))‫))بسمه تعاىل‬
Babylon University-College of Science
Allowed time is two hours only
Chemistry Department
Total mark is 100 marks
Quantum mechanics & Molecular
Spectroscopy
Fourth monthly Examination Questions of fourth stage students
Second Course of Scholar year 2015-2016
Name of student:- ‫االجابة النموذجية المتحان الشهر الرابع‬
Note:-Answer all the questions please.
(Every question 20 marks only)
Q1//Explain the following terms: 1-Molecular Spectroscopy:- The study of structure and properties of Chemical Species (atoms
and molecules) by means of the spectral information obtained from the interaction of
electromagnetic radiant energy with matter . It is the base on which a main class of
instrumental analysis and methods are developed & widely used in many areas of modern
science.
2-Electromagnetic radiation :- Electromagnetic radiation is a form of energy. Wave-particle
duality of electromagnetic radiation . Wave nature - expressed in term of frequency, wavelength and velocity. Particle nature - expressed in terms of individual photon, discrete packet of
energy.
3-Spectrum:- is the display of the energy level of EMR as a function of wave number of
electromagnetic radiation energy. The energy level of EMR is usually expressed in one of these
terms:- 1-absorbance (EMR being absorbed). 2-transmission (EMR passed through). 3-Intensity
. 4- Active Microwave :- Chemical species that’s possess a dipole moment power through the
centers of rotations, to provided the moment of inertia at their dimensions.
1
Q2//Draw Energy change diagram of excited molecules with all needed notation.
• Excitation of a molecule
• The energy levels of a molecule at each state / sub-state are quantised
• To excite a molecule from its ground state (S0) to a higher E state (S1, S2, T1
etc.), the exact amount of energy equal to the difference between the two states
has to be absorbed. (Process A)
i.e. to excite a molecule from S0,v1 to S2,v2, e.m.r with wavenumber v’ must be used
• The values of energy levels vary with the (molecule of) substance.
• Molecular absorption spectra are the measure of the amount of e.m.r., at a
specific wavenumber, absorbed by a substance.
v4
v3
v2
v1
v4
v3
v2
v1
S2
v4
v3
v2
v1
S1
A
T1
absorption
A
v4
v3
v2
v1
S0
2
Q3//Answer one of the following: A-Compare the energies of photons emitted by three radio stations, operating at 80 MHz(AM) ,
120 MHz (FM) and 3400 kHz (MW)?
Ans: Ephoton = hn
80 MHz = 80 x 106 Hz (s-1) =>
E1 = (6.63 x 10-34 Js) x (80 x 106 s-1) = 530.4 x 10-28J
120 MHz = 120 x 106 Hz (s-1) =>
E2 = (6.63 x 10-34 Js) x (120 x 106 s-1) = 795.6 x 10-28J
3400 kHz = 3400 x 103 Hz (s-1)
E3 = (6.63 x 10-34 Js) x (3400 x 103 s-1) = 260585.52 x 10-29J
They found the energy of Am and FM photons are larger than energy of MW photons.
B- The microwave spectrum of 39K127I consists of a series of lines whose spacing is almost
constant at 3634 MHz. Calculate the bond length of 39K127I.
Transformation of frequency into cm-1 units. Calculate the rotation constant (B). calculate the moment of
inertia. Calculate the bond length(rKI=(I*µ)0.5
Q4//Calculate the energetic values of positive ion of hydrogen molecule according to molecular
orbital theory.
Q5// Derive the Schrödinger equation of Helium atom system.
Operator of helium atom system must be consisted from kinetic energy of two electrons and the expressions of
potential energy for interaction of nuclei with electron at the same time of interaction of electron with electron.
High precession calculation methods involved spin coupling –orbital phenomena of angular momentum for both
two electrons. Electronic spin didn’t give up a difference in calculation of energy in this system.
There are three terms for operator:First term is kinetic energy of electrons
A second term is potential energy of nuclei and electrons.
Third term is potential energy of electron interaction with each other.
3
ĤѰ=EѰ
ℎ2
Ĥ=-8𝜋2𝑚 . (∇12 + ∇22 ) + (−
2e2
r1
−
2e2
r2
e2
+ r12)-----1
Schrödinger equation for Helium atom is
ℎ2
-8𝜋2𝑚 . (∇12 + ∇22 )Ѱ + (−
2e2
r1
−
2e2
r2
e2
+ r12) Ѱ = EѰ
------2
e2
At approximate state will ignore the repulsion term of two electrons with each other (r12), at this time equation
2 can be separated into two parts to became as two equations of hydrogen atoms.
Therefore
ℎ2
-8𝜋2𝑚 . (∇12 )Ѱ°1 (−
ℎ2
-8𝜋2𝑚 . (∇22 )Ѱ°2 (−
2e2
r1
2e2
r2
) = E°1Ѱ°1------3
) = E°2Ѱ°2------4
Since equation 3 and 4 represent electron 1 and electron 2 respectively. If we supposed that both wave functions
1&2 are the wave for electrons of helium atoms, so according to the independent particle model we get on
Ѱ=Ѱ˚1+Ѱ˚2 -------5
The approximate total energy system is
E=E˚1+E˚2----------------6
The real total energy is E=E˚1+E˚2+E3--------7
Since E3 is repulsion energy.
Schrödinger equation for first and second electron becomes as follow:8𝜋2𝑚
∇12 Ѱ°1+
ℎ2
8𝜋2𝑚
∇22 Ѱ°2+
ℎ2
. (E°1 +
. (E°2 +
2e2
r1
2e2
r2
) Ѱ°1 = 0------8
) Ѱ°2 = 0------9
Equation 8 is multiple by Ѱ°2 and Equation 9 is multiple by Ѱ°1, to getting up approximate wave function Ѱ in
each state.
Energy value of hydrogen atom is
En=
−2𝜋 2 𝑚𝑒 4 𝑍 2
𝑛2 ℎ 2
since Z2 and n2 equal to one for hydrogen atom therefore
4
E1=
−2𝜋 2 𝑚𝑒 4
=EH =-1/2 h
ℎ2
E˚1=
−2𝜋 2 𝑚𝑒 4 𝑍 2
𝑛2 ℎ 2
= -2h
(h= hartree) (Every 1h=27.21 ev)Energy value in helium atom becomes
because (Z2/n2 =4) and E˚2=
−2𝜋 2 𝑚𝑒 4 𝑍 2
𝑛2 ℎ 2
= -2h
So
E˚1+ E˚2= -4h
--------10
The calculated energy value of helium atom according to independent particle model equal to (-4h), but the
practical value (experimental value) equal to (-2.909h) therefore must be developed the calculated value.
Depending on the values of quantum numbers can be calculating the wave function as following:Ѱ˚1=Ѱ˚1s1=
23⁄2
√𝜋.𝑎˳3⁄2
. 𝑒−
2𝑟1⁄
𝑎˳
--------11
Ѱ˚2=Ѱ˚1s2=
23⁄2
√𝜋.𝑎˳3⁄2
. 𝑒−
2𝑟2⁄
𝑎˳
--------12
The complete solution represented by
Ѱ =Ѱ˚1+ Ѱ˚2 =
23
𝜋.𝑎˳3
. 𝑒−
2(𝑟1+𝑟2)⁄
𝑎˳
--------13
By using Approximation wave function and repulsion term in equation of energy calculation.
E=ʃѰ1s1. Ѱ1s2.Ĥ. Ѱ1s1. Ѱ1s2 𝛛𝜏------14 That's mean
1
2
2
1
E=ʃѰ1s1.Ѱ1s2.[− 2 (∇12 + ∇22 ) + (− r1 − r2 + r12)]. Ѱ1s1. Ѱ1s2 𝛛𝜏------15 Separation the terms of equation 15
E=−
1
ʃѰ1s1.Ѱ1s2.(∇12 + ∇22 )Ѱ1s1. Ѱ1s2 ∂τ
2
1
1
− 2 ʃѰ1s1. Ѱ1s2 (r1 + r2).Ѱ1s1.Ѱ1s2𝛛𝜏
+
ʃѰ1s1. Ѱ1s2
1
(r12)Ѱ1s1. Ѱ1s2 ∂τ------16
In equation 16 ,values of first integral and second integral according to independent particle model equal to (4h), since third terms values equal to (1.25h) and its solution is very complex.
Constants:-h = 6.63*10-34 J.S, C = 3*108 m.s-1, Me =9.11*10-31 Kg, 1J.m-1 =Kg.m.s-2.
With my best wishes
Instructor
Asst. Prof. Dr. Abbas A-Ali Drea
14-4-2016
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