solution set chapter 5 energy levels and radiative

22
SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE PROPERTIES OF MOLECULES, LIQUIDS, AND SOLIDS "LASER FUNDAMENTALS" Second Edition By William T. Silfvast

Upload: others

Post on 30-Apr-2022

1 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

SOLUTION SET

Chapter 5

ENERGY LEVELS AND RADIATIVE PROPERTIES OF MOLECULES, LIQUIDS, AND SOLIDS

"LASER FUNDAMENTALS"

Second Edition

By William T. Silfvast

Page 2: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

CH>

1. Derive all of the allowed frequencies of radiation that might occur among the first four rotational energy levels of the hydrogen molecule H 2 . Assume that H 2 has a

permanent electric dipole moment (which it does not) so that it can radiate. Draw an energy-level diagram of those first four levels, and indicate the possible transi­tions and their frequencies.

Ac.co V'dt't fb ~

&l/1 AT::. Dl +1. se kc17 rr"' r"-- l.e..a

1 fra vr. s i ti' o "'"-::J o ,,.,,_ ~ o cc.u r

13 t..t.., T ,-i... ~ !> P Y' o b / e """'- ~ vt L~ /vi v o / ue s

6. T= - I rr~ tot ) i' n °'4 f) ~ ~ 0 (..(.. V. r JD { 6 W e Y"

h 2 b ::: s e.pa lf"'{l.,.7t-ot.A. 1 4-Tc""'"

M i" ~ {. t> "1-t ~/ ~ ,/ Mat$ 5

I L/ O -., 1;1.J_ I "'I I f='ro""'- ex...~V\Ap~s. t>yt__ Pct_7'2S.

/J h ( ::" (.>, 16 b '>(Io - 2 '- JotA-~)

I _. I

-21 .,-, , ~.V~Xto ~

,2... H ~ V;u = ~. S'D x ID i:-

1/ -2./ ,-

\ I tfQ X 10 .., I I / l.

___ ....;i\ll_~ __ :!io ::; I, 7 S"' 'X IC /1+

Page 3: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

J:: 3

{_H- i;­

'3 ---.---- E ~:::.. 1 L 13 ~c.-=- 2, q(p x 10-'ev = ~. 7Yx10-22

:5'31'. ~

-

-

~ -"' -~ ~

~ ~

'

,,

r-----__...-~- ~---------·----...--.........,.-~ .. - ............. - ....... ( v,:•t = 3.5'<c x11/' H ~ ~?4 :::. 8 't I, '-/ .:;J

- -- --<> - r<i "'"l - ........ ........ '-

~ Q.. Q. Q..

'!I

,j ·

t'i 1

J.,

- - - o. t.!J'-1'4P ev Y'O;f

2..

R(2.) 4,?l~U/ 3

R ( I ) L./, ~ ~ °!; X. I IJ I J

R ( o) 4. ~Si. XIO~J

t:> ( I ) L(, r.; 2 1'iS X I 0 I J

el 2 > '-/,~ I~!- 101S

1:1 ( l) L./ ( ~I)'( x tr/ j,

-. ~ro-=r

32-

~ It - ..._,...,_ o, OD t'i'il e. V ro-r~

'·!I - ....... (),DD0,.,1~3eV #"~ .· ,j;:~. ,,

· .. , . - ..- ~ ~· o. o e V

~,I ~'I

r.o. u~ ~ (p, /~]

~.211..1

f.c. 2 3i0

~t: 2 4S-

Page 4: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

t:,H~

2. Consider the first two vibrational energy levels of .the CO molecule, which we will assume have an energy separation of 0.2 eV. Consider also the first four rotational · . levels of each of those vibrational levels. Derive all of the frequencies of radiation that can occur among all of these levels, including pure rotational transitions and rotational-vibrational transitions. Include an energy-level diagram for those levels, and indicate the allowed transitions and wavelengths. Assume that the separation 1

between C and 0 is 0.111 nm.

F1r~r (..;t;·Vt.s1elr;.,..., pt,,{~ y-oret,t:t>,~ fht"-s;l7fj1tt.s; Er?;.-=:r(J+1)tsh~

.6 Ert>i ( :t ~ r -1) -::: [ J' ( J' + 1) - (_J -1) J J 13 k c.. -::. 2. ::r 6 h c.

13-= h · · 1: ~ /A. l'- h)"­~,,.,., t -

2 b:: (), 111 h~

C. ~ 0 r)«. l

r: = µ (_2 bl ::. e, ~ ~ ~:) '· (, 7 x ,o-21;,,? (a. Ill x ,o-"'; ""'

13 h l -=

:. /.L/IXI0-'1~;k~~"L (,.(,2.)(lo- 3'1.r ... ~)l :: ~'/L/Xl~-i. $ T ~ (f "- /, 'i IX IO~ .. I~ ,;., ~ 'L :: '2. 'f~ 'XI o .. 1

-/ e V

fl.w. s L} Er o:r ( J' ~ J'- i) "'" 2. J B ~ ( ::. 7. ~ " )U b - H :!"' 1'. J'o "-~) 13oR of 14 f1'v-sT lwo v1br~TlDha-P level~ Aa.t.H' ,·~lift~~;..

se-rs tJ f V'o tct.l/o"'t.a...P levels s1'nu ro14,lt"tJ"'t.J /etJel~ tJ~lt/ dee.e. ... t.d upDvt.. T t..01v.-ve. :l.=-o I~ 7l.e lowesT e-"'l.€.~V

Jev~t of e~c.--4- vibyia..flo~J level. H~i.tu ~ -f1~~sT fr;ur v()la...t1·evi.ct..l level~ ft>r bath. V=J a_,,.,,d_ v::2

IA),· 11 b~ ~p~ u.eR a LL~ rp/ ,:.._i TD ~ t:l btJ ue

ft; y HA. tA. l &<_ tl L t- ~ l"-p(_ 1:, 1b :r

Page 5: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

3. Label the molecular electronic energy levels listed in the following table.

s L

1 2 -~A 1 1 - · ~1"1 0 1 - 'fr 0 0 - 'z 1 2 - 36 1/2 1 - '11 ~ (e~t;·o"" Y'tA Lw 3/2 2 - ~D.

1/2 3 - '~ 6A=D)±\ 1 0 - '>~

b. s D -- '-'!ff -3/2 1

3/2 0 - ..,~

1/2 2 - 2A

Identify which levels can have allowed transitions between them, indicating those transitions in the following form: 2 ~ -+ 2 TI.

'3j ~ J4 ;~~ -srr >11-/ ~~

Jrr--=t 1rr 1 rr-/ ~~ I 2_ I

I ~ n-In- -7 n- 'tr-? I~ , 2_ ~ I~ L/T-/ ~

'° 1/--:Y '-1( 'I Lj -/ '111

1 t-?> 1rr '16 -/ 't ~ ? ~-? "2 D.

'1 rr -'?' v A. 1. ~ -/ 2 ~ '1 rr _, 'I~ lt ~ -? Lfrr 3 ~ -7 12- 2 4 -v 'Lrr 2 ~ -7 2l '1 n- -? '-Irr L/ Z -;> '/~

"2.~ -7 "2~

Page 6: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

c_ /-1 !) I

4. Which dyes of Figure 5-11 would make the best lasers at the following wavelengths: . ' (a) the green helium-neon laser wavelength; (b) the yellow copper vapor laser wavelength; (c) the strong atomic sodium emission at 589 nm.

/\f Ct --f. / /.1 o re!...'- a , · ':. ~lj!L

Page 7: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

5. For the dyes given in Figure 5-11, assume that the given range of emission wave­lengths corresponds to the available laser wavelengths. Assume that the dyes radiate with a quantum yield of lOOo/a. Compute the width of the quasi-molecular energy levels of So that are involved in the laser transition.

~~ :: ~ 's- I>'\ w..

Ac vi~,·~ ~ A~ =- (_ ~ 3 o - ~t.>o) vi~ :. 10 1-\ IM.

~ E: h A")) -= h: .... A A

AA=(~!() -~-?o)~""'- ='-ft) I;\~ At)-:..5:'9tJH1t

~1, Jt{J( (f)-1r1x1oi '/O'XI0- "'1 - {),/L/le.V

(_S-fC>)(10·")~ -

~ >-. =- ('S'"~ o ... _,,~)~IN\ -::. 3 £l l.f~ A"= £>-,.15.1-tVt.4.

-1,r- i - ., f. IVX.la 3 )</I) :30 'XIU - Otl'l.S-e v

(';°'"! r X. I 0 ... Pf) L.. .._

7 l+j,di"tJX,~ c.cu"' a.':'! A~ ~ {:t'lo-1fs8)11w.. ~ 'l-tH•"'- >..,~l{'-.1••

1r e -1 A f - 'i,l"I x10 • _J,xl!_ ~.~ x II> ... - 0. I I? e V

- .. ( lf"O x ID_.,) '2... -

Page 8: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

u, c~t) c1-1- ~

I+ ~e ~·\IA.~ h7 lb I/, s iDJ-1.A--p ~v ~ /o !)JCA.., ( ewel .J.)

~ ~ ~~ ~ ::!-rr J_ ===? r:t - __ I --, T;.I r - ~TT A -jlt-t~

IYJ.: Y /+b

l"l d ~ y A-f:r

/Y ~ : 6-/t;_5$

-IV · -IV 2, I 2 x IO S - 3. ~I X I tJ $

- -12 2.c,tA x 10 >L'-

·'Y y, 2.f' 'KI0-14fs _ (,,{J>ZXllJ S

Page 9: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

CH~

6. Compare the emission bandwidths of Nd:YAG and Nd:glass laser materials. De-. I

termine the collisional interaction time (either Ti or T2 ; see Section 4.3) that would '. be required to produce such broadening. Indicate whether this broadening is due to 1

either Ti or T 2.

r L . ~ 1J. .. s. F W..e..l s.s r'~ t~W( .

- :;- r:i-e-c... £ e-s L1 ~ w; d n" Ne1 ,' YA-&

IY.4. : (:;-/e..ss

Vpper /evd // &-tl~

"2. 3 0 11-<- $

A-I Y'Dr>-.. fe._,__pe..,..A-'tU ~ ~ uppev- l~vel // f~fi~ IS

,. .

db M• "'te..tLI! b'/- V'a..diaT/v-e d.uA-tf /11 bon.. ~$;'/;p-11.Js. /)a.

lt>w~v- te.vef het.i:. V\c ir-etdi'A-t·tJc d-LU)_~ (v..or A.I/owe.el),

F rr> IM- (l/ · t./ '-f)

A z{~ : J_ [l ~ A~t; t- ~ ~~ + ~ + ~ + ~J 2 7T A ~ t T, Tl.

-? A.tJ =- D for ho Tl YAf:r + b-t~ss 7

T f ll.t. h VY> et. e;~ 11t ~ ~1 I ? k,i e,r-· ~ 1 .. i14-f bf Tlt.A a. b t; t.re

ex_ p ;..e_ ";. £. t n "\, ) fUlo\. t t {A.) fHL .. { d_ ?ta._ U"e lb b.e.. a._

re~~ 1 e.,;fuvi r,' or Tz., $1&.t..~ w~ d.c J r. ~ C.a.."1.. CA../~/lt.Q_

Vl.&-f ~'Lt.re eu..tJW...iA.. /;,if(:)Y'M-4...t/'~ I

;

I I

fo"' et /4.v r;s ~,

A -, ;. ~ /'- _j_ L +a + 3 ] ~ ~"-we ::::- 2 TT 7; T ~

Page 10: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

I 7. Assume that both Nd:YAG and Nd:glass laser materials are doped to the same Nd. '.

concentration of 1026 /m3 . Assume they are both pumped identically, so that they : have approximately 0 .1 o/o of the Nd 3+ ions in the upper laser level. Compare the

raa1mea power or a cu01c vo1ume iu mm on a s1ae rrom eacn or tnese matena1s. Assume that both materials radiate at approximately the same wavelength of 1.06 µ,m. Compute the radiated power per unit frequency by assuming that all of the power radiated is averaged over the FWHM emission linewidth of each of the transitions.

/\/J CAfl/L ~t ra- ft~ ,--..,_ t-us T

Pear~- I"'- IA.f Pe.¥' /.e.wJ..

wc,VJ"LL~~-\/ f) ( u CM-Q._

I() L7'/A.t 1

{;. I 'lo

.,,..... /,tJ~µw..

3 J (!_b ~~) :::to1~>

11./ d : YA Cr '2 1 tJ J'A- >--

N c(' G-/._,$5 ,,.__ 1 ao µ .s

Page 11: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

8. Make a sketch of the emission spectrum for Cr:BeAb04, Ti:Al20 3 , Cr:LiSAF, . and Cr:LiCaF.broadband laser materials. Compute the collision interruption time T 2 ass'ociated with phonon broadening for each of these materials by using the experimental emission linewidths. Compare these times to the upper laser level lifetimes.

Fro--- le...~ >-2- Ad ~o... .i;.a.. r-70 ( Ttt,~~ I~. n.)

/V{ a_ [U. ' -eJ A....,.~~ '· .AA (~. tM.) · ·· · ~ ~ '1-VIHM T.. (?ed

2.

2, 1ox.10 '3H2-

-1'/

Cr~ Oe All.. IJ~ 757) /20 2"Dp5 j, 2l. ')(/D

Ix 10 'l/ He -If"'

TJ : A ~2. 03 7JO I ~b 3. <l}d 3, 2.)( JD

LV': /_J SA-F 1,5" L.6 C) ft; 7µs !Jfl <J,'~Xld ~

1. ~Y.ID-1>

(v; L,i lAF 1()0 /3 0 /7fJ;us {u~ '-/ ')( /()JJ fl~ ),DXIO-lr

cv:L1 lA F

7eo I I 06 p .. oo

Page 12: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

CH~

9. Look up and compare the electrical conductivity of glass (insulator), silicon (semi-. conductor), and copper (conductor).

-11 - ' < 0,2.XIV (flcl;V.)

/() t> - /6 ~ {_fLliAA.)- 1

kpe,td·~1 tA.f~ t>/.1>p1:t1 U111. ~TrttJict1t.

-I (;, X I D S- (!l. C ~ .... )

Page 13: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

Cff 5"""°

10. From the chart in Figure 5-3l(a), determine the laser wavelength for each of the six different compounds listed in the chart: AlP, GaP, AlAs, GaAs, InP, and InA.s.

APP 2, 5 e V

&a.. p 2,2t;;"eV

L., I 5"'" e \./

/, L/ e V

r"' P /, 3 eV

I~ As (). "/ e V

Page 14: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

CHS-II. Using the chart of Figure 5-32, determine which semiconductor laser materials !

could replace the He-Ne laser operating at 632.8 nm. ·

AP>< &f>..1_~ r,,.,_~P

rnx. Go.. 1_ x As

6-o.. - As 1'=' 1-X X

Page 15: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

12. Assume that an electron is subjected to a potential energy function V of the form

. { Vo = 0 for x = 0 to x = L,

V = oo for x = - oo to x = 0 '

oo for x = L to x = +oo.

Solve the Schrodinger equation to determine the energy levels for this system, which is called an infinite square-well potential. Assuming this is a quantum-well laser, how high would the first energy level occur above the bandgap?

Bo Lt&-(_ d. tt ~ V Uli1-eA-; t# o"'" ~ V= o f -oV' o ~ x ~ L V=- ~ foy - c,r; <.. x < o

fi,v- L < X < +ot1 5 c_ 4 ...- & e d i ~;eV' e2 u..~ tt fr"\

v -i tfJ +- ~ rr '- ~e ( E - V) 4J ::::. o

2 \ i... i. ,_ )._ =- 'llJ!, __ ' ~t" q._.:: -vJ 0 V' d LIJ(x + 4-< '- t.fJ (x) ~ o l'f<- n.. ~

d ~1..

&-~~ salt.(_,1tcrt.\ =:> lf ()(\-= A-~41..x + 13~;;tx ~~ ~ X '::-0 tf (;>() : D ::::;.> l3 ..= 0

~CJ_ '+lx\-=- A ~tk.x.. ~ a:r x =t..., '-PU<1::. o -:::::) 4si"1.4L = o

I> I" k L "- Vl 1')- -= ~ ~ 11~~ . (ff -V) L

b r- [<ii n: ~ (6' - v J] L 2 ::: h '1r ...

- n ,_ h ... rr l.. + v bvtt V= o \...._ ~ -well t___., ::: - L ~ ~ .,.,. .... ~.e r t

h~

E ~ ::. ct Jtl.ie ,_,. n " = , t

Page 16: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

ll+~

13. Suppose that the XeF excimer upper laser level C (as labeled in Figure 5-9) is being populated in an electric discharge and decays in a time of 5 x 10-9 s to the· lower level A in Figure 5-9. Assume the transition takes place at the value of r (the separation distance between the Xe and F atoms) as shown in the figure, and that the molecule subsequently exists temporarily in the A state at that distance. How long would it take for the molecules to be repulsed and thereby separated while 1

in the A state to the point where they are no longer bound to each other? Assume that the molecules are at a temperature of 100°C in the discharge and that their ve­locity is the thermal velocity (it might be higher than that owing to the repulsive force). Compare that time of repulsion to the time the molecule exists in the upper laser level C.

Page 17: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

14. Using the data concerning C02 lasers, compute the wavelength for the P(2) tran·­sition in the V1 symmetric stretch rotational- vibrational mode.

P(2) I vt vo I tre-s

Fov-

'

1V -0::.----0

- . ~ G { t).!71 ~ + q,t,'6x1 o-~ ~· 11) x lo -'f) e I/ -Y, z.... ·7 i , h ::. , . 4. I l )X Io_, 5 e V .$

L71~fJ7 eV '-f, 13 'S"-x,10-''Sev s --

:::- L/, 1 ~ t; x t 6 13 I+~

J x 10 ~ kt/~ -= \ 7. 2 ~/-'.~ L(, I ~O X 10 /) / 5 _.-~ .

Page 18: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

15. The R( 1) rotational-vibrational transition in a nonhomonuclear molecule is mea;- : sured to occur at a wavelength of 12.0 µ,m. The vibrational energy spacing for this ; molecule is 0.10 eV. What is the wavelength of the longest-wavelength pure rota­tional transition in this· molecule?

p_(I)

~f ~,

..,~

I \

,v

~ , ,

I

0

E:\1\

2

D

E T

--

h c. - -~E -

L b i-t.~ s. r r-1> t~:t;· tnt iJ w et l}t_ I~ '1'1-,

I s n..:t a_ s s (; c ( ,4. ~J w /'Tl-_ A E

- 2.t> ..T I.:: vi b - tJ, I e V -:::: I, f.o D z x to ·

/jE = E, - Ev1b L..

-~ ~r:.: 1"2. 0 X/0 ~

- 3 l/ f1 ~I (p. ~ 2 r" x. 10 rs 1 'I, 1 D <;;a

l (, 0 )( 10 - "' ~

~. b 2.S-tPX' /fJ-3'fJ"S 'JX/()Jlt.t../~

~' 7t. x {0-'2..l.. J'""

Page 19: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

16. Consider a hypothetical nonhomonuclear molecule that consists of only the first two vibrational levels, each of which has only the first three rotational levels. The · longest wavelength emission on a rotational-vibrational transition is measured to

occur at 8.00 µm, and the next-longest wavelength transition occurs at 7.90 µm. At what wavelength would the next-shorter-wavelength rotational-vibrational tran-

1

sition occur, what branch would it be associated with, and what is the vibrational : energy spacing (in eV) of the molecule? Assume that the Q-branch transitions are ; not allowed in this molecule. . :

td:&... "At.'";. ll,"L~~

\_- ~'"­/\ 6-V -

Page 20: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

17. A semiconductor laser material has a bandgap of 1.0 eV and an effective mass of the conduction band that is twice that of the effective mass of the valence band. If a recombination transition occurs from an energy that is 0.03 eV above the bandgap, what would be the transition wavelength?

r I

, ' I

\.\ yl..J '2. ~ 'Y\lfi 2 r= = O.D~e.V

-- - -c..

hv = t,o eV + 0,b'leV+ 0,()t.ceV-=-J,o~eV

l,DtteV

- ~ L/,/]S-'tXltJ- 1;, '3Xlu

/, 0'

Page 21: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

C.H>

18. A semiconductor laser material has a bandgap energy of 1.55 eV. This material is / used to form a quantum-well laser having a 5-nm-thick active region. The effec- · tive masses of both the conduction band and the valence band are measured to be greater by a factor of 2 than the mass of an electron. What would be the long­est wavelength of the laser when an electrical current is passed through the active layer?

--

(rr~) L. ~ '-I ) '2.. IT 2._ ( ft;. f.o 3 x I 0- T· ~

({..Tf);_ 2 x"l. ~e L '-

"2_ ~'°-(_~.fa $X I 0 ... ~'i )

/ ~ "'-t 1.1x16 - 3 ' ( S-x 10 - ') ....

_t._fJ -~ v l,Lt>7tpXIO -::- 7.~}X/o e

_,~- ?X' ~,.. J- _ h c.. :::. l/,l 3S-tt>tlo ells~ iD "ti1 LJ ET /, S-IJJ~e V

'= 7. 11..? )( I O .. 7 ~

(), 7 '13 }AW\

Page 22: SOLUTION SET Chapter 5 ENERGY LEVELS AND RADIATIVE

C-lf ~ ;

19. A gallium arsenide quantum-well semiconductor with an active region of 10 nm · is excited by a pulsed laser to produce electrons in the conduction band. That · excitation subsequently decays within the active region back to the valence band via recombination radiation. What would be the longest wavelength transition ob­served during that decay? GaAs has a bandgap energy of 1.43 eV and an index of refraction of 3 .4. The effective mass for electrons in the conduction band is 0.067me , and the effective mass for holes in the valence band is 0.48me.

~

r t

(fr~)~ n '- _ l fr 11)" I t. -

L.. rvt '4f L '- "2.. vY\ c. L '--

-3¥)'2... 1T l I to,, fo l.' x lo _ . c: -i..11

. E ~-= r - ~ J ~ ., ~if- V\'le. -=-lf D ..- l...

f.{ -= {), ()~ 7

V\'\*v~ o.~'l e~

L ::. I 0 lN\

-C. l:: -:;:. I

~,03 XfO-t'L T :- ,,l)fJ '>fl~- 11I= {),(J~tt;ZeV o, () (o 7

-" c -I · -l9, o s x ' o - ~ 'l.- .r -:: /, 2 ii' ~Io- ' 'J = 0' () o? g }e V

o.vf(

b~ 7a.p :::: I. 'i 3 e I/

_ ( I, L/) + D, o!ib2+. o, 6~7tS) e V IJ l= 11> iA-t -

- I I '-/ " ~ e. v Al-~ V\V -

h l. -- -AE

tr Q01f' NP