As long as an organism is alive, it has one carbon-14 (14C) atom for every 7.2 x
ID: 992462 • Letter: A
Question
As long as an organism is alive, it has one carbon-14 (14C) atom for every 7.2 x 1011 carbon-12 (12C) atoms in its body (N14N12=17.2 × 1011). Once it is dead, the carbon-14 is no longer replaced by absorbing more from the environment, so the number of carbon-14 atoms declines. Carbon-12 is stable, so the number of carbon-12 atoms remains the same over time.
a) Suppose a 0.35kg sample of bone has an activity of 0.0015 Ci. How many carbon-14 atoms are there in the bone now?
b) If the bone is pure carbon, how many carbon-12 atoms are there in the bone now?
c) Given that number of carbon-12 atoms (which has never changed), how many carbon-14 atoms were there when the organism was alive?
d) How long ago did the organism die?
Explanation / Answer
(a). Half life of carbon -14 is T = 5730 years
decay constant = ln 2/ T
= 0.693 /5730 yrs
= 1.2096 x10 -4 yr -1
= 1.2096 x10 -4 /(365x24x3600 s)
= 3.8356 x10 -12 s -1
Activity A = 0.0015 x10 -6 Ci
=0.0015 x10 -6 x3.7 x10 10 Bq
= 55.5 Bq
We know A = N
From this N = A/
= 55.5 /( 3.8356 x10 -12 )
= 1.446 x10 13 atoms
(b).Mass of bone m = 0.35 kg = 350 g
Molar mass of carbon 12 is M = 12 g
Number of moles n = m/M
= 350 / 12
= 29.166 mol
We know 1 mole contains 6.023 x10 23 atoms
So, Number of carbon-12 atoms are there in the bone now = n(6.023 x10 23)
= 29.166 (6.023 x10 23)
= 1.756 x10 25 atoms
(c).Number of carbon-12 atoms when the organism was alive is = 1.756 x10 25 atoms
Number of carbon-14 atoms when the organism was alive = (1.756 x10 25 )/(17.2x10 11)
No= 1.020 x10 13 atoms
(d). we know N = No e -t
1.446 x10 13 = 1.020 * 1013 e -(1.2096 x10^-4 yr ^-1 ) t
1.417 = e -(1.2096 x10^-4 yr ^-1 ) t
-1.2096x10 -4 t = ln(1.417)
= 0.3485
t = 0.3485 / (1.2096x10 -4)
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