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\noindent {\bf PHYSICS 1020}
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{\Large \bf Homework set 15\\}
{\large 24 March 1997 \\}
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\item[{\bf [15.1]}] Which force is stronger between two protons separated by
$10^{-15}{\rm m}$ (the size of a small nucleus), the electric or the strong
force? What evidence do you have for your answer?
  \\
Answer:\\
At this distance, it is the strong force which is stronger; if it weren't, the
nucleus would not hold together -- the protons would be repelled by each other,
and the nucleus would fly apart.

\item[{\bf [15.2]}] Which force is stronger between two protons separated by
$10^{-10}{\rm m}$ (the size of an atom), the electric or the strong
force? What evidence do you have for your answer?
  \\
Answer:\\
The strong force is much stronger than the electric force, but it has a 
short range, about $10^{-15}{\rm m}$. At a distance of $10^{-10}{\rm m}$, the
strong force has a negligible effect. If the strong force were stronger than 
the
electric force at this distance, nuclei would be much bigger than they are --
there would be nuclei which are as big as atoms are now.

\item[{\bf [15.3]}] How many protons and neutrons are there in these nuclei:
$$ {\rm ^{13}_{~6}C,~~~~ ^{56}_{26}Fe,~~~~^{90}_{38}Sr,~~~~^3_1H} ~?$$         
Answer:\\
In the symbols for nuclei, the superscript (the upper number) gives the ``mass
number'' of the nucleus, i.e. the number
of nucleons (= number of protons + number of neutrons), while the subscript (the
lower number) gives the ``atomic" number of the nucleus, i.e. the number of 
protons. Thus,\\
$ {\rm ^{13}_{6}C}$ (carbon-13) has 6 protons and 13 - 6 = 7 neutrons,\\
${\rm  ^{56}_{26}Fe}$ (iron-56) has 26 protons and 56 - 26 = 30 neutrons, \\
${\rm ^{90}_{38}Sr}$ (strontium-90)  has 38 protons and 90 - 38 = 52 neutrons,\\
and ${\rm ^3_1H}$ (hydrogen-3 = tritium) has 1 proton and 3 - 2 = 2 neutrons.

\item[{\bf [15.7]}] What do you suppose heated the water in a naturally heated 
hot  spring?
  \\
Answer:\\
The radioactive decay of unstable nuclei in the Earth's interior causes warming
of rock, which in turn heats the water.


\item[{\bf [15.12]}] Use the periodic table to find the residual nucleus in each
of the following:\\
beta decay of $^3{\rm H}$, beta decay of $^{90}{\rm Sr}$, alpha decay of
$^{222}{\rm Rn}$.
  \\
Answer:\\
The decay  reactions are:
$$^3_1{\rm H \to ~  ^3_2He + e^- + \bar \nu_e }$$
$$^{90}_{38}{\rm Sr \to ~  ^{90}_{39}Y + e^- + \bar \nu_e }$$ 
$$^{222}_{~86}{\rm Rn \to  ~ ^{218}_{~84}Po + \alpha }$$
(note that $\alpha \equiv ^4_2{\rm He }$, and $\nu_e $ is an ``antineutrino''.)

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\item[{\bf [15.15]}] Radon has a 4-day half-life. Starting with 1 gram of radon,
how much will remain after 4 days? Starting with 10 radon atoms, how many will
remain after 4 days? Is this a precise prediction? Why? If you have just one
radon atom, will you still have it after 4 days?
  \\
Answer:\\
By definition, the probability of a nucleus having decayed after 
one half-life is equal to 1/2 = 50\%. Therefore,  the half-life is that 
time after 
which approximately half of a  number of  nuclei are expected to have
decayed. Due to the random nature of the decay process, there are 
deviations from the expected fraction, due to random 
fluctuations.
The smaller the number of
nuclei that are being observed,
the more important will be the relative deviations.
 The fraction of decayed nuclei will be very close to 1/2 if the number
of nuclei was very large to start with, but for small numbers of nuclei,
the deviations may be important.
Starting with 1 gram of radon, we expect half a gram to remain after 4 days
(= one half-life), and since one gram represents a very large number of radon 
nuclei ($\approx 3\times 10^{21}$),
it will be rather close to the expected amount.
Starting with ten nuclei, we expect on the average five to have decayed after 
four
days, but there is a significant probability that there will be a few more or 
less: the probability that exactly five nuclei decay is about 17\%,
for 6 it is about 14\%, for one it is about 3.4\%, and the probability that none
has decayed is still about 0.7\%.

\item[{\bf [15.18]}] Can we carbon-date ordinary rocks? Why? Can we carbon-date
objects that are a million years old? Why?
  \\
Answer:\\
Since ordinary rocks do not contain any formerly living material, they cannot be
carbon-dated. Objects that are a million years old, even if they originate from
formerly living things, cannot be carbon-dated either; the reason is that after
this long time, nearly all of the carbon-14 has decayed, and so these objects 
do not contain enough of it to be measurable.

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