Question

An ancient club is found that contains 180 g of pure carbon and has an activity of 5.5 decays per second. Determine its age assuming that in living trees the ratio of (14C12C)(14C12C) atoms is about 1.40×10−121.40×10−12. Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02×10236.02×1023.

Answer 1

Radioactive decay formula can be written as, dN/dt = -N ...(1)
Where N is the number of C-14 atoms in the present state,
= 0.693/T1/2
Wher T1/2 the half life of C-14 atoms
T1/2 = 5700 x 365 x 24 x 60 x 60
= 1.798 x 10¹¹ s
= 0.693/(1.798 x 10¹¹)

N = (dN/dt) / [From (1)]
N = (5.5) / (3.86 x 10^-12)
= 1.43 x 10¹²

Number of C atoms in the beginning = number of C atoms in the present state.
Mass of C atoms = 180 g
12 g C contains Avagadro number of atoms
1 g contains Avagadro number / 12
180 g contains, n = 180 x (6.02 x 10²³) / 12
= 9.03 x 10²⁴

Consider that initially there are No number of C-14 atoms.
Given that No/n = 1.40 x 10^-12
No = n x (1.40 x 10^-12)
= (9.03 x 10²⁴) x (1.40 x 10^-12)
= 12.642 x 10¹²

N = No exp(-​​​​​​t)
exp(-t) = N/No
= (1.43 x 10¹²) / (12.642 x 10¹²)
= 0.1128
Taking ln on both sides,
-t = ln(0.1128)
= - 2.182

t = 2.182/
= 0.693/5700
= 0.000121605 year^-1
t = 2.182/ (0.000121605)
= 17942.71 years