Problems on spectrophotometry
Lab 8
Week 9
Problems on spectrophotometry
E=h
E = h c/
wavenumber = 1/ , cm-1 .
Frequency (, nu) is the number of waves in 1 second
(it is expressed in Hz = 1 cycle/s)
Frequency (, nu) = C / 
Wavenumber ( , nu par), gives number of waves per unite distance
(it is expressed in cm-1).
wavenumber = 1/ , cm-1 .
E = Energy in joules (J) or ergs (erg)
 = Frequency (Hz)
h = Plank’s constant (6.625 X 10-34 J.Sec. or 6.625 X 10-27 erg. s)
 = Wavelength (cm) = C/ 
c = Speed of light (3 x 1010 cm/s in vacuum =  
Different units of length are used to express wavelengths and their amplitudes
1 m = 102 cm = 103 mm = 106  = 109 nm = 1010 o.
Beer’s - Lambert’s Law:
A=abC
Units of absorptivivity
1- Absorptivity (a):
A =abc
( a is expressed in L.g-1cm-1) where concentration (c) is as gram / Liter.
A = ε bc
2- Molar absorptivity (ε , Epsilon):
(ε is expressed in L.mol-1cm-1) where conc. (c) expressed as molar
3- A1%
1cm
A one percent one centimeter
1%
A1cm
=
1%
A
bc
A=
1cm
It is expressed in g% -1 cm -1 if c is expressed in g/100 mL
1%
A1cm
units
= ε x 10 / Mol.Wt.
10 a
Transformation between units
• 1cm = 10 mm= 10000 m = 10000000 nm = 100000000 A
• 1 cm = 10 3 m = 107 nm = 10 8 A
To convert g/L → g/ ml : we have to multiply by 10 3
To convert g/ ml → g/L : : we have to multiply by 10 - 3
Transformation between units
Different units of length are used to express wavelengths and their amplitudes
1 m = 102 cm = 103 mm = 106  = 109 nm = 1010 o.
gm
mg
𝜇g
ng
1
103
106
109
m
1
cm
102
L
1
𝜇g
106
mm
103
ml
103
A∘
1010
nm
109
𝜇l
106
Examples of problems with answers
1- Calculate frequency and wavenumber of light beams with a
wavelength 200 nm?
C = 
velocity of light C = 3 x 10 10 cm/sec.
•  = c/  = 3 x 10 10 / 2 x 10 -7 =1.5 x 10 17 cycle /sec.( or Hertz)
• ύ = 1/  = 1/ 2 x 10 -7 = 50000 cm -1
• -------------------------------------------------------------------------------------------2- Calculate the wavelength and frequency of light beams with
wavenumber of 5X 10 5 cm -1
 = 1/ ύ = 1/ 5 x 10 5 = 2 x 10 -6 cm
• ύ = 1/ 
•  = c/  = 3 x 10 10 / 2 x 10 -6 = 1.5 x 10 16 Hz
•
3 Express the wavelength 2500 A as micrometers and nm.
2500 A = 2500 / 10 = 250 nm
= 2500 x 10 -4 = 0.25 m
-------------------------------------------------------------------------------------------4 What is the energy of photons with wavelength equal to 100 nm.
E = h  = hC /  = 6.625 x 10 -27 x 3 x 10 10 / 1 x 10 -5
= 1.987 x 10 -11 erg.
5- A molecule absorb energy of 3.3125 x 10
wavelength in nm of its spectrum ?
-12
erg. what is the
E
= hC / 
3.3125 x 10 -12 = 6.625 x 10 -27 x 3 x 10 10 / 
= 6 x 10 -5 cm = 6 x 10 -5 x 10 7 = 600 nm
•

•
• (1cm = 10 mm= 10000 m = 10000000 nm)
• 1 cm = 10 7 nm
6- What is the wavelength and wavenumber of photons having a
frequency = 5 x 10 16 Hertz.
•  = c /  = 3 x 10 6 / 5 x 10 16
= 6 x 10 -7 cm
• ῡ= 1/  = 1 / 6 x 10 -7 = 1666666.667 cm -1
--------------------------------------------------------------------------------• 7- What is the wavelength and wavenumber of photons having an
energy = 5 x 10 -15 erg.
• E = hC / 
 = hc / E
•  = 6.625 x 10 -27 x 10 10 / 5 x 10 -15 = 0.03975 cm
• ύ = 1/  = 1 / 0.03975 = 25.157 cm -1
8-Convert the wavelength 4000 A  , 250 nm, 35 m and 5.5 mm into
frequencies and wavenumber.
• 400 A  = 4 x 10 -5 cm
• ῡ = 1/ λ = 1/ 4 x 10 -5 = 25 000 cm -1
• γ = c/ λ = 3 x 10 10 / 4 x 10 -5 = 7.5 x 10 14 cycle/ sec. or Hz.
9-Find the absorbance of the following solution, in all cases 1 cm cells
are used:
• Solution A of  = 1500 and C= 0.0001 M.
• A = bc = 1500 x 1x 0.0001 = 0.15
• Solution B of A1% = 200 and C = 0.003 g%
• A = A1% b c = 200 x 1 x 0.003 = 0.6
• Solution C of a = 30 and c = 0.015 g/ L
• A = abc = 30 x 1 x 0.015 = 0.45
• Solution E of  =5000 and c = 0. 001 g% (Mwt – 100)
• A 1% =  x 10 / Mwt = 5000 x 10 / 100 =500
• A = A 1% x b x c = 500 x 1 x 0.001 = 0.5
10- A drug of Mwt . 250, its absorbace is 0.8 in 1 cm cell at 550 nm and
its concentration is 40 g/ml .Calculate its  , A1%. And absorptivity.
40 g/ml → x 10 -3 = 0.04g/L
A=abc
0.8 = a x 1 x 0.04
absorptivity or a = 0.8 / 0.04 = 20 L. g -1. cm -1
 = a x Mwt = 20 x 250 = 5000 L. mol-1 cm -1
• A1%
= 10 a = 10 x 20 = 200
• 11- Drug of molecular weight 350 has molar absorptivity =1400 at 26 0 nm.
Calculate its absorptivity and A 1% 1 cm at that wavelength. If this drug
shows an absorption = 0.7 in 1 cm cell, find its concentration in g/ml.
•  = a x Mwt
• a =  / Mwt = 1400 / 350 = 4
• A 1% 1 cm = 10 a = 10 x4 = 40
• A = abc
• 0.7 = 4 x 1 x c
• c = 0.7 / 4 = 0.175 g/L = 0.175 x 10 6 / 10 3
•
= 175 g / ml
•
12- A drug has A1% of 100 at 300 nm. Calculate its molar absorptivity
at that wavelength if its Mwt is 280 . Then if an aquous solution of
this drug exhibits an absorbance of 0.4 in 0.5 cm cell. Find its molar
concentration.
 = a x Mwt = A1% x Mwt / 10 = 100 x 280 / 10 = 2800.
A = bc
0.4 = 2800 x 0.5 x c
c = 0.4 / 2800 x 0.5 = 0.0002857 M
• 13- A compound of Mwt. 450, its absorbnce is 0.33 in 2 cm cell at 400
nm and its conc. is 20 g/ml. calculate its  , A1%.
• C = 20 g/ml = 20 x 10 -3 g/L
•A = abc
• 0.33 = a x 2 x 20 x 10 -3
• a = 0.33 / 2x 20 x 10 -3 = 8.25 L.g-1 cm -1
•  = a x Mwt = 8.25 x 450 = 3712.5 L.mol -1 cm -1
• A1%.= 10 a = 10 x 8.25 = 82.5