1. Radiocarbon dating Radiocarbon dating (or simply carbon
dating) is a radiometric dating technique that uses the decay of
carbon-14 ( 14 C 6 ) to estimate the age of organic materials, such
as wood and leather, up to about 58,000 to 62,000 years Before
Present (BP, present defined). radiometric dating carbon-14organic
materialsBefore Present Before Present (BP) years is a time scale
used mainly in geology and other scientific disciplines to specify
when events in the past occurred. Because the "present" time
changes, standard practice is to use time
scalegeologyscientific
Slide 3
Radiocarbon dating (Continued) 1st January 1950 as commencement
date of the age scale, reflecting the fact that radiocarbon dating
became practicable in the 1950s.radiocarbon dating Carbon dating
was presented to the world by Willard Libby in 1949, for which he
was awarded the Nobel Prize in Chemistry.Willard LibbyNobel Prize
in Chemistry Since the introduction of carbon dating, the method
has been used to date many items, including samples of the Dead Sea
Scrolls, the Shroud of Turin, enough Egyptian artefacts to supply a
chronology of Dynastic Egypt, and tzi the IcemanDead Sea
ScrollsShroud of Turin EgyptianchronologyDynastic Egypttzi the
Iceman
Slide 4
Radiocarbon dating (Continued) The Earth's atmosphere contains
various isotopes of carbon, roughly in constant proportions. These
include the main stable isotope ( 12 C 6 ) and an unstable isotope
( 14 C 6 ). Through photosynthesis, plants absorb both forms from
carbon dioxide in the atmosphere. When an organism dies, it
contains the standard ratio of 14 C 6 to 12 C 6, but as the 14 C 6
decays with no possibility of replenishment, the proportion of
carbon 14 C 6 decreases at a known constant rate. The time taken
for it to reduce by half is known as the half-life of 14 C
6.isotopes carbon dioxidehalf-life
Slide 5
Radiocarbon dating (Continued) The measurement of the remaining
proportion of 14 C 6 in organic matter thus gives an estimate of
its age (a raw radiocarbon age). However, over time there are small
fluctuations in the ratio of 14 C 6 to 12 C 6 in the atmosphere,
fluctuations that have been noted in natural records of the past,
such as sequences of tree rings and cave deposits. These records
allow fine- tuning, or "calibration", of the raw radiocarbon age,
to give a more accurate estimate of the calendar date of the
material. One of the most frequent uses of radiocarbon dating is to
estimate the age of organic remains from archaeological
sites.sequences of tree ringscave depositscalibration
Slide 6
Radiocarbon dating (Continued) Calculating ages : While a plant
or animal is alive, it is exchanging carbon with its surroundings,
so that the carbon it contains will have the same proportion of 14
C 6 as the biosphere ( 12 C 6 ). Once it dies, it ceases to acquire
14 C 6, but the 14 C 6 that it contains will continue to decay, and
so the proportion of radiocarbon in its remains will gradually
reduce. Because 14 C 6 decays at a known rate, the proportion of
radiocarbon can be used to determine how long it has been since a
given sample stopped exchanging carbonthe older the sample, the
less 14 C 6 will be left. The equation governing the decay of a
radioactive isotope is
Slide 7
Radiocarbon dating (Continued) where N 0 is the number of atoms
of the isotope in the original sample (at time t = 0), and N is the
number of atoms left after time t. The mean-life, i.e. the average
or expected time a given atom will survive before undergoing
radioactive decay denoted by , of 14 C 6 is 8,267 years, so the
equation above can be rewritten as
Slide 8
Radiocarbon dating (Continued) The ratio of 14 C 6 atoms in the
original sample, N 0, is taken to be the same as the ratio in the
biosphere 12 C 6, so measuring N, the number of 14 C 6 atoms
currently in the sample, allows the calculation of t, the age of
the sample.
Slide 9
Radiocarbon dating (Continued) The half-life of a radioactive
isotope (the time it takes for half of the sample to decay, usually
denoted by T 1/2 ) is a more familiar concept than the mean-life,
so although the equations above are expressed in terms of the
mean-life, it is more usual to quote the value of 14 C 6 half-life
than its mean-life. The currently accepted value for the half-life
of radiocarbon is 5,730 years. The mean-life and half-life are
related by the following equationhalf-life
Slide 10
Radiocarbon dating (Continued) For over a decade after Libby's
initial work, the accepted value of the half-life for 14 C 6 was
5,568 years; this was improved in the early 1960s to 5,730 years,
which meant that many calculated dates in published papers were now
incorrect (the error is about 3%). However, it is possible to
incorporate a correction for the half-life value into the
calibration curve, and so it has become standard practice to quote
measured radiocarbon dates in "radiocarbon years", meaning that the
dates are calculated using Libby's half-life value and have not
been calibrated.
Slide 11
Examples (Radiocarbon dating) (Continued) A fossil bone is
found to contain 1/1000 the original amount of 6C14 Determine the
age of the fossil. We have the equation governing the given
phenomenon as under, which gives the amount of y(t) = y0. where y0
is the initial amount of the radioactive substance. For t = 5730
years ( the half age of 6C14 ), this equation gives, y(t) = y0
/2.
Slide 12
Radiocarbon dating (Continued) Thus from the above equation we
can determine the value of the constant k as under, y0/2 = y0. or -
5730 k = In (1/2). Thus we get, k= In(1/2)/(-5730) 0.000120
Therefore the above equation becomes, y(t) = y0. Thus, when y(t) =
y0 / 1000, as given, we get, y0/1000 = y0.
Slide 13
Radiocarbon dating (Continued) On taking log-natural of both
sides this equation gives, -0.00012097 t = In (1/1000) = - In (
1000 ). So that, t = In ( 1000 ) /0.00012097 = 57136 years. Note :
In the above method radioactive Carbon dating much of the accuracy
of result depends upon the chemical analysis of the fossil. In
order to obtain better estimations of 6C14 present in the fossil,
destruction of large samples of the specimen are required. The same
may not always be possible from archival point of view. In view of
the same the age estimation of the fossil in the above example is
not very accurate.
Slide 14
Radiocarbon dating (Continued) In the recent years geologists
have shown that age estimation of the fossils by the above
mentioned method may be out by as much as 3500 years in certain
cases. One of the possible reasons for this error is the fact that
6C14 levels in the air very with time. They have devised another
method for the purpose, based on the fact that the living organisms
ingest traces of Uranium. By measuring the relative amounts of
Uranium and Thorium (the isotope into which Uranium decays ), and
by knowing
Slide 15
Radiocarbon dating (Continued) the half-lives of these
elements, one can determine the age of the fossil. By this method
one can estimate the ages of even 5000,000 old fossils. However,
this method is not applicable to marine fossils. Some other
techniques can also be used for the purpose including the use of
Potassium 40 and Argon 40 (which can estimate ages up to millions
of years), and non-isotopic, methods based on the use of amino
acids.
Slide 16
2. Newton's Law of Cooling Newton's Law of Cooling states that
the rate of change of the temperature of an object is proportional
to the difference between its own temperature and the ambient
temperature (i.e. the temperature of its surroundings). Newton's
Law makes a statement about an instantaneous rate of change of the
temperature. We will see that when we translate this verbal
statement into a differential equation, we arrive at a differential
equation. The solution to this equation will then be a function
that tracks the complete record of the temperature over time.
Slide 17
Newton's Law of Cooling (Continued) Crime Scene A detective is
called to the scene of a crime where a dead body has just been
found. She arrives on the scene at 6:00 pm and begins her
investigation. Immediately, the temperature of the body is taken
and is found to be 85 o F. The detective checks the programmable
thermostat and finds that the room has been kept at a constant 72 o
F for the past 3 days.
Slide 18
Newton's Law of Cooling (Continued) After evidence from the
crime scene is collected, the temperature of the body is taken once
more and found to be 78 o F. This last temperature reading was
taken exactly three hour after the first one (i.e., at 9:00 pm).
The next day the detective is asked by another investigator, What
time did our victim die? Assuming that the victims body temperature
was normal (98.6 o F) prior to death, what is her answer to this
question? Newton's Law of Cooling can be used to determine a
victim's time of death.
Slide 19
Newton's Law of Cooling (Continued) The governing equation for
the temperature, T of the body is where T = temperature of the
body, To = ambient temperature t = time in hrs k = constant based
on thermal properties of body and air
Slide 20
Newton's Law of Cooling (Continued) The characteristic equation
of the above differential equation is
Slide 21
Newton's Law of Cooling (Continued) Let the particular solution
is Substituting this particular solution into the ordinary
differential equation The complete solution is
Slide 22
Newton's Law of Cooling (Continued) Given is where B = time of
death, We get
Slide 23
Use equations. (1) and (2) to find A and K, we get Substituting
the values of A and K into equation (3), to find B The time of
death is 3.221 hrs, that is 0.3221*60 = 13.326 minutes after 3 pm.
Time of death = 3:13 pm Newton's Law of Cooling (Continued)
Slide 24
3. Exponential Population Growth Bacterial growth Suppose the
population of bacteria doubles every 3 hours. What exactly does
that mean? Imagine you inoculate a fresh culture with N bacteria at
12:00 pm. At 3 pm, you will have 2N bacteria, at 6 pm you will have
4N bacteria, at 9 pm you will have 8N bacteria, and so on. If these
cell divisions occur at EXACTLY each of these time points the cells
are said to be growing synchronously. If this were the case, the
growth process would be geometric.
Slide 25
Exponential Population Growth(Continued) A geometric growth
model predicts that the population increases at discrete time
points (in this example hours 3, 6, and 9). In other words, there
is not a continuous increase in the population.
Slide 26
Exponential Population Growth(Continued)
Slide 27
However, this is not what actually happens. Imagine you take a
small sample of the culture every hour and count the number of
bacteria cells present. If bacterial growth were geometric, you
would expect to have N bacteria between 12 pm and 3pm, 2N bacteria
between 3 pm and 6 pm, etc. However, if you perform this experiment
in the laboratory, even under the best experimental conditions,
this will not be the case.
Slide 28
Exponential Population Growth(Continued) If you go a step
further and make a graph with the number of bacteria on the y-axis
and time on the x-axis, you will get a plot that looks much more
like exponential growth than geometric growth.
Slide 29
Exponential Population Growth(Continued) Why does bacterial
growth look like exponential growth in practice? The answer is
because bacterial growth is not completely synchronized. Some cells
divide in fewer than 3 hours; while others will take a little
longer to divide. Even if you start a culture with a single cell,
synchronicity will be maintained only through a few cell divisions.
A single cell will divide at a discrete point in time, and the
resulting 2 cells will divide at ABOUT the same time, and the
resulting 4 will again divide at ABOUT the same time.
Slide 30
Exponential Population Growth(Continued) As the population
grows, the individual nature of cells will result in a smoothing of
the division process. This smoothing yields an exponential growth
curve, and allows us to use exponential functions to make
calculations that predict bacterial growth. So, while exponential
growth might not be the perfect model of bacterial growth by binary
fission, it is the appropriate model to use given experimental
reality.
Slide 31
Exponential Population Growth(Continued) Problem -Calculate the
number of bacteria in a culture at a given time How many bacteria
are present after 51 hours if a culture is inoculated with 1
bacterium? Use the model, N(t) = N o e kt, and assume the
population doubles every 3 hours. (N(t) is the population size at
time t and k is a constant.)
Slide 32
4. Radioactive Decay In nature, there are a large number of
atomic nuclei that can spontaneously emit elementary particles or
nuclear fragments. Such a phenomenon is called radioactive decay.
This effect was studied at the turn of 19-20 centuries by Antoine
Becquerel, Marie and Pierre Curie, Frederick Soddy, Ernest
Rutherford, and other scientists. As a result of the experiments,
F.Soddy and E.Rutherford derived the radioactive decay law, which
is given by the differential equation
Slide 33
Radioactive Decay(Continued) where N is the amount of a
radioactive material, is a positive constant depending on the
radioactive substance. The minus sign in the right side means that
the amount of the radioactive material N(t) decreases over time.
The given equation is easy to solve, and the solution has the
form
Slide 34
Radioactive Decay(Continued) To determine the constant C, it is
necessary to indicate an initial value. If the amount of the
material at the moment t = 0 was N 0, then the radioactive decay
law is written as The half life or half life period T of a
radioactive material is the time required to decay to one-half of
the initial value of the material. Hence, at the moment T:
Slide 35
Example ( Radioactive Decay ) (continued) In a certain
radioactive substance the rate of decrease in mass is proportional
to the current mass. If the mass present is reduced by half, in
half, an hour, what percent of the original mass is expected to
remain present at the end of 0.9 hours?
Slide 36
Radioactive Decay (continued) Let the mass at time is y(t).
Then the given problem can be represented by dy/dt=ky, where k is
constant of proportionality. On solving at we get, y= c e-kt where
c is the constant of integration. If y0 is the mass at t =0, the
above equation becomes, y0 = c e0 = 12 y0 = c e-k/2 = y0 e-k/2
Thus, 12 = e-k/2,
Slide 37
Taking log-natural of both sides we get 12 k = In(1/2)= In (2),
or k = In 4. Therefore, the desired solution is, y = y0 e-(In4)t
The above solution can be put as : y = y0 / et(In4) = y0/4t.
Therefore, when t = 0.9 (given ), then the above equation gives: y
= y0 / 40.9 = 0.2871745, y0 = 29% of y0 ( approx) Thus at the end
of 0.9 hours about 29 percent of the original quantity of the given
radioactive substance would be left. Radioactive Decay
(continued)
Slide 38
5. Compound Interest Compound Interest: Non-Continuous P =
principal amount invested m = the number of times per year interest
is compounded r = the interest rate t = the number of years
interest is being compounded A = the compound amount, the balance
after t years
Slide 39
Compound Interest(Continued) Notice that as m increases, so
does A. Therefore, the maximum amount of interest can be acquired
when m is being compounded all the time - continuously.
Slide 40
Compound Interest(Continued) Compound Interest: Continuous P =
principal amount invested r = the interest rate t = the number of
years interest is being compounded A = the compound amount, the
balance after t years Compound Interest: Continuous
Slide 41
Compound Interest(Continued) EXAMPLE EXAMPLE (Continuous
Compound) Ten thousand dollars is invested at 6.5% interest
compounded continuously. When will the investment be worth $41,787?
SOLUTION We must first determine the formula for A(t). Since
interest is being compounded continuously, the basic formula to be
used is Since the interest rate is 6.5%, r = 0.065. Since ten
thousand dollars is being invested, P = 10,000. And since the
investment is to grow to become $41,787, A = 41,787. We will make
the appropriate substitutions and then solve for t. This is the
formula to use. P = 10,000, r = 0.065, and A = 41,787.
Slide 42
Compound Interest(Continued) Therefore, the $10,000 investment
will grow to $41,787, via 6.5% interest compounded continuously, in
22 years.