. We simply solvethe system without the added potential and then treat the E-M potential as a small perturbationon the solved system.

Let us assume that there are two states in the system that are accessible and that the totalsystem solution is a linear combination of these two states. Thus

where a0 and a1 are the coefficients whose

weighting give the probability of the system beingin a given state.

Ψ t( ) a0 t( ) Ψ0 t( ) a1 t( ) Ψ1 t( )=

After proper treatment, we find that a1(t) has an interesting form given as::

a1 t( )i q Eoz H10

2

e

i t

hbarh υ ΔE( )

1

h υ ΔE

e

i t

hbarh υ ΔE( )

1

h υ ΔE

=

Interpretation: This coefficient represents the transition from the lower to the upper quantumstate using E-M radiation. If the coefficient is non-zero, then the transition is "allowed". If it iszero, then it is "forbidden". The result of the analysis for each given system results in a set of"selection rules".

1

In order for a transition to occur, in general, we see from above that

1.) q must not be zero. That is for molecules, there must be a temporary orpermanent dipole moment!

2.) The transition matrix H10 must be non-zero. We will examine this in a moment.

3.) h must be close to or equal to E.

The first point deserves some embellishment.

A molecule under the influence of an E-M must have the electric-magnetic properties in order tointeract with it. For example, one can not move a non-magnetic item with a magnet. In order forthe magnet to have an effect on the object, both must possess these properties.

This is the same with molecules. A rotating molecule that has a dipole moment with thus have anoscillating dipole and if that oscillation is at the same frequency as the radiation, the absorptioncan occur. So

In a rotational spectrum, the molecule must possess a permanent dipole moment.

Vibrationally, a dipole is also necessary. Consider the molecule NaF. A strong dipole oscillateswhen the molecule vibrates.

2

Now, consider the carbon dioxide molecule.

This molecule, having no permanent dipole, will not have an oscillating dipole. Therefore, thesymmetric stretching motion is transparent to the radiation.

Consider a bend, however.

A vibrating molecule can absorb radiation if

a.) It possesses a permanent dipole or

b.) It produces a temporary dipole

Now let us return to our calculated spectra.

3

h 6.626 1034

J s ΔE 6.626 1021

J hbarh

2 π IF 1 ps 10

12s

a1_Real t υ( )

cosh υ ΔE( ) t

hbar

1

h υ ΔE

cosh υ ΔE( ) t

hbar

1

h υ ΔE

a1_Imag t υ( )

sinh υ ΔE( ) t

hbar

h υ ΔE

sinh υ ΔE( ) t

hbar

h υ ΔE

Inten t υ( ) a1_Real t υ( )2

a1_Imag t υ( )2

υ 5.01 1013

s1

4.999 1013

s1

5 1013

s1

t 0.10 ps

40 20 0 20 400

2 1041

4 1041

6 1041

8 1041

Inten t υ( )

h υ

1021

J

Notice that there are two peaks at energies corresponding to +E and -E. This is representativeof an emission and an absorption peak. Both, apparently, are in process. One might wonder:Why do the two E-M waves, one going in and one going out, not cancel each other? The answeris forthcoming, however, let us examine the absorption peak for a moment.

4

Other aspects of the appearance of the spectral peak

It is important to note that t is the time after the time at which the EM wave is turned on. In otherwords, we have only collected data for 0.10 picoseconds. If we collected data for a longer time,let us examine the effects.

First, let's keep the radiation on for a longer time, say 1.0 ps. Calculating...

υ 0.0 1013

s1

0.0003 1013

s1

4.000 1013

s1

t 0.10 ps

0 5 10 15 200

2 1041

4 1041

6 1041

8 1041

1 1042

Inten t υ( )

h υ

1021

J

Notice, now, that the FWHM is now much narrower, in fact, 10 times narrower! (clue)

This indicates an inverse relationship to the time of measurement or, more specifically, to thelifetime of an excited species. The time of measurement cannot not exceed the lifetimeof the excited species. This important relationship shows that the relationship is inverselyproportional to plancks constant...it is a wave effect related to the Heisenburg UncertaintyPrinciple. It is a cause of line broadening.

ΔE Δthbar

2=We find that

exactly for a quantum system. Imperfections in the taking of a spectrum makes thisexpression approximate.

5

Selection Rules: One of the other features of the preceding analysis is that of the transitionmatrix, H10. This is a quantum mechanical "couplling" effect that necessitates that the origin and

destination quantum states must both have some probability density. This creates somewhat of a"quantum path" by which the molecules may transition. The applications are complex, however,some generalized results may be presented.

Application to H.O. and Rotators

Which are allowed?

For both rotating and vibrating systems, the quantum mechanics shows that the states that arecoupled radiatively are those that involve the transitions:

Vibration n -> n +/- 1 and Rotation: J -> J +/- 1

All other transitions are forbidden.

Peak Intensities:

Several factors contribute to the heights of the peaks as observed in the lab. Among them, theprinciple factors are:

1.) The population of the quantum states.2.) The energy density of the radiation3.) The efficiency of the transition, stimulated and spontaneous.4.) The concentration and path length of the sample

The first three factors can be tied together nicely.

The net observed transition rate at equilibrium should exactly balanced between excitation andde-excitation rates. We can take these expressions and combine them to get an overall netrate of molecule transition between states as...

Absorption Term Emission Terms

Spontaneous

Wnet NjBj ρ T ν( ) N

j 1 Bj ρ T ν( )8 π h ν

3

c3

=

Stimulated

6

In the lab...practical considerations

The previous expression is applicable for exacting work. Moreover, it applies to a measuredintensity of E-M radiation through the sample and collected.

In real lab situations, it is more often the case that the radiation due to de-excitation ismeasured. This is because it is radiated out at all angles. As a result, only the loss in intensity ofincident radiation is measured, leaving us with only the leading term, or

Wnet NjBj ρ T ν( )=

As such, we need only concern ourselves with the excitation in a conventional instrument.

With this, we may predict the relative intensity of two adjacent bands as Nj 1

Nj

The population ratio, Nj 1

Nj

is already known and is given by the Boltzmann distribution.

Nj 1

Nj

e

Ej 1 Ej

kb T= e

ΔE

kb T= e

h ν

kb T=

So, we can now write that the relative heights of adjacent peaks can be calculated using thisexpression.

Sample Concentration

Finally, for experimental factors, we know that the reduction in the incident radiation intensity isproportional to the concentration and path length, L. The result is

logI

Io

ε L C=

Also, the ratio of I/Io is called the transmittance. Combining..we now have

log T( ) ε L C=

7

This expression predicts a logarithmic relationship to the concentration which isinconvenient. TO make it linear, we define the Absorbance as A = -log(T) and write

A log T( )= ε L C= which is, of course, the Beer Lambert Law.

For %T, A 2 log %T( )=

8

Peak Locations

Now comes the question as to where to look for a peak in a spectrum. It is important torecognize that the wavelength of light absorbed is, as seen above, equal to the energydifference between quantum states.

We know the quantum states and the energies. Therefore, all we need to do is to determinethe energy difference between these stats. BUT....we also have selection rules to guide us.These will turn out to help us here in limiting the absorbance peaks that we will observe.

Example: Consider that we have solved a quantum system and have found that the energies

are given as E = Ks2, where K is a constant that has a value of 1000 cm-1 and s is aquantum number that has integer values of 0, 1, 2, 3...etc.

We also have found that the selection rules determine that only transitions from s to s + s + 1are allowed.

Where might we look for an absorption peak?

To solve this, we need to know the energy difference between quantum states subject to theselection rules. Given this, the allowed energy differences are

ΔE E s 1( ) E s( )= K s 1( )2

K s2

= K s 1( )2

s2

= K s2

2 s 1 s2

=

or ΔE 2 s 1( ) K=

For K = 1000 cm-1. s ΔE

0 K 1000 cm1

1 3K 3000 cm1

2 5K 5000 cm1

3 7K 7000 cm1

4 9K 9000 cm1

So we need to look in the 1000 cm-1 to 10000 cm-1 range which is the infrared region.

9

:Application to Known Systems.

Rotation

Calculate the peaks in a rotational spectrum of HCl.

For rotation, we found that... EJ J J 1( )hbar

2

2 I=

The location of the rotational peaks are the differences between the rotational enregy states.

BUT!! The selection rules for rotation are J -> J + 1. So the only peaks we will see arethose between adjacent energy levels. We calculate the spacing as..

ΔE E j 1( ) E j( )= j 1( ) j 2( )hbar

2

2 I j j 1( )

hbar2

2 I= j 1( )

hbar2

I=

So these are our peak locations. Now let's gather molecular information...

Na 6.022 1023

mol1

Ang 1010

m c 2.998 1010

cm

s

ForHCl..

mH 1.007825gm

mol m35Cl 34.97790

gm

mol Re 1.2746 Ang

Calculating the moment of Inertia...

muHClmH m35Cl

mH m35Cl

1 kg

Na 1000 gm IHCl muHCl Re

2 IHCl 2.643 10

47 m

2kg

Now we are ready to go. The energy difference and, thus peak locations are now

j 0 1 8 ΔEj

j 1( )hbar

2

IHCl

10

The Rotational Constant

We commonly use the valuehbar

2

2Iin our calculations.

Further, we commonly use wavenumbers in our work. To make our work easier, we definethe rotational constant as:

Bhbar

2

2 I h c=

h2

8 π2

I h c= or simplifying. B

h

8 π2

IHCl c

ΔEj

j 1( ) 2 BNow in wavenumbers we have

calculating ....

0 2 1026 4 10

26 6 1026 8 10

26 1 1027

0.999

0.9995

1

1.0005

1.001

1

ΔEj

h c cm1

ΔE

21.184

42.369

63.553

84.737

105.922

127.106

148.291

169.475

190.659

cm1

Here we see where the peaks are located. From an EM spectrum, we see that rotationalinformation lies in the microwave region.

These peaks give us location. Now recall that the intensity of the peaks are given by thepopulation of the quantum states. The more molecules that are in a given state, the larger theabsorption peak.

So, I return to our statistical mechanics discussion and compute the heights of the peaks fromthe Boltzmann populations.

11

Now, we use the population of the states in order to calculate the peak heights. From our studyof statistical mechanics, we found that

NJ

2 J 1( ) expJ J 1( ) h

2

8 π2

IHCl kb T

Zrot= If we set Z = 1, we get a good relative picture

of the peak heights. Executing..

T 300 K kb 1.381 1023

J

K IP

j2 j 1( ) exp

j j 1( ) h2

8 π2

IHCl kb T

0 2 1026 4 10

26 6 1026 8 10

260

1

2

3

4

IPj

ΔEj

h c cm1

12

Peak Spacing:

Notice that the peaks are evenly spaced. The spacing between the peaks can be shown to be..

Δ ΔE( ) J 2( )hbar

2

I J 1( )

hbar2

I=

hbar2

I=

And, since I μ Re2

= we find that Rehbar

2

μ Δ ΔE( )=

Example: The first two peaks in a microwave spectrum of N=O are given below. Fromthis, calculate the equilibrium bond distance for N=O.

Peak2 6.81 cm1

Peak1 3.41 cm1

space Peak2 Peak1

μ

14.011gm

mol 15.998

gm

mol

14.011gm

mole 15.998

gm

mol

kg

1000 gm Na

μ 1.24 1026

kg

Rehbar

2

μ space c h Re 1.152 Ang Relit 1.151 Ang

13

Peak Location - Non-Rigid Rotor

The foregoing analysis was based on a rigid-rotor system, that is, one in which it is assumed thatthe bond length never changes. We know this is not strictly true as the molecule can stretch in its'bond. As such we introduced a correction term to account for this.

As the rotational energy levels become higher, the rigid approximation becomes worse. Thiscan be thought to be due to centrifugal distortion. Inclusion of a correction term yields...

EJ J J 1( ) B J2

J 1( )2

D= where D is known as the centrifugal distortion constantand is usually fairly small for light molecules.

Now, as before, we would like to calculate the energy of transition from one J state to thenext. For a simple Rigid rotor we found that..

ΔE j 1( ) 2 B=

When examining the energy differences between the corrected rigid rotor, we get...

ΔE j 1( ) j 2( ) B j 1( )2

j 2( )2

D j j 1( ) B j2

j 1( )2

D=

ΔE 2 B j 2 B 4 D j3

12 D j2

12 D j 4 D= 2 B j 1( ) 4 D j3

3 j2

3 j 1 =

or

ΔE 2 B j 1( ) 4 D j 1( )3

=

The expression above can be made linear if we observe that

ΔE

j 1( )2 B 4 D j 1( )

2=

Thus a plot of ΔE

j 1( )vs. j 1( )

2 will yield a straight line of slope 4Dand intercept 2B

14

Example: B and D for hydrogen gas are given as: 60.80 cm-1 and 0.0463 cm-1 ,respectively. Demonstrate that a plot as described is indeed linear.

B 60.80 cm1

D 0.0463 cm1

ΔE j( ) j 1( ) 2 B 4 D j 1( )3

0 20 40 60 80 1001.05 10

4

1.1 104

1.15 104

1.2 104

1.25 104

ΔE j( )

j 1( )

j 1( )2

15

Harmonic and Anharmonic Oscillator:

Recall that for the H.O., the energy levels are given by...

En n1

2

h υo= where n = 0, 1, 2, 3,... and υo

1

2 π

k

μ=

Now consider an excitation from n -> n+1. Then the energy difference..

ΔEn n1

2 1

h υo n

1

2

h υo= h υo=

A spectral peak occurs when radiation corresponding to an energy equal to the energy oftransition is absorbed, resulting in a transition between states. In this instance, regardless ofthe originating state, the energy difference is exactly the same. Thus, only 1 peak should beobserved!

What about n -> n+2, n -> n+3 etc. transitions? These will be found in a subsequent section tobe quantum mechanically "forbidden". The only "allowed" transitions are n - n+1. More later.

Recall that a vibrating molecule is only approximated by the H.O. model. As the molecule isexcited to higher states, the approximation deviates more significantly. As we discussed, animproved potential model is the "Morse" potential which results in an energy expression given by..

En n1

2

h vo

h2vo2

4 Den

1

2

2

=

most frequently, this expression is given in terms of wave number. Dividing by hc gives...

En n1

2

υo n1

2

2

xυo= where xυo

h υo

4 c De=

anharmonicity constant.

Important! The constants υo and xυo are now in wavenumbers!

16

Now,

ΔE cm1 n

1

2 1

υo xυo n1

2 1

2

n1

2

υo xυo n1

2

2

=

ΔE cm1 υo 2 xυo n 1( )=

Now, E is a function of n and will result in different peak locations. These energy differenceswill converge to 0 as the molecule dissociates. Also, the selection rules begin to break downfor anharmonic molecules, so "overtones" can usually be seen. We can use these overtonesin order to find the vibrational wavenumber and anharmonicity

Example: The vibrational wavenumber and anharmonicity constant for HCl are givenbelow. Calculate the location of the n = 0 -> n = 1 vibrational peak and the overtone peakof n = 0 -> n = 2.

υo 2989.7 cm1

xυo 52.05 cm1

ΔE1 n( ) υo 2 xυo n 1( ) n = 0 -> n = 1

ΔE1 0( ) 2885.6 cm1

ΔE2 n( ) 2 υo 2 xυo 2n 3( ) n = 0 -> n = 2

ΔE2 0( ) 5667.1 cm1

So we see that the overtone band is not quite double of the principle band. It will also have amuch smaller intensity.

Isotopic effects. As the vibrational frequency is given as: υo1

2 π

k

μ=

We can see that the location of the peak will shift with isotopic substitution

υo_2 υo_1

μ2

μ1

=υo_2

υo_2

1

2 π

k

μ2

1

2 π

k

μ1

=μ2

μ1

=

17

Polyatomic Molecules:

If a molecule possesses more than 2 atoms, then there are more ways in which it may vibrate.We've seen this many times so far in this course and we've even determined the number ofvibrations that a molecule may possess. These are:

3N - 5 for linear molecules and

3N - 6 for non-linear molecules.

Each of these motions are true vibrations of the molecule around the center of mass and aredetermined by mathematical methods involving matrices and symmetry. Each of the motionsdetermined from this analysis are called "normal modes" and will have a vibrational frequencyassociated with it, although some may be "degenerate" and have the same frequency. Somemotions are given below for a linear and a nonlinear molecule

CO2 = 3 (3) - 5 = 4 Normal Modes

18

H2O = 3 (3) - 6 = 3 Normal Modes

Methanaldehyde = 3 (4) - 6 = 6 Normal Modes.

19

Poly-Atomic example

Here is a vibrational-rotational spectrum of CO2.. Notice that the main center is at the

asymmetric stretch. The accompanying peaks are rotational in origin.

The rotation of the linear form of CO2 follows the same quantum mechanics as diatomics. In

this case, however, we need the moment of inertia for a polyatomic molecule. This is moredifficult to calculate, however, can be calculated from:

Ang 1010

m

Ia MA RAB2

MC RBC2

MA RAB MC RBC 2

MA MB MC= for a linear molecule A-B-C

h 6.626 1034

J s hbarh

2 π Na 6.022 10

23 mol

1 c 2.998 10

10

cm

s

RAB 1.20 Ang RBC 1.20 Ang MA

16gm

mol

Na

MB

12gm

mol

Na

MC

16gm

mol

Na

Ia MA RAB2

MC RBC2

MA RAB MC RBC 2

MA MB MC Ia 7.652 10

46 m

2kg

The rotational constant, B in wave numbers is then Bhbar

2

2 Ia

1

h c 0.366 cm

1

20

The first peak location should therefore be at 2B or: Peak1 2 B 0.732 cm1

The second peak should be located at 4B, or Peak2 4 B 1.463 cm1

The peak spacing, then, should be 2B Space Peak2 Peak1 0.732 cm1

What if the molecule had the structure: O=O=C? Could we distinguish this in a spectrum?

1.) The molecule would possess a symmetrical stretch!

2.) The peak spacing would change. Note

RAB 1.21 Ang RBC 1.20 Ang MA

16gm

mol

Na

MB

16gm

mol

Na

MC

12gm

mol

Na

Ia MA RAB2

MC RBC2

MA RAB MC RBC 2

MA MB MC Ia 6.667 10

46 m

2kg

Now the rotational constant, B in wavenumbers is Bhbar

2

2 Ia

1

h c 0.420 cm

1

The first peak location should therefore be at 2B or: Peak1 2 B 0.84 cm1

The second peak should be located at 4B, or Peak2 4 B 1.68 cm1

The peak spacing, then, should be 2B Space Peak2 Peak1 0.840 cm1

Peak spacing is larger!

21

Co-Transitions among Spectra

Rotational-Vibrational Spectra

We know from vibrational spectra, that rotational lines accompany each vibrational excitation.Thus we know that molecules vibrate and rotate. We can calculate the positions of the lines bycombining the two expressions for each motion as..

Evib_rot B J J 1( ) D J2

J 1( )2

n1

2

υo xe υo n1

2

2

αe n1

2

J J 1( )=

The last term is found to account for additional peak spacing differences and is called thevibrational-rotational coupling constant.

Now we have additional transitions that are possible. These are best illustrated with an example.

HCl Vibrational-Rotational Spectra:

HCl Spectral Parameters mH 1.007825 m35Cl 34.97790 m37Cl 36.97754

Fundamental Vibrational frequency

Anharmonicity Constant

Rotational constant

Centrifugal Distortion constant

Vib-rot coupling constant

Equilibrium Bond distance

vlit 2989.7

xvlit 52.05

Blit 10.591

Dlit .0005

alit 0.3019

Relit 1.2746 1010

Na 6.022137 1023

mu35mH m35Cl

mH m35Cl

1

Na 1000 T 310 kb 1.38066 10

23

c 2.99792458 1010

h 6.626075 1034

I mu35 Relit2

22

Here is the Energy expression combined for rotator and vibrator.

E n j( ) vlit n1

2

xvlit n1

2

2

j j 1( ) Blit j2

j 1( )2

Dlit alit n1

2

j j 1( )

Now that we have the energy values, it is of interest to examine the most likely vibrationaltransition, n = 0 -> n = 1. Others are less likely since the molecule is most likely going to be inthe ground state. (Recall the statistics.) However, since the molecule is also in different rotationalj states, we need to consider these as well. Let us begin with the transition...

E 1 0( ) E 0 0( ) 2885.6

The transition from one rotational j state to another has been found previously to be allowed onlyif j = j +/- 1. In this transition, the n increases in value, however, j does not. Thus it does notsatisfy the transition requirements and as such, the transition is forbidden and will not occur.We are left, then to consider j = j+1 and j = j-1 transitions.

j -> j+1 transitions. These are generally referred to a P-Branch Transitions.

j 0 1 10 n 0

R n j( ) E n 1 j 1( ) E n j( ) IR n j( ) 2 j 1( ) expj j 1( ) h

2

8 π2

I kb T

These will be the spectral lines. This is the proability function for molecules indifferent rotational states. More molecules in a givenj state will result in more molecules being excitedand will result in a more intense spectral absorptionline.

Not only can j states go up, they can go down in number as well. Thus...

j -> j-1 transitions. These are generally referred to a R-Branch Transitions.

j 1 2 10 n 0

P n j( ) E n 1 j 1( ) E n j( ) IP n j( ) 2 j 1( ) expj j 1( ) h

2

8 π2

I kb T

Now that we have the locations and intensities of the spectral lines, let us place them...

23

2600 2700 2800 2900 3000 31000

1

2

3

4

IP n j( )

IR n j( )

P n j( ) R n j( )

Notice some of the features of this calculated spectrum. a.) The R Branch lines are moving closer together. b.) The intensities look very Boltzmann. c.) There is much rotational information even though the transition is in the infrared (vibrational)region of the light spectrum.

Accompanying transitions can be found in many other spectra as well. For example, RAMANspectra uses lasers to carry out en electronic excitation. Accompanying the electronic excitationare vibrational transitions. Thus vibrational spectra (well similar. transition probabilities are differentbecause the mechanism of transition is different) can be obtained from excitations in a laserfrequency. This allows study of vibrational information in conditions that would occlude a IRspectrum such as water solutions.

24

Vibronic Spectra

Vibronic Spectra are similar to the ro-vibrational spectra, except that vibrational transitionsaccompany electronic excitations. Consider the following I2 gas-phase spectrum.

25

26

Electronic Transitions and the Franck-Condon Principle

You will notice that, in an electroni spectrum, there are regions of high and low absorption. In thegas phase, as shown above, these transitions expose vibrational information. The transitions aremostly from single vibrational states. Tn these instances, the intensities are not determined by aBoltzmann distributions. Instead, another principle is involved.

When a photon is absorbed, the molecule usually is not merely transferred into an excitedelectronic state, but also acquires some vibrational energy. According to the so-calledFranck-Condon principle, the absorption of a photon is a practically instantaneous process,since it involves only the rearrangement of practically inertia-free electrons. James Frankrecognized the obvious: the nuclei are enormously heavy as compared to the electrons. Thus,during light absorption, that occurs in femtoseconds, electrons can move, not the nuclei. Themuch heavier atomic nuclei have no time to readjust themselves during the absorption act, buthave to do it after it is over, and this readjustment brings them into vibrations. This is bestillustrated by potential energy diagrams, such as that shown below

27

Immediately following absorption of a photon, several processes will occur with varyingprobabilities, but the most likely will be relaxation to the lowest vibrational energy level of the firstexcited state. This process is known as internal conversion or vibrational relaxation (loss of energyin the absence of light emission) and generally occurs in a picosecond or less. Because asignificant number of vibration cycles transpire during the lifetime of excited states, moleculesvirtually always undergo complete vibrational relaxation during their excited lifetimes.

This process can allow other processes to occur leading to relaxation to other excited states onthe way to the ground state. The figure below shows the electronic processes that lead to thefluorescence of carotene.

28

Overall, many such processes can occur as the molecule undergoes de-excitation. The ratesvary and processes that can occur depend very much on the surroundings. (i.e. gas phase,high/low pressure, solvent system, etc. Such a summary is given below.

Finally, if the transitions are executed via a voltage source, then the radiation emitted can becollected for the development of a laser system.

29