Continued from previous installment:
4.
Nuclear Fusion Reactions:
In general a nuclear fusion
reaction is one in which two light nuclei combine (fuse) to form a heavier
nucleus with positive energy given off (the Q of the reaction). Nuclear fusion is demonstrated in its most
compelling form in the case of stellar energy. Exhaustive investigations in
this regard, eventually led to the realization that fusion was the only
practical energy by which stars could be sustained over long periods of time,
such as billions of years.
In
the Sun, for example, two distinct nuclear fusion processes occur: 1) the
proton-proton cycle, and 2) the carbon-nitrogen cycle.
In the first of these (the easier one because
it has fewer reactions):
1H + 1H + e- ®
2 H + n + 1.44 MeV
2 D + 1H
®
3 He + g +
5.49 MeV
3 He + 3 He ®
4 He + 1H + 1H + 12.85 MeV
The
top line shows two protons fusing to yield deuterium (heavy hydrogen) with a
positron and neutrino (n)
emitted, along with 1.44 MeV of energy. Empirical evidence of this reaction is
obtained from gallium detectors, of the neutrinos given off, which are within
1-2% of what theoretical models predict. In the
second fusion reaction, the deuterium combines with a proton to give the
isotope helium 3, along with a gamma ray (g)
and 5.49 MeV energy. In the final fusion, two helium-3 nuclei combine to yield
one helium-4 nucleus, along with two protons, and 12. 85 MeV energy. Note that
the two ending product protons commence the cycle anew, so that the generation
of nuclear energy is ongoing.
The
ending quantities on the right sides of each part of the cycle denote the Q of
the reaction for that part. Let us check
the Q for the first and simplest part. We know the hydrogen mass =
1.007825 u and for deuterium we have (from atomic tables): : 2 D = =
2.01410 u. Then:
Q
= [ 2(1.007825 u) – 2.01410 u] c2
Q
= [ 2.01565 u – 2.015941u] c2
Q
= [2.01565 – 2.01410] 931.5 MeV/u
Q
= [0.00155] 931.5 MeV/u = 1.44 MeV
The effect of ongoing fusion reactions such as
this, means that the central core of the
Sun becomes heavier and heavier, as more and more helium is produced. This
despite the fact that the Sun as a whole is losing an amount of mass of roughly
4 x 106 metric tones per second
Insight
Problem:
If
the atomic mass for helium 3 (3 He)
is equal to 3.01603 u, then verify the other Q-values for the last two parts of
the proton-proton cycle. A simplified,
compressed “net reaction”:
1H
+ 1H +1H + 1H ®
4 He + Energy
Is sometimes used to evaluate
the total energy released in the proton-proton cycle. Compute this energy and
compare to the value obtained for the total energy released in the earlier
example. Can you account for the difference?
Nuclear
Fusion Reactions in the Aging Sun:
At some stage, when nearly the entire
solar core is helium a new helium fusion phase will be ushered in (at higher
temperature), such that the following reaction series, known as the ‘triple
alpha’ process, kicks in:
4He + 4He ®
8Be + g (- 95 keV)
8Be + 4He ® 12C + g
+ 7.4 MeV
Here, the two alpha particles (helium
nuclei) first fuse to give unstable beryllium and a gamma ray (g), with 95 keV
energy absorbed. Then the beryllium
fuses with a helium-4 to give carbon–12 plus a gamma ray and 7.4 MeV energy
given off.
In
this way a new cycle commences, leading to a heavier molecular weight core.
Each successive burning phase, however, is less efficient than its predecessor,
as can be seen by comparing the energy given off in the triple alpha process to
the energy given off in the proton-proton cycle. The key thing to bear in mind
in terms of a stable phase (i.e. ‘Main sequence’) star like the Sun is that it
is in pressure-gravity balance. The outer gas pressure balances the weight of
its overlying layers. Any condition likely to disrupt this balance is therefore
of paramount interest.
The stable lifetime of the Sun depends on how
long before it consumes ninety percent of the hydrogen in its core. Theoretical
investigations using data from nuclear reaction rates and cross sections
suggest the Sun’s Main Sequence lifetime at 8-10 billion years. Since it
already has spent 4.5 billion of those years, there are anywhere from 3.5 to
5.5 billion years remaining.
Once the triple-alpha process gets underway and
the energy balance declines, the Sun will have to compensate for the lost
energy to sustain any kind of balance. Thus, the Sun’s core must contract and
convert gravitational potential energy into thermal energy. Meanwhile, ignition of hydrogen burning in
the Sun’s outer layers will create radiation pressure that forces the outer
layers to expands. The Sun will then become a “Red Giant” and its new
larger surface will be expected to engulf all the planets up to and including
Mars.
Example
Problem:
If the atomic mass of beryllium 8 (8Be) = 8.00531 u, verify that the first
part of the triple-alpha fusion process is endothermic and has the value given.
Solution:
We
have:
Q
= [ 2(4.00260 u) – 8.00531 u] c2
Q
= [ - 0.00011] 931.5 MeV = 0.102465 MeV = - 102.4 keV
Of
course, not taken into account here is the gamma ray (g) which also comes off.
Hence we will have:
(-102.4
keV) + (E (g))
= -95.7 keV
So
that:
E
(g) =
hc/ l =
6.7 keV
Is the missing energy of the gamma ray photon, with the
difference factored in yielding 95.7 keV.
Aside:
The Problem of the Coulomb Barrier in Solar Fusion
The problem of the Coulomb repulsive barrier
to solar nuclear fusion was first highlighted and explored by Prof. Martin Schwarschild
in his excellent monograph 'The Structure
and Evolution' of the Stars’ (Dover,
1958).
Schwarzschild once calculated that the
probability of any one proton fusion in the Sun’s core would ordinarily be about once
every 14 billion (14 x 109) years. Since the universe itself is
only 13.8 billion years old this means it could never occur unless another
factor was present to enable it.
The reason for this has to do with the
Coulomb (electrostatic) repulsion between the potentially fusing H-nuclei.
Thus, each proton, having (+) charge tends to repel any other proton within a
discrete sphere or distance around it. (Recall from your basic physics, like
charges repel, unlike attract - and that's what essentially obtains here)
In order for thermonuclear fusion to
be realized, the Coulomb barrier must be overcome. Fortunately quantum
mechanics allows for a certain non-vanishing probability that a particle (say
proton) of kinetic energy K, can overcome a barrier of energy V ("barrier potential"), via the process
of "quantum tunneling".
Note that tunneling is a general
feature of low mass systems, such as single proton (H) states.
Consider a deBroglie (matter) wave arising from a single proton (p+) of form:
U(x) ~ sin(kx)
Where x is the particle's linear displacement (e.g. in 1-D) and k, the wave
number vector(k= 2π/l),
where l denotes the wavelength.
Though the associated kinetic energy K < V (the barrier
"height") the wavefunction is *non-zero* within the barrier, e.g.
U(xb)~ exp(- cx)
So, visualizing this behavior as shown below:
Fig. 4: Tunneling through Coulomb barrier potential to allow nuclear fusion
with
the "barrier" at height V, so we can visualize the particle of lesser
energy K, moving from the left side of the E-axis "tunneling" through to the right side where it may have
wave function, U(x) ~sin (kx + φ), where φ denotes a phase angle.
Note that if the barrier is not too
much higher than the incident particle energy, and if the mass is small, then
tunneling is significant.
It's important here to point out that
the penetration of the barrier is a direct result of the wave nature of matter.
In effect, this wave nature - which is uniquely quantum mechanical in origin-
allows a higher energy barrier to be penetrated by a lower energy particle,
something totally without parallel in classical, Newtonian physics!
Even given tunneling, an
"offset" is required to reduce the low penetration probability ,
since clearly the Sun and other stars are shining by fusion.
This 'offset' arrives via enormously
high density of protons, e.g. in the core, which: i) increases the probability
enormously, since so many more protons are in extremely close proximity, and
(ii)enhances temperatures to the point they can be sustained, and continue -
thereby building up other fusion reactions to finish the initial one.
The idea here being that a particle of
relatively low incident energy (of kinetic energy K, say) can actually
penetrate a higher potential energy barrier, say of energy V(x) > K. Note that the penetration of the barrier is a direct result
of the wave nature of matter! (The matter wave form changes in the process of
transmission through the barrier, say from an exp(-ikx) function to a sin
(kx + f) where f
denotes phase angle). In effect, this wave nature - which is uniquely quantum
mechanical in origin- allows a higher energy barrier to be penetrated by a
lower energy particle, something totally without parallel in classical,
Newtonian physics! Note that if the barrier is not too much higher than the
incident energy, and if the mass is small, then tunneling is significant. It
was insights such as this that paved the way to apprehending how much subtler
nature was than hitherto realized, and how many more technological advances
could be achieved when the wave nature of matter was factored into
designs.
5. The
Quantum Treatment of the Deuteron.
The deuteron is perhaps the most
basic nuclear system to confront. As we know the deuteron consists of one
proton, one neutron and the electron – with the first two comprising the
nucleons.
To proceed, we write the usual Schrodinger
equation for the hydrogen atom but let F and Q = const. so that their derivatives are
zero. Then, substitute in the reduced mass:
m’= m n m p/ (m n
+ m p)
So
we obtain the new form of the Schrodinger equation:
1/r2
d/dr (r2 dR/dr) + 2m’/ ħ2 [E – V] R = 0
This
can be further simplified by letting:
U(r) = rR(r)
which results in the equation:
d2U/dr2 + 2m’/
ħ2 [E – V] U = 0
In
this equation we find that V, the potential, has two values:
V
= - Vo and V = 0 outside the
well. The diagram below shows how we are treating the deuteron in terms of the
function V(r).
Fig.
1: The potential well associated with deuteron
There are then two solutions
we can designate:
i) u I for r <
r o
and
ii)
u II
for r > r o
Inside the potential well:
d2 u I /dr2 + 2m’/
ħ2 [E + Vo] u I = 0
and we let: a2 =
2m’/ ħ2 [E + Vo]
so that:
d2 u I /dr2 + a2 u I = 0
For which we can show:
u I = A cos (ar)
+ B sin(ar)
Since R = u/ r the cosine solution must be discarded, lest
we get an unwanted infinity. This leaves:
u I = B
sin(ar)
Outside the potential
well V = 0 so that:
d2 u II /dr2 + 2m’/
ħ2 [E] u II = 0
Let:
b2 =
2m’/ ħ2 [- E ]
Since the total energy of the neutron is negative, i.e.
being bound to the proton. Then:
d2 u II /dr2 - b2
u II = 0
Which can be shown to have the
solution:
u II = C exp (-br) + D (exp(br))
For consistency we demand u ® 0 as r ®¥, so D = 0 and:
u II = C exp (-br)
For continuity at r = r o and u I = u II :
B sin(ar) = C exp (-br)
Thence:
du I /dr = d u II /dr Þ aB cos (a r o) = -bC e-b r o
Now, divide the solution on
the left side by the one on the right side, e.g.:
a B cos (a r o)
/ -b C e-b r o
so:
tan (a r o) = - a/b
In effect, the deuteron
problem cannot be solved analytically, only graphically.
Quantum Numbers for Deuteron:
Since
there are two particles, each with intrinsic spin ½ in
the deuteron, the total intrinsic spin angular momentum can only have the
values: S = s1 + s2 = 0 or 1.
The orbital angular momentum quantum number, L (describing the motion in
space of the proton and neutron relative to each other) can assume the values
L= 0, 1, 2 (i.e. S, P, and D states).
The
total angular momentum of the deuteron has been measured and the total angular
momentum quantum number J has been found to be 1. This must be the vector sum
of the orbital angular momentum (L) and the total spin momentum (S).
Problems:
1) Calculate the wavelength of the gamma ray photon
(in nm) which would be needed to balance the endothermic part of the triple
–alpha fusion equation. (Recall here that 1 eV = 1.6 x 10 -19 J)
2) Verify the second
part of the triple-alpha fusion reaction, especially the Q-value. Account for
any differences in energy released by reference to the gamma ray photon coming
off and specifically, give the wavelength of this photon required to validate
the Q.
3) The luminosity or
power of the Sun is measured to be L = 3.9 x 1026 watts. Use this to estimate the mass (in kilograms)
of the Sun that is converted into energy every second. State any assumptions
made and reasoning.
4) In a diffusion
cloud chamber experiment, it is found that alpha particles issuing from decay
of U238 ionize the gas inside the chamber such that 5 x 10 3 ion
pairs are produced per millimeter and on average each alpha particle traverses
25 mm. Estimate the energy associated with each detected vapor trail in the
chamber if each ion pair generates 5.2 x
10-18 J.
5)
When 118 Sn 50
is bombarded with a proton the main fission fragments are:
24
Na 11 and 94
Zr 40
The
excitation energy necessary for passage over the potential barrier is:
e > 3 ke2
Z2/ 5
Where the right hand side denotes the height of the
Coulomb barrier.
a) What must this value be?
(Take ke2 = 1.44 MeV/ fm)
b)
Find the energy difference between
the reactants and the products. (Take c2 = 931.5 MeV/ u)
6) a) Show that u I = A cos (ar)
+ B sin(ar)
In a solution of the reduced Schrodinger equation for the deuteron:
d2 u I /dr2 + a2 u I = 0.
Show
why the cosine solution needs to be discarded.
b) Show that u
II = C exp (-br)
Is
a solution of the other reduced Schrodinger equation for the deuteron:
d2 u II
/dr2 + b2
u II
= 0?
Explain.
If you move the detector, you are changing the detector relative to the photons that it is detecting — which were emitted by the source long ago.
This is a nice example of one of the many classic mistakes people make in misunderstanding relativity. You have to pay very close attention to what Einstein’s equations do and do not say — otherwise you will get yourself profoundly confused.
What Einstein’s equations say is that if you have a source moving toward the detector FOREVER it is the same as a detector moving toward the source FOREVER. If the source and detector are initially stationary and you CHANGE the situation, then you have to calculate much more carefully what the effect is.
Incidentally particular mistake this has nothing to do with Einstein’s relativity. You can make the same mistake with Galilean relativity using sound waves.