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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6
de Broglie hypothesis
Dual nature for de Broglie
- de Broglie reasoned that if light can display this waveparticle duality, then matter might also display wave like
properties under certain conditions.
- de Broglie was able to put his idea into a quantitative
scheme
- Einstein had shown from relativity theory that the
momentum of a photon is
p
h

- de Broglie assumed that both light and matter obey the
equation

h
p
(1-32)
This equation predicts of mass m moving with a velocity v
will have de Broglie wave length given by
h

mv
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6
Example 10:
Calculate the de Broglie wavelength for a baseball 0.14 kg
traveling at 40 m/s.
Solution:
The momentum of the baseball is
P= m v=0.14x40=5.6 kg. m. s-1
The de Broglie wavelength is
h 6.626 1034
 
 1.2 1034 m
p
5.6
- The de Broglie wavelength of the baseball is so small as to
be completely undetectable. The reason for this is the large
value of m.
Example 11:
Calculate the de Broglie wavelength of an electron
traveling at 1% of the speed of light. The mass of an
electron is 9.11 x 10-31kg.
Solution:
One percent of the speed of light is
v= 0.01x3x108= 3x106m/s
The momentum of the electron is given by
P=m v= (9.11x10-31)(3x106)=2.73 x10-24 kg.m.s-1
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6
The de Broglie wavelength of this electron is
h 6.626 1034
10
 

2.43

10
m
24
p 2.73 10
= 2.43Å
Example12:
Through what potential must an electron initially at rest
fall in order that λ=10-10 m?
Solution:
The momentum associated with this wavelength is
6.626 1034
24
1
p 

6.626

10
kg
.
m
.
s

1010
h
The energy of an electron with this momentum is
24
2
6.626

10


1
p
2
E  mv 

 2.411017 J
31
2
2m 2  9.1110
2.411017
 150eV
= 1.602 1019
,and hence the voltage
through which the electron must fall is 150 V.
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6
Heisenberg uncertainty principle
- The Heisenberg uncertainty principle states that it is not
possible to specify both the position and the momentum of
a particle simultaneously with infinite precision
- This is means that if we located an electron within a
distance Δ x, then this causes uncertainty in the
momentum of the electron Δ p
x p  h
(1-33)
This equation called uncertainty principle for Heisenberg.
Example 13:
Calculate the uncertainty in the position of a baseball, where
uncertainty in the momentum is 5.6x10-8 kg.m.s-1.
Solution:
8
p  5.6 10 kg .m .s
1
The uncertainty in the position of the baseball is
h
6.625 1034
26
x 


1.2

10
m
8
p
5.6 10
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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 6
Example14:
What is the uncertainty in momentum and in speed if we
located an electron within an atom, Δx=50 x10-12m?
Solution:
h
6.626 1034
23
1
p 


1.3

10
kg
.
m
.
s
x
50 1012
Because p=mv and the mass of electron is 9.11x10-31kg,
p 1.3 1023
7
1
v=


1.4

10
m
.
s
m 9.111031
This is very large uncertainty in speed.
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