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%TCIDATA{Created=Thu Apr 01 12:54:11 1999}
%TCIDATA{LastRevised=Thu Apr 15 13:57:22 1999}

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\SetTitle{Assignment 11} \SetAuthor{} \Setdate{} \TitlePage{}

\textbf{1. Solar temperatures.} Here we will try to estimate some
relevant temperatures in the sun. Assume you start with an
initially diffuse cloud of hydrogen and helium atoms (initially at
rest), which subsequently collapse under its gravitational
attraction.

(a) Use dimensional analysis to show that the total energy
released in the gravitational collapse is about
\begin{equation}  \label{energy}
E_{\odot} = \frac{G~M_{\odot}^2}{R_{\odot}}
\end{equation}
Calculate this energy.

(b) How does this energy compare to the total energy that could be
released by (hypothetical) chemical reactions in the sun? To the
total energy that could be released by converting all the hydrogen
into helium? (You can assume that the cloud has the ``primitive''
abundance of helium---it does not make much difference as long as
you have mostly hydrogen).

(c) Assuming that all of the energy (\ref{energy}) is converted
into heat (a rather dubious assumption), estimate the interior
temperature of the sun. Compare with the accepted value of the
temperature at the center of the sun (look it up).

(d) The intensity of solar radiation has a peak at a wavelength $490~\mathrm{%
nm}$. What is the surface temperature of the sun? As you go from
the surface
of the sun toward the center, what is the approximate temperature gradient $%
dT/dx$? How can this temperature gradient be maintained?

\vspace{1cm}

(a) The energy it takes to bring two point masses $m_{1}$ and
$m_{2}$ from infinitely far away to within a distance $r$ is
\[
E=-G\frac{m_{1}m_{2}}{r},
\]
where $G=6.6726\times 10^{-11}~\mathrm{N\cdot m^{2}/kg^{2}}$, this
is the gravitational potential energy. Dimensional analysis
suggests that the gravitational energy associated with collapsing
a diffuse cloud to an object the size and mass of the sun is
\[
E=-G\frac{M_{\odot }^{2}}{R_{\odot }}
\]
(The negative sign indicates that it does not require energy,
instead energy
is gained.) For the sun which has a mass $M_{\odot }=1.99\times 10^{30}~%
\mathrm{kg}$ and radius $R_{\odot }=6.96\times 10^{8}~\mathrm{m}$,
we get
\[
E=-3.8\times 10^{41}~\mathrm{Joules}.
\]

\textbf{A digression.} If we assume the density is constant, we
can calculate the numerical coefficient. Let us take $m_{1}$ to be
a sphere of radius $r$ and mass density $\rho $ (it looks like a
point mass so long as we stay outside its radius), $m_{1}$ is
\[
m_{1}=\rho \frac{4\pi r^{3}}{3}.
\]
Let us take $m_{2}$ to be a spherical shell which after it is
brought in
from infinity has a radius $r$, a thickness $dr$ and the same mass density $%
\rho $, then $m_{2}$ is
\[
m_{2}=\rho \,4\pi r^{2}dr.
\]
The change in energy required to bring in the shell is thus
\[
dE=-G\frac{\rho \frac{4\pi r^{3}}{3}~\rho 4\pi r^{2}dr}{r}.
\]
So the energy to make a sphere of radius $R$ is
\begin{eqnarray*}
E(R) &=&-\frac{16\pi ^{2}G\rho ^{2}}{3}\int_{0}^{R}r^{4}dr \\
&=&-\frac{16\pi ^{2}G\rho ^{2}}{15}R^{5}
\end{eqnarray*}
Now expressing $\rho $ in terms of the total mass $M$
\[
\rho =\frac{M}{\frac{4\pi R^{3}}{3}}
\]
and substituting we find
\[
E(R)=-\frac{3GM^{2}}{5R}=-2.28\times 10^{41}~\mathrm{Joules},
\]
if the density were constant.

\medskip

(b) The proton-proton cycle essentially turns four protons and two
electrons into an alpha particle and gives off $26.7~\mathrm{MeV}$
in the process. Assuming the sun starts off entirely as protons
and electrons and is converted entirely into helium (alpha
particles), then the total energy would be
\[
\frac{1.99\times 10^{30}~\mathrm{kg}}{1.67\times
10^{-27}~\mathrm{kg}}\times
\frac{2.67\times 10^{7}~\mathrm{eV}}{4}\times \frac{1.60\times 10^{-19}%
\mathrm{Joules}}{\mathrm{eV}}=1.3\times 10^{45}~\mathrm{Joules},
\]
which is the number of protons divided by four multiplied by the
energy released in a proton-proton cycle. A hypothetical chemical
reaction would have an energy $10^{5}$ to $10^{6}$ times smaller.
Thus, the hypothetical chemical energy is smaller than the
gravitational energy which is smaller than the fusion energy.

(c) Assuming all of the energy calculated in part (a) is converted
into heat (thermal kinetic energy), we get
\[
\langle K.E.\rangle _{total}=-E_{dim}.
\]
(In a similar problem given by Kittel and Kroemer, one is told to
assume that \textit{half} of the gravitational potential energy is
converted into thermal kinetic energy in accordance with the
\textit{virial theorem}, but we are only concerned with order of
magnitude here.) Let us suppose this is the average kinetic energy
everywhere in the entire sun. Then the average kinetic energy per
particle
\[
\langle K.E.\rangle _{particle}=\frac{\langle K.E.\rangle
_{total}}{N},
\]
where $N$ is the total number of particles. Again assuming the sun
is composed purely of hydrogen, let $N$ be the number of hydrogen
atoms
\[
N=\frac{M_{\odot }}{m_{p}}=\frac{1.99\times
10^{30}~\mathrm{kg}}{1.672\times 10^{-27}~\mathrm{kg}}=1.19\times
10^{57},
\]
so then
\[
\langle K.E.\rangle _{particle}=\frac{3.8\times 10^{41}~\mathrm{Joules}}{%
1.19\times 10^{57}}=3.2\times 10^{-16}~\mathrm{Joules}.
\]
If we take the equipartition theorem for a monatomic gas then
\[
\langle K.E.\rangle _{particle}=\frac{3}{2}k_{B}T,
\]
(where we have assumed the temperature is the same throughout,
which is not consistent with what we find later). But anyway this
leads to
\[
T=\frac{2\dot{3}.2\times
10^{-16}~\mathrm{Joules}}{3\dot{1}.38\times
10^{-23}~\mathrm{J/K}}=1.5\times 10^{7}~\mathrm{K}.
\]
\textit{Melissinos} (p. 193) lists the temperature at the interior
of the sun as $T_{\odot }^{int}\sim 1.6\times 10^{7}~\mathrm{K}$.

\medskip

(d) The electromagnetic radiation from the sun is determined by
its surface temperature assuming it radiate as a black-body.
\textit{Wien's displacement law} says that $\lambda _{m}T$ is
constant, where $\lambda _{m}$ is the wavelength at which the
blackbody radiation intensity is maximal and $T$ is the
temperature of the blackbody. More specifically,
\[
\lambda _{m}T=2.898\times 10^{-3}~\mathrm{m\cdot K}=\frac{hc}{4.96511~k_{B}}%
.
\]
where $4.96511$ is the root of the equation $e^{x}(5-x)=5$. If
$\lambda _{m}=490~\mathrm{nm}$ then $T_{\odot }^{sur}\approx
5900^{\circ }\mathrm{K}$.

So we go from a temperature of $T_{\odot }^{int}=1.6\times
10^{7}~\mathrm{K}$ at the center to $T_{\odot
}^{sur}=5900~\mathrm{K}$ at the surface. If the gradient is
constant, it is
\[
\frac{dT}{dx}=\frac{T_{\odot }^{int}-T_{\odot }^{sur}}{R_{\odot }}=0.023\,%
\mathrm{K/m}
\]
This temperature gradient is maintained and further gravitational
collapse is prevented by the radiation pressure exerted by the
energy issuing from the fusion processes in the interior.

\vspace{0.5cm}\hrule\vspace{0.5cm}

\textbf{2. List three radioactive nuclides} that naturally occur
(in easily detectable amounts) on Earth today. What are the
half-lives of these nuclides? How did they originate? At least one
of them should have a half-life of less than a thousand years.
Explain how it is possible to find nuclides with a half-life so
much shorter than the age of the Earth.

\vspace{1cm}

Several nuclides have lifetimes of the order of 10$^{9}$ or
10$^{10}$ years, which is about the age of the earth, and they are
still around today. Among them are Uranium-238 ($^{238}U$) and
Uranium-235 ($^{235}U$). Many of their decay products are also
radioactive and have much shorter lifetimes. For instance:

\begin{itemize}
\item  Radon-222 ($^{222}Rn$) is present in the atmosphere and collects in
caves and basements. It has a half-life of $3.8~\mathrm{days}$ and
is
created when ($^{238}U$) decays. $^{238}U$ has a long half-life of $%
4.51\times 10^{9}~\mathrm{years}$ and is found in some rocks. With
such a long half-life it serves a slow but steady source of
$^{222}Rn$.

\item  Iodine-131 ($^{131}I$) which has a half-life of $8.02~\mathrm{days}$
is produced when $^{235}U$ fissions. While this isotope occurs
naturally, most of it is man-made.
\end{itemize}

Carbon-14 ($^{14}C$) is also present in the earth's atmosphere; it
has a half-life of $5730~\mathrm{years}$. It is created when
cosmic rays enter the atmosphere producing a shower of particles,
some of which are neutrons.
Nitrogen ($^{14}N$) can capture a neutron and give off a proton, leading to $%
^{14}C$.

\vspace{0.5cm}\hrule\vspace{0.5cm}

\textbf{3. Particle decays.}

(a) Name the particles involved in the process $n\rightarrow p+e+\bar \nu _e$%
. Which are hadron, mesons, baryons, leptons? Complete the
following table

\begin{tabular}{lllll}
& $\,\,\,n\,\,\,$ & $\,\,\,p\,\,\,$ & $\,\,\,e\,\,\,$ &
$\,\,\,\bar \nu _e\,\,\,$ \\ Electric charge &  &  &  &  \\ Baryon
number &  &  &  &  \\ Electron number &  &  &  &
\end{tabular}

(b) Some of the particles in this process are made out of quarks.
Describe the process in terms of the transformation of the
constituent quarks.

(c) What are these decays forbidden?

i. $n\rightarrow \pi ^{+}+e^{-}+\bar \nu _e$

ii. $n\rightarrow e^{+}+e^{-}$

iii. $n\rightarrow p+\mu ^{-}+\bar \nu _\mu $

iv. $n\rightarrow \bar p+e^{+}+\nu _e$.

\vspace{1cm}

(a) The particles in the process $n\rightarrow p+e+\bar \nu _e$
are the neutron ($n$), the proton ($p$), the electron ($e$), and
the electron antineutrino ($\bar \nu _e$). Hadrons are made of
quarks that are held together by the strong interaction (that is,
by exchanging gluons). The neutron and the proton are hadrons.
Hadrons come in two varieties: mesons which are made of a quark
and an antiquark bound together and baryons which are made of
three quarks bound together. (So what? A meson has integer spin
and is bosonic, while a baryon has half-integer spin and is
fermionic.) The neutron and proton are baryons. The electron and
electron antineutrino are leptons.

\begin{tabular}{lllll}
& $\,\,\,n\,\,\,$ & $\,\,\,p\,\,\,$ & $\,\,\,e\,\,\,$ & $\,\,\,\bar{\nu}%
_{e}\,\,\,$ \\ Electric charge & $0$ & $+$ & $-$ & $0$ \\ Baryon
number & +1 & +1 & $0$ & $0$ \\
Electron number & $0$ & $0$ & +1 & $-1$%
\end{tabular}

(b) The neutron and proton are baryons made of up and down quarks,
$u$'s and $d$'s. The $u$ has a charge of $+2/3$ and the $d$ has a
charge of $-1/3$. From this we can deduce that the neutron, which
is neutral, is made of one up and two down quarks, while the
proton, which has charge $+1$, is made of one down and two up
quarks. Thus the process $n\rightarrow p+e+\bar \nu _e$ can be
seen as $d\rightarrow u+e+\bar \nu _e$ where the $d$ on the
left-hand-side is bound to a $u$ and a $d$, making an $n$, and the
$u$ on the right-hand-side is also bound to a $u$ and a $d$, this
time making a $p$.

(c)

i. $n\rightarrow \pi ^{+}+e^{-}+\bar \nu _e$ violates baryon
number conservation.

ii. $n\rightarrow e^{+}+e^{-}$ also violates baryon number
conservation.

iii. $n\rightarrow p+\mu ^{-}+\bar{\nu}_{\mu }$. Charge is
conserved; baryon number is conserved; muon number is conserved.
However, in the rest frame of the neutron, it has an energy
$939.573~\mathrm{MeV}$, while the proton has energy
$938.280~\mathrm{MeV}$ and the muon $105.7~\mathrm{MeV}$. Thus
unless something were to bash into the neutron and give it
sufficient energy the above process could not take place (except
perhaps virtually) because of energy Conservation.

iv. $n\rightarrow \bar p+e^{+}+\nu _e$ violates baryon number
conservation.

\vspace{0.5cm}\hrule\vspace{0.5cm}

\textbf{4. Smoke detectors.} Read about them in
\textit{Bloomfield}. The one we looked at in class uses an
$^{241}$Am source with an activity of 1 $\mu
\limfunc{Ci}=37~\mathbf{kBq}$.

(a) Find out something more about $^{241}$Am. How is it
manufactured? Is it cheap and plentiful? How does it decay,
mostly?

(b) How many decays per second is 1 $\mu \mathrm{Ci}$? How much
$^{241}$Am, in grams, gives out 1 $\mu \mathrm{Ci}$? After one
year, how much is the activity of the decay product(s), compared
to that of the remaining americium? (Is it significant?) Why is it
important that the isotope used in a smoke detector not decay into
radioactive gas such as radon?

(c) If you kept 1 $\mu \mathrm{Ci}$ of $^{241}$Am in your mouth
for an hour, would you get a bad dose of radiation? You can find
out in Melissinos what is the permissible quantity of an ingested
isotope, but here we have to worry about a local burn, rather than
a whole-body effect.

\vspace{1cm}

(a) I found my information on $^{241}$Am from the webpages at
\newline http://www.uic.com.au/nip35.htm \newline and \newline
http://www.in-search-of.com/frames/periodic/elements/95.html

One starts with $_{\,\,\,92}^{238}$U in a nuclear reactor. It
captures a neutron giving $_{\,\,\,92}^{239}$U, which in turn beta
decays (i.e gives off an electron) yielding $_{\,\,\,93}^{239}$Np,
which beta decays again leading to $_{\,\,\,94}^{239}$Pu. This
isotope of plutonium captures a
neutron ($_{\,\,\,94}^{240}$Pu) and then another producing $%
_{\,\,\,94}^{241} $Pu. Finally this last isotope of plutonium beta
decays to yield $_{\,\,\,95}^{241}$Am. According to the first
website above, a gram of Americium oxide, AmO$_2$ sells for \$1500
(but this amount is enough to make
5000 smoke detectors). $_{\,\,\,95}^{241}$Am alpha decays, which leads to $%
_{\,\,\,93}^{237}$Np, which has a half life of $2.14\times 10^6~\mathrm{years%
}$, and thus is relatively stable compared to
$_{\,\,\,95}^{241}$Am which has a half life of
$432~\mathrm{years}$.

(b) A becquerel ($1~\mathrm{Bq}$) is a decay per second $\mathrm{decay/sec}$%
. So
\[
1~{\mu \mathrm{Ci}}=3.7\times 10^4~\mathrm{Bq}=3.7\times 10^4~\mathrm{%
decays/sec}.
\]
The number of decays per unit time is proportional to the number
of particles at the particular time in question
\[
\frac{dN}{dt}=-\frac 1\tau N,
\]
where $\tau $ is the constant of proportionality and has
dimensions of time. The minus sign appears because the number of
particles is decreasing. The solution of this differential
equation is
\[
N(t)=N_0\mathrm{e}^{-t/\tau },
\]
where $N_0$ is the initial number of particles. We know the half-life $%
t_{1/2}$ (the time at which the number of particles, in this case $%
_{\,\,\,95}^{241}$Am, is half of its initial value is
$432~\mathrm{years}$, so
\[
N(t_{1/2})=\frac{N_0}2=N_0\mathrm{e}^{-t_{1/2}/\tau },
\]
or
\[
\tau =\frac{t_{1/2}}{\ln 2}=\frac{t_{1/2}}{0.693}.
\]
So $\tau =623~\mathrm{years}$. The rate of particles undergoing
decay is
\[
\frac{dN}{dt}=-\frac{N_0}\tau \mathrm{e}^{-t/\tau }.
\]
We want the decay rate (at least initially) to be $3.7\times 10^4~\mathrm{%
decays/sec}$, so
\begin{eqnarray}
3.7\times
10^4~\mathrm{decays/sec}&=&\frac{N_0}{623~\mathrm{years}}\left(
\frac{1~\mathrm{year}}{365.24~\mathrm{days}}\right) \left( \frac{1~\mathrm{%
day}}{24~\mathrm{hours}}\right) \left( \frac{1~\mathrm{hour}}{60~\mathrm{%
minutes}}\right) \left(
\frac{1~\mathrm{minute}}{60~\mathrm{seconds}}\right) \nonumber \\
&=&\frac{N_0}{1.965\times 10^{10}~\mathrm{sec}}. \nonumber
\end{eqnarray}
Therefore $N_0=7.27\times 10^{14}\,\mathrm{particles}$ which
corresponds to
\[
7.27\times 10^{14}\,\mathrm{particles}\left( \frac{1~\mathrm{mole}}{%
6.022~\times 10^{23}\,\,\mathrm{particles}}\right) \left( \frac{273~\mathrm{%
gm\,}\text{AmO}_2}{1~\mathrm{mole}}\right) =3.3\times
10^{-7}\,\,\mathrm{gm.}
\]
Recall that $1~{\mu \mathrm{Ci}}$ is the activity of $1~{\mu
\mathrm{gm}}$ of radium (and its daughters). Because Am has a
shorter half life (the half life of $^{226}$Ra is $1.6 \times 10^4
~\mathrm{years}$), it requires less of it to have the same
activity.

The activity ($dN/dt$) after one year is
\[
\frac{dN(1\,\,\mathrm{year)}}{dt}=-\frac{N_0}\tau \mathrm{\exp (}-1\,\,%
\mathrm{year}/623\,\,\mathrm{year)=-0.9984}\frac{N_0}\tau ,
\]
that is, the activity is $99.8\%$ of its initial value, hardly
changed at all.

In terms of the operation of the smoke detector, a reaction in
which the products are themselves radioactive is not a problem. Of
the decay products, it is the alpha particles that ionize the air
which carry the current in the detector. So we want a reaction or
series of reactions in which alphas are maximized and gammas are
minimized. If the further decay of the products increased the
relative alpha yield, it would be useful, and we could simply
reduce the amount of material we use. But if the further decay
increases the amount of gamma radiation, it is undesirable.
Furthermore, it is desirable that any decay products remain
chemical inert and in the solid form. Radon is particularly bad
because it is a gas.

(c) According to the first webpage above ``Americium-241 is ... a
potentially dangerous isotope if it is taken into the body in
soluble form. It decays by both alpha activity and gamma emissions
and it would concentrate in the skeleton. However, swallowing the
radioactive material from a smoke detector would not lead to
significant internal absorption of Am-241, since the dioxide is
insoluble. It will pass through the digestive tract, without
delivering a significant dose.''

\textit{Melissinos} (p. 190) gives a conversion of $\mathrm{Ci}$ to $\mathrm{%
rem/hr}$ for $^{60}$Co valid if the tissue in question is
$1~\mathrm{m}$ from the source
\[
1~\mathrm{Ci} \ \mathrm{of} \ ^{60}Co \ \mathrm{at}\ 1~\mathrm{m} = 1.3~%
\mathrm{rem/hr} .
\]
Since the source is in the mouth and we are interested to damage
to the mouth, we will use the distance of $1~\mathrm{cm}$. The
intensity varies like $1/r^2$, so the conversion will have a
factor of $10^4$ because of the closer proximity. So the
conversion for a microgram of $^{60}$Co is
\[
1~\mu \mathrm{Ci} \ \mathrm{of} \ ^{60}Co \ \mathrm{at}\
1~\mathrm{cm} = 0.013~\mathrm{rem/hr} .
\]
Now let us assume a similar conversion for $^{241}$Am; this is
probably overkill as the gamma rays in the decay of Am are low
energy. Nevertheless, exposure for an hour would yield
$0.013~\mathrm{rem}$. This is fairly low; however, there may still
be some local burning.

\vspace{0.5cm}\hrule\vspace{0.5cm}

\textbf{5. Expansion of the universe.}

(a) Does the expansion of the universe imply that the earth is
getting bigger? The solar system? Our galaxy? The distance from
our galaxy to the Virgo cluster?

(b) If Hubble's constant is $70~\mathrm{km/sec/Mpc}$, how far are
galaxies that are moving away from us at one third the speed of
light? At even greater distances, the galaxies eventually move
away from us at a speed exceeding the speed of light, if Hubble's
law is valid. Is this possible?

(c) If Hubble's constant is $70~\mathrm{km/sec/Mpc}$, how old is
the universe in the simplest model in which the ``radius''
increases in proportion to $t^{2/3}$? Justify your answer.

\vspace{1cm}

(a) The size of the earth is determined by the kind and amount of
materials it is made of and the various processes it is
undergoing. Its size is locally determined and is unaffected by
the expansion of the universe. The same is true of the solar
system and galaxy, where the size is determined by the
gravitational attraction of the constituent bodies. The distance
from our galaxy to the Virgo cluster is a another story. They are
not bound together by gravity, and thus the distance between them
grows as the universe expands. In fact, this distance was among
the original measurements of Hubble which gave credence to the
idea of an expanding universe.

(b) Hubble found that on average the galaxies were moving away
from ours with velocities proportional to the distances to them
\[
v=H_{0}R,
\]
where $H_{0}=70~\mathrm{km/sec/Mpc}$. According to this relation,
the distance to a galaxy moving away with one third the speed of
light is
\[
R=\frac{v}{H_{0}}=\frac{1\times 10^{8}~\mathrm{m/sec}}{7\times 10^{4}~%
\mathrm{m/sec/Mpc}}=1.4\times 10^{3}~\mathrm{Mpc}.
\]
(This is close to the size of the known universe.)

An object cannot move through space at a speed faster than that of
light, but two objects can move away from each other at speeds
exceeding the speed of light if the space between them is
expanding. The analogy is ants crawling on a balloon while it is
being blown up. There is a maximum speed at which the ants can
crawl with respect to the balloon, but the distance separating the
ants can grow faster. Note that while it is possible for an object
in an expanding universe to move away with a speed greater than
that of light, it is impossible for us to \textit{see} it.

(c) The first thing to note is that Hubble's constant is not
constant. It varies with time. If we note that $v=dR/dt$, we have
\[
\frac{dR}{dt}=H(t) R,
\]
or
\[
\frac{dR}{R}= H(t) dt,
\]
which has a solution
\[
R(t)=R(t_0) \mathrm{exp}\left[ \int_{t_0}^t ~H(s) ~ds \right].
\]

By using a simple ``escape velocity'' criterion, one is lead to a
another relationship between $R$ and $t$:
\[
\frac{R(t)}{R(t_{0})}=\left( \frac{t}{t_{0}}\right) ^{2/3}.
\]
See the handout on ``The Thermal History of the Universe''. These
two formulas can be reconciled if
\[
H(t)=\frac{2}{3~t}.
\]
Note that $H(t)$ has dimensions of reciprocal time.

If we know the Hubble ``constant" for the present, we can calculate from it $%
t$
\[
t = \frac{2}{3~H(t)}.
\]

Now all we have to do is transform $H$ from these strange units in
which it is given. One $\mathrm{parsec}$ is
\[
1~\mathrm{parsec}=3.26~\mathrm{light~year},
\]
so
\[
H=\frac{7 \times 10^{4}~\mathrm{m/sec}} {10^6 \cdot 3.26 \cdot 3\times 10^8~%
\mathrm{m/sec} \cdot \mathrm{yr}} = 7.16 \times
10^{-11}~\mathrm{yr}.
\]
Thus
\[
t= \frac{2}{3~H}=9.31 \times 10^{9}~\mathrm{yr}.
\]

\end{document}
