Babylon University
College of Science
Chemistry Department
Allowed time is 2 hour only
Scholar year 2015-2016
total mark is 100 marks
‫اإلجابة النموذجية‬
Third Monthly Examination of Quantum Mechanics
B.S.c. Students
Name of Student:Note: - Answer all the Questions, and every question 20 marks.
Q1/ Can you classify the particles according the nature of the spins for elementry particles
that’s known either from experiments or from theories could be up to 2.
The classification of particles according the nature of the spins for elementry particles can be
done the spins for elementry particles known either from experiments or from theories could be
up to 2.
 Spin 0 particles:Higgs
 Spin 1/2 particles: electron, protons,neutrons,neutrinos, quarks .
 Spin 1 particles : mesons, photons(no 0 state though),gluons.
 Spin 2 particles: gravitons
 Half integer spin particals are fermions.
 Integer spin particales are bosons.
Q2// Calculate the wave function of particle motion in box according to variation theory. If you
supposed the proposed wave function is √2/𝑎. 𝑠𝑖𝑛 ∝ 𝑥 for the lowest energy value, since the
Hamiltonian operator is Ĥ=-√ℎ2 /8𝜋 2 𝑚.(
𝛿
).
𝛿𝑥2
a
Ĥ=-√ℎ2/8𝜋2𝑚.(
𝛿
)
𝛿𝑥2
E=
ʃΨ∗ĤΨ ∂τ
ʃΨ∗Ψ ∂τ
∫0
to get up E =
ℎ2
a
∫0
1
𝛿
2
√2/𝑎.𝑠𝑖𝑛∝𝑥.(−√8𝜋2𝑚.(𝛿𝑥2)).√𝑎.𝑠𝑖𝑛∝𝑥.∂𝑥
2
√2/𝑎.𝑠𝑖𝑛∝𝑥..√𝑎.𝑠𝑖𝑛∝𝑥.∂𝑥
To calculate ĤѰ must be do the following ĤѰ= =-
ℎ2
8𝜋2𝑚
2 𝜕
.√ .
𝑎 𝜕𝑥
(∝ 𝑐𝑜𝑠 ∝ 𝑥) ĤѰ=
ℎ2
8𝜋2𝑚
ℎ2
.
𝜕
8𝜋2𝑚 𝜕𝑥2
2
. √ .sinαx
𝑎
2
.√ . (∝ 2. 𝑠𝑖𝑛 ∝ 𝑥)
𝑎
By substituted the ĤѰ in general equation Ex=
𝑎
ℎ2
2 2 .
.√ .√ .∝2 ∫0 (𝑠𝑖𝑛∝𝑥)2𝜕𝑥
8𝜋2𝑚 𝑎 𝑎
2
𝑎
2 . 𝑎
∫ (𝑠𝑖𝑛∝𝑥)2𝜕𝑥
𝑎 0
√ .√
Ex=
ℎ2
8𝜋2𝑚
. ∝2 , α= √
8𝜋2𝑚𝐸𝑥
ℎ2
Therefore the proposed wave function of particle in box as following:Ѱx= B. sinαx ,
Ѱx= B.sin√
8𝜋2𝑚𝐸𝑥
ℎ2
x
Since B represented constant quantity called the normalization constant. Can be calculated by
normalization the Ѱx . B value equal to √2/𝑎 , therefore the proposed wave function equal to
2
8𝜋2𝑚𝐸𝑥
𝑎
ℎ2
Ѱx=√ . sin(√
.𝑥
There are same two results during the treatment of particle in box according to Schrödinger
equation.
Q3// What do you mean by approximation methods, at last explain the conditions of Variation
method that's can be used to solve chemical problems?
More complicated atoms, let alone molecules, have Schrödinger's Equations which cannot
be solved exactly. Such a problem commonly occurs when it is too costly either in terms of
time or complexity to compute the true function or when this function is unknown and we
just need to have a rough idea of its main properties. So, what good is the Schrödinger's
Equations approach? Well, fortunately, a number of approximation methods have been
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developed to allow us to calculate energies for example to, almost, any level of desired
accuracy. The two most widely used methods are Variation Theory and Perturbation Theory.
 Variation method
 Perturbation method
 Hartree –Fock self-consisted field method
Variation method can be used to solve chemical problems according to following conditions:1. Postulate the initial function, since this function must be do not closer to true function of
system.
2. Conformation the original function of the system to give up the best function of lowest
energy, that’s calculate by eq(1).
3. Postulate the mathematical operator of one parameter or more than. Controlling the value
of these parameters to obtain the best function of choosing functions.
Q4// Drive the Differential Schrödinger’s equation of Harmonic oscillator system in three
dimensions .
All components of Schrödinger formalism are treatment in three dimensions.
T=
H=
(𝑥𝑃2 .𝑦𝑃2 .𝑧𝑃2 )
, V=1/2K(x2+y2+z2)
2𝑚
(𝑥𝑃2 .𝑦𝑃2 .𝑧𝑃2 )
+ 1/2K(x2+y2+z2)
2𝑚
At this time the represented equation according to operators of quantum is
ℎ2
⟦−
8𝜋2𝑚
.(
𝜕2
𝜕𝑥2
+
𝜕2
𝜕𝑦2
+
𝜕2
1
) − 2 . 𝐾(𝑥 2 + y 2 + z 2 ) ⟧Ѱx,y,z =EѰx,y,z
𝜕𝑧2
Or
𝜕2𝜑
⟦. (
𝜕𝑥2
+
𝜕2𝜑
𝜕𝑦2
+
𝜕2𝜑
𝜕𝑧2
+
8𝜋2𝑚
ℎ2
1
) 𝐸 − 2 . 𝐾(𝑥 2 + y 2 + z 2 ) ⟧Ѱx,y,z=0 ----25
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By some approximations
8𝜋2𝑚
(
ℎ2
)𝐸 = 𝜆
𝜕2𝜑
⟦. (
𝜕𝑥2
+
𝜕2𝜑
𝜕𝑦2
+
,
4𝜋2𝑚𝑘
𝜕2𝜑
𝜕𝑧2
ℎ2
+
= 𝛼 2 the equation becomes
8𝜋2𝑚
ℎ2
) + 𝜆 −∝2 𝑥 2 +∝2 y 2 +∝2 z 2 ) ⟧Ѱx,y,z=0
The solutions of this equation come out by separation of the mathematical terms by using single
wave functions depend on single variables for each them..
Since Ѱx,y,z= Xx . Yy. Zz
By the differential for the wave equation in two time for each sub function with constant the
other two sub function and repeating the process for all. The result substituted in equation and
divided the resultant equation on X.Y.Z and rearrangement the final equation
1 𝜕2𝑋
1 𝜕2𝑌
1 𝜕2𝑍
−∝2 𝑥 2 ) + ( .
−∝2 y 2 ) + ( .
−∝2 z 2 )) + 𝜆 = 0
(( .
𝑋 𝜕𝑥2
𝑌 𝜕𝑦2
𝑍 𝜕𝑧2
The equation is symmetrical equation (equatorial terms) and because of the real symmetry, can
be separated into three sub equation all of them equal to zero.
𝜕2𝑋
+ (𝜆𝑥 −∝2 𝑥 2 )𝑋 = 0
𝜕𝑥2
𝜕2𝑌
𝜕𝑦2
+ (𝜆𝑦 −∝2 𝑦 2 )𝑌 = 0
𝜕2𝑍
𝜕𝑧2
+ (𝜆𝑧 −∝2 𝑧 2 )𝑍 = 0
4
The total energy of system is λ=(λx+λy+λz),it’s the Eigen value of the Eigen wave function
(Ѱx,y,z).
(harmonic oscillator for one dimension. The solution of equations is done by the same way of
equation and the final solution is come by the multiple the products of equation . The Eigen
value (energy value of system) equal to E=h[(vx+1/2)vx +(vy+1/2)vy +(vz+1/2)vz]
Q5// Calculate the relative probability of electron foundation in hydrogen atom (H) at energy
levels (2s) for 0.3 nm and 0.053 nm distance, what the two results represented the true distance
of electron from the nuclei in these levels?
Ψ(2s)=Ψ(2,0,0) =1/4√2π (z/ao)3/2 .(2 - zr/ao).e-zr/2ao
The relative probability is Ψ2 .r2
Ψ(2s)2 .r2=(0.3)2.1/32𝞹.(1/0.059)3.(2-0.3/0.059)2e(-0.6/2*0.059)
Ψ(2s)2 .r2=0.281 when r=0.3 nm
Ψ(2s)2 .r2=2,71*10-5 when r=0.053nm
As they see from two results that first result has larger relative probability than the second result
by 28100/2.71=103691 times ,that mean the 0.3 nm distance is the true distance of electron
from the nuclei in 2s orbital for hydrogen atom.
Constants: (R=8.31451 J.mol-1.K-1, C=2.997 *108 m. s-1, h=6.62608 *10-34J.s, NA =6.02214*1023 mol-1).
With my best wishes
A. Prof. Dr. Abbas A-Ali Drea
Instructor
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