LECTROSCOPY 1199

tchng or bending vibrations characteristic of particular functional group in any molecule. Such fre-
ies, are known as group frequencies. For example, the absorptions at 3,050 cm and
cmin the IR spectrum of acetophenone (CH,COCH) are typical of the stretching modes
H(aromatic) andC=0 groups respectively. Thus, the IR spectra permits recogniation of the
offunctional groups present in organic molecules.
stantNote
for a: The roupbond
particular frequencies remain
is constant, and constant from molecule
(b) the reduced mass (u)toismolecule, becausein :big
nearly constant (a) organic
the valuemolecules.
of force
Table 4. Characteristic infrared absorption frequencies for different groups.
Sond Type of compound Frequency (cm ) Bond | Type of compound Frequency (cm l)
-H Aliphatic 2,800-3,000 C=0 Esters (aromatic) -1,230 -1,650
-H Aromatic (Ar - H) 3,000 3,100 C=0 Acids -1,650
-H Cycloalkanic 2,920 -3,100 C=0 Amides 1,790 - 1800
-H Olefenic 3,000 3,100 O-H Acid chloride -3,045
-H Acetylenic -3,300 O-H Primary alcohols -3,630
-C Open chain aliphatic 750 - 1,100 O-H Secondary alcohols -3,620
Aromatic 1,460- 1,600 O- H Tertiary alcohols -3,610
Olefenic 1,620 1,670 N-H Phenols 3,500
Acetylenic 2,100 -2,250 N-H Primary amines -3,450
=0 Aldehyde (aliphatic) 1,720 - 1,740 C-N Secondary amines 1,180 - 1,360
=0 Aldehydes (aromatic) 1,495 1,715 C=N Amines 1,650
=0 Ketones (aliphatic) 1,705 1,705 C=N 2,210 -2,260
Ketones (aromatic) 1,720 1740 - NO, Nitro compounds 1,565 1,585
=0 Aldehydes (aliphatic) 1,720- 1,740 C-Cl 600- 700
=0 Aldehvdes (aromatic) 1,695 - 1,715 C- Br 500 600
=0 Esters (aliphatic) 1,725 1,745 C-F 1,000 1,350
Applications for IR spectroscopy : (1)In establishing the identity of the compounds :The
spectrum of the compound is compared with that of known compounds and from the resem

C-H
C-M

2000 600-1S00
3300

N-M C=c
CEC

2100
H
1600
2200 1700
3300
3500

2500-3600 1600 1000

1000 1900

1000 1000 1400 1200 1000

4000 3500 3000 2500 2000
Wavenumberlom
most common groups appear on an intrared spectrum
rig. 10. The diagram shows where some of the
1200

blance of the two spectra, the nature of the compound can be

established. ENGINEERING CHEMISTRY
particular group of atoms give rise to characteristic absorption bands in the This is
1R
sketopneectrusmbchoeow,mcapuosuieen,d
particular group strongly absorbs light of certain wavelength IR spectrum, no matter
in
belongs. For example, IR spectra of both benzaldehyde and phenyl to
nonounced absorption peak at 1,700cm. This is indicative ofthe presence methyl ohich
the compounds.
(2) In detectingimpurities in asample : IR spectra of impure sample will
absorption bands. By comparison with IR spectra of pure compound, presence of
ofc:=0grshowoup int
ro

sorne rxr
detected.
NO energy
absorbed
impurity canbe

CH Rock
jmpuity
CH
C=N stretch wag csC
strech stretch CH2 rock stretch twist
Allenergy 3 5 6 7 9 10 11 12 13 1 15
absorbed

Wavelength (micrometers)

Fig.11. Infrared spectrum of acrylonitrile, CH2 = CHCN, absorption bands occurs at those
wavelengths, where the frequencies of infrared radiation correspond to
the natural vibrational frequencies of the acrylonitrile molecule.

(3) To ascertain hydrogen bonding in a molecule : Generally, it is not possible to distingu

between intra and inter molecular hydrogen bonding. This can be ascertained by taking aseries
IR spectra bf the compound at different dilutions. As the dilution is increased, the absorption ba
due to internolecular hydrogen bonding dininishes ; while that due to intramolecular hydrogen bon
remains unchanged.
(4) Provides valuable infor1nations of nolecular synnetry, dipole monments, bond lengtis
(5) Distinguishing positional isomers of acompound.
mixture ofcom
(6) IR spectra may also be employed for rapid quantitative analysis of a
pounds, in pollution detection, in milk analysis, etc.
wh
benzene is shown1in Fig.12,
Examples of IR spectra:() Benzene (CçH): Spectrum of occurin
additio
is fairly complex, because absorption due to the the benzene molecule as a whole
of certain
vibration
to thöse characteristics of particular bonds. Various peaks in the spectrum
the molecule and the major one have been marked. twoal
absorbance(the
wavelength(incm
Note : IR spectra are'normally plots of percentage transmittance andofnot
the
related inversely). In addition, the wavernumber (V), which is the reciprocal E).
employed and this quantity is related directly to the energy of the transition (i.e., Voc
ECTROSCOPY 1201

100 -
90
80

C-H bending
(in plane)
C-H stretching C-H bending
10 C-C stretching (out of plane)
3000 2500 2000 1800 1600 1400 1200 1000 800 600
Wavenumber / cm-1
Fig. 12. IRspectrum of benzene (in liquid film)

(ii) Propanone (CH,COCH,) spectrumis shown in Fig, 13, with characteristic peaks marked.
100

C-H Fig. 13. IR spectrum of propanóne.

stretch C=0
stretch

3000 2500 2000 1800 1600 1400 1200 1000 800

Wavenumber / cm!
(ii) Etlhanol (CH;CH,OH) spectrum is shown in Fig. 14 with thrce characteristic peaks.
100r
/ 90
%
Iransmittance
80
70H0-H
stretch
50
40 C+H stretch Fig. 14. IR spectrum of ethanol.
30

10

3500 3000 2500 2000 1800 1600 1400 1200 1000 800
Wavenumber / cm )

cm, 2950 cm l
(77) Pent-1-ene (CH,(CH,),CH = CH,). The bands appearing at 3080 and

H cnt are due to =1C- H. -C-H and -C=C-stretching, respectively. (see Fig. 15).
1202

100
ENGINEERING CHEM STRY

3000 2500 2000 1800 1600 M00 1200 1000

600
vlcm-!
Fig. 15. IR spectrumo» pent-1-ene.

(i) Ethyl ethanoate ((CH;CO,CH,CH) spectrunm is shown in Fig. 16 with

three
peaks marked.
characteristi
C-H stretchirg

20
C=0 stretching U
3000 2500 2000 1800 1600 1400 1200 1000 800
Wavenumber/cm-!
Fig. 16. Spectrum of ethyl ethanoate (in liquid film).

(vi) 1-butanol,(CH,CH,CH,CH,OH). Broad band is due

Broadening of the band suggests internolecular lydrogen bonding tobetween
the stretching of the O-Hgrou
group of one molecule with the oxygen of the hydrogen of the hydroxd
hydroxyl group of another molecule.

Transmittance%

200
100

Fig. 17. IR spectrum of l-butanol.

De
CTROSCOPY 1203

ii) Dicthylamine ((C,H;),NHÊ.Like alcohols, amines have a broad band, although at lower
umber region (3350 cm ),due to N -H stretching. The broadening suggests that the hydro
atomtakes part in intermolecular hydrogen bonding.
100

-N-H

3000 2500 2000 1800 1600 1400 1200 1000 800 600
V/cm-1

Fig. 18. IRspectrum of diethylamine.

(vi) Ethyl acetate (CH,COOC,H,). The IR spectrumn of ethyl acetate (see

Fig. 19) has intense
daround 1700 cm, which is
characteristic of C=0 stretching.
100

3000 2500 2000 1800 1600 1400 1200 1000 00

V/cm

Fig. 19. IR spectrum of ethyl-acetate.

Note : The carbonyl compounds (acids, esters, ketones and aldehydes)\show the prominent band o.
cm).
FRANCK-CONDON PRINCIPLE
Franck-Condon principle is important for the interpretation of electronic spectra: According
1"since electrons move much nnore rapidly than nuclei, so it is a good approxination to assume that in
clectronic transition, the uuclei do not change their positions.' Therefore, an electronic transition may
represented by a vertical line. Fig. 20(a) shows the potential energy curves for the ground state
hd first excited stat of a diatomic molecule XY. The molecule lhas adifferent electronic structure in tlhe
Ctted state, and the potential energy of the excited state is higher. The equilibrium internuclear Separation
1208

Table 5. Some common chromophores.

ENGINEERING CHEMISTRY
Chromoplhore Example Typeof excitation
max (nm)
CH,OH
183

CGHSH
224

CH,CI
173

-Br: CH,Br 1
204

-1: CHI 258

-N (CH),N 227

H,C=CH, T 171

-C=C HC= CH 173

SC-ö: (CH,),CO 189

11 279

CH,NO, 201
274

Instrumentation : In visible-UV spectrometer, a beam of light is split into two equal halves.
One-half of th beam (the sanple beam) is directed through a transparent cell containing a soluton
of the compound being analyzed and one-half (the reference beam) is directed through an
cell that contains only the solvent. The instrument is identical
so-designed that it can compare the nte
Recorder
Sample
Source Detection
Monochromator
unit
|Reference
Beam
splitter
Fig. 25. Schematic diagram for a typical
visible-UV spectrometer.
ECTROSCó
1245
The right hand spectra of vinyl chloride show
abundant having amass of 35 amu, and that chlorine is alsc composed of two
wmpositionof chlorine and bromine is : the minor isotope a mass 37 amu. The isotopes, the
(a) Chlorine : 75.77% *Cl and 24.23% precise isotopic
(b) Bromine:50.50% Br "CI.
and 49.50% Br.
Note: The presence of chlorine or
as of ions differing by 2 amu. brominein amolecule or ion is easily
In the case of detected by noticing the intensity
188 amu and their methylene chloride, the molecular ion consists of three
Rp. Loss of a chlorine diminishing intensities may be calculated from the naturalpcaks at m/z = 84, 86
atom gives
orporating asinglechlorine atom. two isotopic fragment ions at n1/z = 49 and abundances given
51 amu, clearly
-ed Examples

Example 1. Calculate the frequency of radiations lhaving

vavelength 5,000 A, tohere c= 2.996 x 10" cm/s.
Solution. Frequency, v= 2.996 x 100 Cms l
=6x 1o!6 1
5,000 x 10- 10 Cm
Example 2. Calculate the energy per mole of light having
wavelengths of : (1) 85 n1n and (ii) 300 nm.
Solution. (i) E= N¡ hc 6.023 x 10¬° molx6.625 x10 4Jsx3.0x 10 ms1
(PT, May 04)
85 x 10 m
= 1.408x 10° J mol or
1.408 x 10° kJ mol.
(ii) AE= 6.023 x 10 mol'x6.625 x104 Jsx3.0x 10 ms1
300 x 10 m
=3.990× 10 J mol or 3.990 x 10 kJ mol
Example 3. Calculate : (a) AE (in k< mo! ), (b) v, (c) V
Solution. (a) To calculate the energy per mole, corresponding to ' = 200 nm.
multiply hc/ by Avogadro constant
(NA). Thus:
AE= N hc 6.023 x 10 moix6.625 x 10
200 x 10-9
|sx3.0x 10 ms
m
= 5.981 x 10 J mnol=598.1 kI mol.
3.0 x 10* ms! =1.499x 10!5 or Hz.
200 × 10 m
(c) v= 1 1
=5x 10 m'=5 x10* cm
200 x 10-9 m Im'=(10 cm)l= 10 cm ')
Example. 4. Calculate the reduced mass of C.I molecule,
40067 annu. given atomic weights of C=12.011 and
Solution. Reduced mass is given by :
MË 2g/kg
(nn1 + )
where m and m, are actual masses in g/kg.

(n1/N ) (m,/N ) 8 here mand m, in amu.

(my/N) (my/N )
(12.011) (14.0067) 8 =1.0737x 10g=1.0737x 100 kg.
(M1 + 2) N (12.011 + 14.0067) x (6.023 x 10)
1246
10
ENGINEERING CHEMISTRY
The internuclear distance of NaCl is 2.36 x1 10 m. Calculate3the red1uced mass and
Example 5. x10 kg mo 1
Cl= 35 x 10kg mol and Na = 23
inertia of NaCl. Given atomic nasses
Solution. () Reduced mass of
of
NaCl, moment of

(35 x 10 kg mol) (23 x 10 kg mol ') = 2.304 >x 10

mo l ko.
m +m (35+ 23)x10°kg molx (6.023 x 10
(ü) Moment of inertia,
(2.36 x 10 10 m'=1.283 x 10 kg m".
I= ur=(2.304x 10 kg)
Example 6. Calculate the force constant for carbon monoxide, if this compound absorbss at 2.143 x10 m
10 ° kg. and
its reduced mass is 1.139 x
Solution. We know that

VÍ=
1 k or k=4r v!
2C
=4x3.142x [3 x10° ms l'x[2.143 x 10 mx1139x 1026 kg
=1,857 kg m's (m =1,857 (kg ms (m}
=1,857 Nm 1
[: 1kgms'=1N
Example 7. Match each absorption bandwith the follotving groups:
-N-H, =C-H, -CaC-; -0-H
Functional group -C=0,
V/cm1 1,700, 2,050, 3,020, 3,350, 3,400
Solution.
Functionalgroup -N-H =C-H
-C=C -0-H
-C=0
V/cm-1 1,700 3,350 3,020 2,050 3,400

Example 8. Calculate the frequency (in Hz and cm ') of O- Hband, if the force constant and reducd mass
ofthe atom pair are 770 Nm and 1.563 x 10 kg respectively.
1 770 Nm1
Solution. (Ù) v=V
2. ! 2x 3.142 1.563 x 162 kg
-1

2x3.149 XV
770 kg m
sxm =1.117 x 10 s or Hz.
1.563 x 1027 kg
V=1117 x10'4s-!
(ii) 3,726cm1
C
3x 10 me1
Example 9. "H°CI has a force constant (k) value of 480 Nm Ifrequencyandis
Calculate the fundamental
wavenumber.

Solution. () u of 'HCI= 1x35 103

("1 + my) (1+35)..o23 kg= 1.615 X10 * kg
1 1 480 N n
Vo = = 8.676 >x 10" Hz.
27 2x3.142 1.615 x 10-27 kg
V 8.676 x 1o!3 s-1
(ii) V=
C
2.998 x10 ms1 =2.894 x 10 m1
= 2.894 >x 10
mx(10cm m)=2,894 cm.
sPECTROSCOPY 1247

Example 10. The value of force constant is same for 'HCI and D*CI. If the fundamental frequency of
iy'Clis 2890c1 1 Calculate the fundanental freq1uency offDCI.

Solution.
Vo(DCI) k(DCI) k (HCI)
Vo(HCI) 2r u(DCl) u(HCI)
(HCI)
=

(DCI) [: k(HCI)= k (DCI)]

1x35 10 kg
(1+35) 6.023 x 1023 35 x 37 37
= =0.7169
2x 35 10 kg 36 x 70 72
(2 +35) 6.023 x 1023

Hence, Vo (DCI) =Vo (HCI) x0.7169 =2,390 cmx 0.7169 =2,072 cm 1

Example 11. Calc1ulate the force constant of the CO molecule, if its fundamental vibrational frequency is
140 cm . (At. nmass of: C=1.99 x 10° kg and O = 2.66 x 10 2° kg.)
Solution. Reduced mass of CO,
1.99 ×10 20 kg x 2.66 x 10 2 kg
(1.99 + 2.66) x 10 26 kg 2=1.14>x 10 -26 kg
Now V= 1
2Tc
:. Force constant,
k= 41c (v'u =4 (3.14) (3x 10 cm s ly' (2,140 cm '
-2
(1.14 x 1026 kg)
=1,855 kg s2L,855 kg ms =1,855 Nm1
m

Example 12. From the IR spectrum given below identify the

eaks indicated by the arrows. possible functional groups corresponding to the
100
Transmittance%

4000 3000 2500 2000 1600 1600 e00 I000

P/cm-"

Solution. The IR spectrum has peaks at 3300 cm.2100 cmand l640 cm thereby
resence of =C-H, -C=C - and - C=C- groups.
ndcating the
Example 13. Tlhe wave number of the fundamental vibration of Br-Br is 323.2 cm
Stant of the bond. (Mass of "Br =78,9183 amu and Br=80,9163 am) Calculate the force US
(BPUT, Dec. 03)
 1248

Solution. Reduced mass of"BrBr, ENGINEERING CHEMMSTR

78.9183 g molx80.9163 g mol
(78.9183+ 80.9163) gmolx6.023 x 102 mol1
78.9183 x
80.916386.6333 x 10
159.8346 x 6.023 x 10* g=6.6333 x 10"2 kg
1
Now V=
2TC

k=4(3.14)' (3 x 100 cmsy (323.2 cmx (6.6333x 102h kg)

=246.095 kgs = 246.345kgs m= 246.095 Nm 1

Example 14. The spacing between lines in rotational spectra of HBr is 16.94 cm-1
of the molecule (H = 1, Br = 80). calculate the bond
Solution. Spacing between the lines,
leng
2 B= 16.94 cm or B=8.47 cm 1

I 100cm-1 6.625 x 10*Js

'8Bc 8n (8.47 cm ') (100 cm m }(3x10 ms) =3.30 x 10 kgm'.
or 1x 80g = 1.640 >x 10 *g=1.640 x 10
(m + m,) N 81x 6.023 x 10* kg

Hence, 3.302 x 10* kg r = 1.423 x 10

1.640 x 10 kg m=142.3 pm.
Example 15. Which of the following molecules will be rotationally active and why ?
() H, (ii) NO, (ii) HCI and (iv) Fz.
(BPUT, ec. 03)
Solution. NO and HCl.
Reason :Both these molecules possess permanent dipole moment, so these are
other hand, H, and F, are homonuclear molecules and these do not possess rotationally active. On the
dipolemoment.
Example 16. Microwave spectrum of gaseous HCl molecule exhibits a series of equally spaced lines t
interspacing of 20.7 cm. Calculate the internuclear distance of HCl molecule.
Solution. Here 2B= 20.7 cm=0.207 m or B=0.1035 x10 m'.
Now 6.625 x 10 kg m² sl
8rBc 8 (3.14) (0.1035 x 10 m ) (3x 1o ms ) =2.702 × 10 kg m.
and u= 1gmol x35.5 g mol1 1.561 x10 kg.
MË+ m2 (1+35.5) g mol x6.023 x 10. mol =1.561x10 g=
r= 2.702 x10 kg m² =1.422 x 10 0 N.
1.561 x 10-27
kg
of ico. Ta
Example 17. Calculate the frequency of the J =3’ 2 transition in the pure rotational.spectrum
(BPUT,Dec.0
equilibrium bond length is 112.81 pm.
Solution. We knowthat : vå ’+)=2B (J + 1)cm
V23=2B (2+ 1)cm=6B cm-l
SPECTROSCOPY 1249

Reduced mass of CO,

12 g molx16 g mol
1
(11 + m2) (12 + 16)g molx6.023x 10 mo
12 x16g =1.138x 10-23 g=1.138 × 10 " kg.
28 x 6.023 x 1023
Now h h
Or B=
8n Bc 8r²lc 8rurc
B= 6.625 ×10 kg m²s1 1.93 cm 1
8x (3.14) × 1.138 x 1026 kg x(112.81 x 10 12 m)x3× 10" Cm s
Hence, V2-3=6x1.93 cm= 11.58 cm 1
Example 18. Frequencies used in NMR spectra are of the order 60 MHz.
n kcal mor 1 Calculate the corresponding energy
Solution. E= Nalv (per mole)
= 6.023 >x 10° molx6.626 × 10 Is x 60 x
10 s =2,395 x 10J mol
=2.395 x 105kI mol-12.395 x10 kJ mol1 =5.723 x 10 0 kcal mol!
4.184 kJ kcal1
Example 19. At what frequency shift from TMS wvould a group of nuclei with &= 1.00 resonate in a
perating at 500 MHz ?
spectronmeter
Solution. Now: 8sVo
x 106 or Vs-Vo=8 v10-6
v,- Vo =1.00x 500 MHz × 10 =1.00 x
500 × 10° Hz x 10= 500 Hz.
Example 20. How many NMR signals are there in
in) CH,-CH, - Cl, (iv) CH, -CHCI CH3, (v) CçH,CH3, (vi) : (0) CH3- CH3, (ii)
CH - CH, CH
1 1 CGHsCH,CH, ?
Solution. () CH3- CH3 One signal, because of only 1
1 2 1 kind of protons.
(ii) CH3 -CH,-CH; Two signals, because of 2
1 2 kinds of protons.
(ii) CH, - CH, -Cl Two signals, because of 2
1 2 1 kinds of protons.
(iv) CH¡ CHCI CH, Two signals, because of 2 kinds of
2 1 protons.
() CGHs-CH, Two signals, because of 2 kinds of
1 2 3 protons.
(vi) CH;CH,CH, Three signals, because of 3 kinds of
Example Predict the high-resolution NMR spectrum of :
21. protons.
() CH,Cl-CHClh, (ii) CH3- CO-CH3.
1 2
Solution. (1) CH,CI-CHCI, i. Two signals
For photons '1', n =1 (of CHCI2), SO 1t gives a otblet, because n
+1=1+1=2.
For proton, 2, n=2 (of CH;Cl), so it gives a triplet, because H+l =24l-2
1 1
(ii) CH, -C- CH,. . One signal protons 1' are cquivalent, so only one line is
lution NMR spectrum. observed in
high-s-
1250
Example 22. Predict the high-resolution'H NMRspectrum of: () CH,CH,CHCI,
ENGINEERING CHEMASTR
(ii) CH,CHCICH,CI, (ii) CH,CC,CH (iv) CH,CICH,CH,CI, (v) CH,CH,OH.
1 2 3
CH,CH,CHCI, has three sets of equivalent protons, SO number of
Solution. (1)
ton) with relative intensits 3:2:2.
signals =3
"1, n=2 (due to -( CH,-
Sphtting of signals (igh resolution) : For protons n=3 (due to group), so there
(W+)=3 with relative intensities of 1:2:1. For protons "2, CH¡- is a
Therefore, there are group)
five and triole
-CHCI, group). Since the chemical environment is not different.
Telative intensities 1:4:6: 4:1. For protons '3, n =2 (due to -CH2- group), so there is a ines (quintet) with
intensities 1:2:1. triplet with relative
1 2
() CH,CHCICH,CI has three sets of equivalent protons,so number of signals = =3(low.
relative intensities 3:2:1.
For proton 1, n=1 (due to 2 CHCI groups) so there aretwo with relative
resolution) with
intensities
2, n=3 +2, so there aresix lines (sixtet) with relative intensities 1:5:10 :5:1. For proton1:1.FFor
-CHCI -group), so there are two lines (doublet) with relative intensities 1 :1.
1 1
(in) CH,-CCl,-CH, has only one set of equivalent protons, so number of signals
3,n=1(dpureotons
to

Splitting of signals (higlh resolution):For proton»1', n =0(due to Cl,- group), so thereis =1(loonewlinreesoltion)
1 2
(i7) CH,CI-CH, -CH,CI has tw0 sets of equivalent protons, so number of signals =22
(singlet
with relative intensities of 4:2 or 2:1. (low resolution
Splitting of signals (igh resolution) : For protons '1', n=2, so there are three line (triplet)
intensities of 1:2:1..For protons'2', n=2+2, so there are five lines (quintet) with relative with relative
:4:1. intensities 1:4:6
of
1 2 3
(v) CH3, -CH,-OH has three sets of equivalent protons number of signals = 3 (low
relative intensities 3:2:1.
resolution) with
Splitting of signals (high resolution) : For protons ´1', n=2, so there is a triplet with relative intensities ]
:2:1.
For protons "2, n =3, so there are = 4 lines (qiuartet) with relative intensities 1:3:3:1.
For protons '3,n =2, so there is a triplet with relative intensities 1:2:1.

Fig. Proton resonance spectrum of

ethylalcohol at 40 MHz. The signal
from the radiofrequency receiver is
plotted vertically,and the magnetic
field strength is plotted horizontaly.
The field increases linearly
from left to right.

Example 23. Give the high resolution "H NMR spectrum of methyl ethyl (PT,June04)
1 2 3
ether.
Solution. H,C-0-CH, -CH, has three types of protons, So there are three signals with relative
intensities 3: 2:3.
For protons 1", n=0, so there is a singlet (n +1=0+1=1). For protons 2,n=3(of adjoining (it),so
there is a quartet (n +1=3+l=4) with relative intensities 1:3:3:1. For protons '3, n=2 (of adjvning
CH), so there is a triplet (n +1=2+1=3) with
relative intensities 1:2:1.
 Practice
-ercises
For Hz ectrons) spectrum. SR
ntains .ndi
4. 2 1.
3. PECTROSCOPY
The the The Solution. Example Solution. Example iv=BNH Solution.Example (l1 :Solution.Example Solution. Example (b)Solution. Example line
and (Planck's
WhatCalculate =6.6262 has no
wavelengthVividh unpaired was
(iii)
wavelength is tvo observed
the the 29. 28. 27. 26. 25. 24.
constant, the HÍ = x
electrons N, Outer Vanti-Stokes (a)
Bharti V= V=
energy energy Show Calculate Calculate
10moleculeWhichelectron O1ut A AVRaman A
energy =2.8 =4.2 (in
of &BNHo hv sampleat
of &PNHÍ *s electronic
in h station
the x that out
of Å) 4,447
x ;g=5.585 1unpaired
of a kJ/mnol,=6.63
beam 10 10° (6.6262 Cu = =
one electromagnetic of 1 th e
(5.585) the (14 of ; was
s=2.80 whereas ion cm! inV 4,358 4,358 A.
1 of G=2.80 s= precessional
5.585 5.585 magneticelectrons) N2 10 108 excited
of of xmole All in and structure and 108 At
its light 10
associated (5.0508 10 x molecular ; +460 what
photon. India 10
6.62625.0508 x 42,000
>x MHz. x B O, d Cu 4,447
* of 5.0508
is Js)
photons x field system 10 by
2.80 10° 6.6262frequency Js) =5.0508 will 2.34110x wavelength
radiationRadio, x has ofion, the
{Ans. Hz x 10° × 10(60 strength show 10
x with 10 10" x orbitals. all Cut which em cm
1 4,358
10 of Hz >x x x contains
= 10 the =2.295
() electromagnetic Delhi 2.80 of JG 10° 10 ESR (Z=28) 460cm=
monochromatic m. enitted JGx1Gs JGx will in A
2.80 = electrons )s') required electrons 4,272 Å
Calculate broadcasts MHz. 42,000 s Hence, spectrum line
x =14,094G.
JG). one show x ¢
104 15,000 is 10* vould of
by MHz. unpaired d" mercury.
in to O, ESR cm'+
m; its the a (paramagnetic) give paired. ? and the
light G
15,000 spectrum anti
(ii) : radiation
transmitter, on a
() a precessional On d-electron. Cu* 460
1.07
wavelength radiation frequency Calculate
Stokes
G the
field. (Z= cm
10" × ?
with other line
of Hence, 27) =2.341× the
Hz; frequency of frequency will appear
in
wavelerlgth 1,368 hand, is Raman
(ii) cm, [Ans. show
Cu d'.
7.06
()requency, 523 219.3
[Ans.
m]kHz. Now 10* ?
for ESR O, shift
x [Ans.
9.3)] 5x will cm
kJ225 Calculate protonmolecule
1021 meo) 10+ spectrum. d 1 in
given
nm? system cm
Hz. of
60 (16 1251
an if
as