carbon decay question?

Question

A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5730 years) for every 7.70×1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 23.6 mg of carbon. When the sample is placed inside a shielded beta counter with 82.0% counting efficiency, 821 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth's atmosphere have not changed appreciably since the sample was formed, calculate the age (in years) of the sample.

Answer 1

If we could determine exactly why Isotopes of elements exist and why they were made unstable we could understand The decay of carbon 14. Presently all decay models are probablibalistic.

Answer 2

23.6 mg of carbon = 2.36e-2 gm = (2.36e-2/12.011) moles C
= (2.36e-2/1.2011e1)* No = (2.36/1.2011) e-3 * No
= 1.965e-3 * (6.022e23) = (1.965*6.022)e20
= 1.183e21 atoms of carbon

A fraction, X, of these are Carbon-14, so the number of Carbon-14 atoms is:
1.183e*X

If the rate of decay is
dN/dt = -a*N
N(t) = N(0)exp(-at)
and so N(T) = N(0)/2 when
exp(-aT) = 1/2
aT = ln(2)
a = ln(2)/T
So a = ln(2)/(5730 years) = ln(2)/(5730*365*24*60*60)
= ln(2)/(1.807e11 seconds) = (0.6931/1.807)e-11
= 3.836(e(-12))/s

Therefore,
dN/dT = -3.836e(-12) N(t)

By measurement, dN/dt = 821/week = 821/(7*24*60*60)
= 821/604800 = 1.357e-3 ; but we increase this rate because of the non-perfect counting efficiency, by the ratio
1/.82 = 1.2196
so dN/dt = -1.2196*1.357e-3 = -1.655e-3

So, since the rate is 1.655e-3, according to the decay equation,
N(t) = -(1/a)*dN/dt = 1.655(e-3)/3.836e(-12)
= (1.655/3.836) e9 = 4.3157e8

But this is equal to:
1.183e19*X, where X is the current fraction of Carbon-14. So
X = 4.3157e8/1.183e21 = (4.3157/1.183)e(-13) = 3.648e(-13)

Now the "fresh" ratio of Carbon-14 is
1/7.7e11 = (1/7.7)e-11 = 1.299e-12

So the current ratio compared to the "fresh" ratio is:
3.648e-13/1.299e-12 = (3.648/1.299) e(-1)
= 2.8083e-1 = 0.28083

The age of the specimen should therefore be found from:
exp(-a(age)) = 0.28083
age = -ln(0.28083)/a = -ln(0.28083)/(3.836e-12) = 3.31e11 s

since 1 year = 365*24*60*60 = 3.154e7 seconds,
the age of the specimen is
3.31e11/3.154e7 = (3.31/3.154)e4 = 1.049e4
= 10,496 years.

To be honest: There could be a mistake here somewhere, because there are an awful lot of steps to this calculation. But I hope that the method is at least clear: I am much more sure of the general approach than I am of all the arithmetic in-between!