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Modern Physics
Part III : Quantum Mechanics
Ch.# 41
…………………………: ‫مدرس المادة الدكتور‬
1431 – 1430
‫ادة‬
1
1. An interpretation of Quantum Mechanics
Lecture # 1
Ch. 41
Quantum mechanics is a theory. It is our current “standard model” for
describing the behavior of matter and energy at the smallest scales (photons,
atoms, nuclei, quarks, gluons, leptons, …).
Probability: a quantity that connects wave and particles.
Probility
N
  I E 2
V
V
Electromagnetic waves: Probability per unit volume of finding
a photon in a given region of space at an instant of time is
proportional to the number N of photons per unit volume V at
that time which its proportional to the intensity of the radiation
I or the Square of the electric field amplitude E.
2
2. A particle in a Box
2.1 The quantum particle under boundary conditions
Problem: A particle is confined to a one-dimensional
region of space with length L (“box”). It is bouncing
elastically back and forth.
Potential energy:
0 x L
0
U ( x)  
Otherwise

Wave function:
1) The walls are impenetrable: ψ(x) = 0 for x < 0 and x > L.
2) The wave function is continuous: ψ(0) = 0 and ψ(L) = 0.
These are boundary conditions.
3) The wave function:
2x


nx



(
x
)

A
sin

2L
2L 
L
Boundary conditions 
 n   

n  n is the quantum number
 ( x)  A sin
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Graphical representation:
 ( x)  A sin
• Only certain wavelengths for the particle are allowed:
nx
L

2L
.
n
• |ψ|2 is zero at the boundaries as well as some other locations depending on n.
• The number of zero points increases when the quantum number increases.
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Momentum and energy:
The momentum is restricted to
p
The energy is just the kinetic energy:
22

h


1 2 p2
h
2
2

n

,,
En  mv 
 n 
22 
mL 
2
2m
88mL
h


nh
.
2L

nn 11,,22,,3,,
The energy of the
particle is quantized.
Energy level diagram:
• Ground state: The sate having the lowest
allowed energy.
• Excited states: En = n2E1.
• E = 0 is not an allowed state since ψ(x) = 0.
The particle can never be at rest. Zero energy
also means an infinite wavelength.
Note that the energy levels increase as
n2, and that their separation increases as
the quantum number increases.
Lowest level n = 1, energy not zero why ?
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Example 41.2
An electron is confined between two impenetrable walls 0.200 nm apart.
Determine the energy levels for the state n=1.
h2n2
E
8ml 2
n=1
h=6.63x10-34 J.s
m=9.11x10-31 kg
l=0.2x10-9 m
E=1.51x10-18 J
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3. An electron is confined to a one-dimensional region in which its
ground-state (n = 1) energy is 2.00 eV. (a) What is the length of the
region? (b) How much energy is required to promote the electron to its
first excited state?
(a)
E1  2.00 eV  3.20  1019 J
h2n2
E
8ml 2
, For the ground-state, n=1
L
(b)
h2
E1 
8m eL2
h
 4.34  1010 m  0.434 nm
8m eE1
For the first excited state, n=2
 h2   h2 
E  E2  E1  4

 6.00 eV
2 
2
 8m eL   8m eL 
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8. A laser emits light of wavelength λ. Assume this light is due to a
transition of an electron in a box from its n = 2 state to its n = 1 state. Find
the length of the box.
hc  h2  2 2
3h2
2  1  
E 

2 
 8m L2
  8m eL 
e
so
L
3h
8m ec
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12. A photon with wavelength λ is absorbed by an electron confined to a
box. As a result, the electron moves from state n = 1 to n = 4. (a) Find the
length of the box. (b) What is the wavelength of the photon emitted in
the transition of that electron from the state n = 4 to the state n = 2?
h2
n2
(a): The energies of the confined electron are En 
2
8m eL
h2
2
2
4

1
Its energy gain in the quantum jump from state 1 to state 4 is
8m eL2
and this is the photon energy:
h215
hc

hf


8m eL2
Then 8m ecL  15h
2
and
 15h 
L
 8m c
12
e
(b): Let λ’ represent the wavelength of the photon emitted:
hc
h2
h2
12h2
2
2

4 
2 
2
  8m eL2
8m eL
8m eL2


h215 8m eL2
hc

5 and

Then


 hc 8m eL212h2
4
9
   1.25


Homework
Ch. 41, Problems: # 7, 11.
7. A ruby laser emits 694.3-nm light. Assume light of this wavelength is
due to a transition of an electron in a box from its n = 2 state to its n = 1
state. Find the length of the box.
11. Use the particle-in-a-box model to calculate the first three energy
levels of a neutron trapped in a nucleus of diameter 20.0 fm.
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3. Schrödinger Equation
Building on de Broglie’s work, in 1926, Erwin Schrödinger devised a theory
that could be used to explain the wave properties of electrons in atoms
and molecules.
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3.1. The wavefunction
Each particle is represented by a wavefunction Ψ(position, time) such that Ψ Ψ*
= the probability of finding the particle at that position at that time.
•Wavefunction Ψs are normally complex (real and imaginary parts)
•Probability density | ψ |2 always positive.
We write
| ψ |2 = Ψ Ψ*
Where
Ψ* is the complex conjugate of Ψ
If Ψ = A+ iB then Ψ* = A-iB
and | ψ |2 = A2 + B2
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3.2. Schrödinger Equation
In 1926 Schrodinger wrote an equation that described both the particle and wave
nature of the e-. The Schrödinger equation plays the role of Newton's laws and
conservation of energy in classical mechanics - i.e., it predicts the future behavior
of a dynamic system.
Wave function (Ψ) describes :
1. energy of e- with a given Ψ
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must
approximate its solution for multi-electron systems. The equation is of the form :
H Ψ= E Ψ
Ψ: wavefunction
E: Energy for the system
H: Hamiltonian operator
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Philosophic Implications; Probability versus
Determinism
The world of Newtonian
mechanics
is
a
deterministic one. If you
know the forces on an
object and its initial
velocity, you can predict
where it will go.
Quantum mechanics is
very different – you can
predict what masses of
electrons will do, but
have no idea what any
individual one will.
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