Download Solution Set # 3 Wavelike properties of particles

Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
Solution Set # 3
PH-102
September, 28, 2009
Wavelike properties of particles
Answer 1.
(a)
Start with the expression that tells you where maxima in scattering occur,
nλ = 2d sin θ .
Knowing that sin θn = 0.18 for n = 1, we want to find the value of d . Rewriting the above
equation as,
d =
λ
λ
=
·
2 sin θ1
2 × 0.18
Thus to find the separation (d) we need the wavelength of electron, which we can find out
by using the expression,
K =
p2
h2
=
,
2m
2 m λ2
where K is the kinetic energy of the electron. Therefore,
λ2 =
h2
(6.63 × 10−34 J s)2
=
= 1.37 × 10−20 m2
2mK
2 (0.91 × 10−30 Kg)(110 eV)(1.6 × 10−19 J/eV)
λ = 1.17 × 10−10 m.
In turn, this gives,
d =
1.17 × 10−10 m
= 3.14 × 10−10 m.
2 × 0.186
This will be the spacing between the crystal planes.
(b)
In the given problem, knowing the location of the first peak, we want to count the total
number of peaks. The above equation can be written as,
nλ
= sin θn
2d
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
As,
sin θn ≤ 1
nλ ≤ 2d
2d
n ≤
λ
Substituting values yields,
2 × 3.14 × 10−10
1.17 × 10−10
n ≤ 5.4
n ≤
We will see at the most 5 diffraction maxima.
Answer 2.
Given is the interatomic spacing d = 2.81Å, n = 1, θ = 20◦ . We have to find the energy of
thermal neutrons. For a first order Bragg’s reflection,
n λ = 2 d sin θ
λ = 2 (2.81Å) sin 20◦
λ =
2 (2.81Å) (0.342)
λ = 1.92Å.
Using the De Broglie’s relation,
h
p
h
p =
λ
6.63 × 10−34 J.sec
=
1.92 × 10−10 m
p = 3.45 × 10−24 Kg m/sec.
λ =
Kinetic energy will be calculated using the non-relativistic expression,
(3.45 × 10−24 Kg m/sec)2
p2
=
2m
2 × 1.67 × 10−27 Kg
K = 3.57 × 10−21 J .
K =
(b)
If we treat the neutrons as point particles, this means that the neutrons have only kinetic
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
energy and no potential energy. Since the neutrons can move in three orthogonal directions
(x, y and z), there are three degrees of freedom. This means that the average kinetic energy
is,
K = 3×
1
3
kB T = kB T.
2
2
2K
3 kB
2 × 3.57 × 10−21 J
=
3 × 1.38 × 10−23 J/K
T = 172 K.
T =
This is the approximate temperature of the thermal neutrons.
Answer 3
The De Broglie wavelength λ of a matter wave is given by,
hc
h
=
p
pc
hc
1243 eV nm
pc =
=
λ
0.5 × 10−15 m
p c ≈ 2480 MeV.
λ =
k
E
pc
moc 2
moc 2
Using the relativistic energy momentum relationship,
E 2 = (p c)2 + (mo c2 )
= (2480 MeV)2 + (938 MeV)2
E ≈ 2660 MeV.
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
Since the total energy is the sum of kinetic and rest-mass energy, therefore,
E = K + Eo
K = E − mo c2
= (2660 − 938) MeV
= 1720 MeV.
Answer 4
At time t = 0, the uncertainty in proton’s position is ∆ xo , then the uncertainty in the
momentum will be,
∆p ≥
~
·
2 ∆ xo
Since v ¿ c, then momentum uncertainty is,
∆ p = ∆(m v)
= m ∆v.
Then the uncertainty in proton’s velocity is,
∆v =
∆p
~
≥
·
m
2 m ∆xo
The distance x the proton covers in time t cannot be known more accurately than,
∆x = t∆v ≥
~t
·
2 m ∆ xo
Here ∆ x is inversely proportional to ∆ xo , the more we know about the proton’s position
at time t = 0, the less we know about its later position at t > 0.
The value of ∆ x at time t = 1.00 sec is,
(1.054 × 10−34 J sec) (1.0 sec)
2 (1.67 × 10−27 Kg) (1 × 10−11 m)
≥ 3.15 × 103 m,
∆x ≥
which is 3.15 Km!
The uncertainty increases with time. As time proceeds, the position of the particle becomes
more imprecise.
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
Answer 5
Given is the wavelength of emitted radiation λ = 589nm. ∆ t = 20 × 10−9 .
Using the Heisenburg’s uncertainty principle,
∆E ∆t ≥
~
2
Since E = h f which means ∆ E = h ∆f . But
c = fλ
c
f =
λ
−c ∆λ
∆f =
λ2
c ∆λ
|∆ E| = h 2 ·
λ
Therefore, the whole expression becomes,
h c ∆λ
~
∆t
≈
λ2
2
~ λ2 1
∆λ ≈
2 h c ∆t
h
1 λ2
≈
2 π 2 h c ∆t
1 λ2
≈
4 π c ∆t
1
(589 × 10−9 )2 m2
≈
4 π (3 × 108 m/sec) 20 × 10−9 sec
≈ 4.60 fm.
To find out the length of photon, we use the expression,
∆p = ~ ∆k.
From uncertainty relation,
~
2
~
∆x (~ ∆k) ≈
2
1
·
∆x ≈
2∆k
∆x ∆p ≈
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
∆k can be evaluated by using the expression,
2π
λ
2π
| ∆k| = 2 ∆λ
λ
k =
2π
4.60 × 10−15 m
(589 × 10−9 )2 m2
| ∆k| = 0.08 m−1 .
=
Therefore, the length of photon will be,
1
1
≈
2∆k
2 × 0.08 m−1
∆x ≈ 6.25 m.
∆x ≈
Answer 6
When the uncertainty in the velocity of a particle is equal to its velocity, i.e. ∆v = v. This
means that,
∆ p = m ∆v = m v = p
Therefore, the uncertainty in its position will be given by,
∆x ≥
~
~
h
=
=
·
∆p
p
2πp
The De Broglie wavelength is related to the momentum by,
λ =
h
·
p
Hence, the uncertainty in the position of the particle is,
∆x ≥
λ
·
2π
Answer 7
The wave function in real space ψ (x, 0) is related to the wave function in momentum space
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
A(k) through the given Fourier transform. Hence,
1
A(k) =
2π
=
=
=
=
Z
a/2
1
√ e−ikx dx
a
−a/2
¯a/2
1 1 e−ikx ¯¯
√
2π a −ik ¯−a/2
µ
¶
1
−ika/2
ika/2
√
e
−e
−2πik a
µ ¶µ
¶
1
1
−ika/2
ika/2
√
e
−e
πk a 2i
1
√ sin(ka/2)
πk a
Now the probability of finding the wave number k = p/~ is the modulus squared of A(k).
|A(k)|2 =
1
π2k2a
sin2 (ka/2)
1 a sin2 (ka/2)
π 2 4 k 2 a2 /4
a sin2 (ka/2)
=
4π 2 (ka/2)2
=
The curve represents the probability of measuring the particle with momentum k. This
curve is called a sinc function. The most probable value of the momentum is zero.
Answer 8
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
The position wavefunction is,
Z
∞
ψ(x, 0) =
δ(k − ko )eikx dk
−∞
= eiko x .
We have used the property of the Dirac delta function.
Z ∞
δ(k − ko ) f (k) dk = f (ko )
−∞
δ(k − ko ) is zero everywhere, except at k = ko , where it is non-zero, and the area under
it becomes unity. Therefore the integral vanishes everywhere except at k = ko and the
integral equals the value of the function at k = ko .
Now finding the probability,
|ψ(x, 0)|2 = e−iko x · eiko x = 1.
Hence the particle has a uniform probability of existing “anywhere“ in the space. The
position uncertainty ∆x is practically infinity.
Answer 9
The energy and momentum of a relativistic particle (in this case, electron) are given by,
mo c2
E = m c2 = q
·
v2
1 − c2
mo v
p = mv = q
·
2
1 − vc2
(1)
(2)
The energy in terms of the relativistic momentum p and the rest mass mo can be obtained
from the expression we have now encountered several times,
E 2 = p2 c2 + m2o c4 ,
p
E = c p2 + m2o c2 .
The corpuscular features (energy and momentum) of an electron are connected to its wave
characteristics (wave frequency and number ) by the relations,
E = ~ω
Chapter 4 of Eisberg and Resnick.
and
p = ~ k.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
Therefore the group and phase velocities will become
dω
dE
=
,
dk
dp
E
E
=
=
·
p
p
vg =
vph
and
From Equation (1) and (2), we find that p2 + m2o c2 =
m2o c2
2
(1− v2 )
. so, the phase velocity is
c
vg
µ p
¶
dE
d
2
2
c p + mo c
=
=
dp
dp
pc
= p
2
p + m2o c2
q
2
mo v c/ 1 − vc2
=
E
c
2
=
mo v c
q
E 1−
2
v2
c2
mo v c
= q
×
v2
1 − c2
q
1−
v2
c2
mo c2
·
Hence vg = v.
This shows that the speed of relativistic particle is equal to its group velocity.
Similarly, the phase velocity of relativistic particle can be calculated as
vph =
=
=
=
=
=
Hence vph =
Chapter 4 of Eisberg and Resnick.
E
p
p
c p2 + m2o c2
p
s
m2 c2
c 1 + o2
p
s
v2
m2 c2
c 1 + 2o 2 × (1 − 2 )
mo v
c
r
v2
c2
c 1 + 2 (1 − 2 )
v
c
r
c2
c
.
v2
¡c¢
c
.
v
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
As c > v, this means vph > c, predicting that the phase velocity for the relativistic particle is
greater than the speed of light c. This appears to be a violation of the postulates of special
theory of relativity. Actually, the phase velocity does not represent the physical velocity of
the particle, rather it is the group velocity which represents the speed of propagation of the
particle.
Answer 10
The resolving power of an electron microscope is far better than that of an optical microscope
because the wavelength associated with an electron is thousand times shorter than the
wavelength of visible light. From the theory of single-slit diffraction, we know that spatial
resolution δθ =
λ
,
d
where d is the diameter of the lens or the slit. If λ is small, spatial
resolution δθ is small, so two closely spaced objects can be resolved. It is the wave nature
of electrons which makes the electron microscope to show minor details, not visible with
optical microscope.
Answer 11
The energy of harmonic oscillator is given by,
E =
p2
1
+ mω 2 x2 .
2m
2
(3)
Here we assume that ∆p ∼ p and ∆x ∼ x . Using the uncertainty principle,
∆p ∆x ≥
~
2
p = ∆p ≈
~
~
≈
·
2 ∆x
2x
Writing Equation (3) in terms of x,
~2
1
+ mω 2 x2
2
4x 2 m
2
2
1
~
+ mω 2 x2 .
E =
2
8 mx
2
E =
Chapter 4 of Eisberg and Resnick.
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Solution set 3: Pair production, Bragg’s reflection, De Broglie’s Hypothesis and
Uncertainty principle
To find out the minimum energy, we set ∂E/∂x = 0, so,
µ ¶
¡ 2¢
~2 ∂ 1
1
2 ∂
∂E/∂x =
+
mω
x
= 0
8 m ∂x x2
2
∂x
µ ¶
~2 −2
1
+ mω 2 (2x) = 0
8m x
2
2
−~
+ mω 2 x = 0
4mx
mω 2 x =
mω 2 x2 =
x2 =
x =
x =
~2
4 mx
~2
4m
~2
2 2
4m
rω
~2
±
4m2 ω 2
~
±
·
2mω
At this position, the energy of the harmonic oscillator will be minimum. If one tends to
decrease x further, i.e., to squeeze the particle further, the energy shoots up again. To find
the minimum, energy we substitute the expression in the above Equation (3),
µ
¶
~2
1
p2
2
+ mω
E =
2m
2
4 m2 ω 2
p2
~2
E =
+
·
2m
8m
This is the required minimum energy of harmonic oscillator.
Chapter 4 of Eisberg and Resnick.
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