Question

An ancient club is found that contains 150 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 14C/12C atoms is about 1.00 x 10^-12 . Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02 x 10^23 .

Answer 1

We know that radioactive decay is given by:

N(t) = N0*exp(-t)

= decay constant = (ln 2)/T_half

T_half = half-life of carbon-14 = 5700 yr

= (ln 2)/5700 =1.216*10^-4 per yr

Since 1 yr = 3.156*10^7 sec, So

= 1.216*10^-4/(3.156*10^7) = 3.853*10^-12 per sec

Given that

N_14/N_12 = 1.00*10^-12

N_12 = number of atoms of 12C carbon

Given that club contains 150 gm of pure carbon (which is 12C)

N_12 = [150 gm/(12 gm/mol)]*(6.023*10^23 atmos/mol)

N_12 = 7.5287*10^24 atoms

Now

N_14 = N_12*1.00*10^-12

N_14 = 7.5287*10^24*1.00*10^-12 = 7.5287*10^12 atoms = N0

Now using above equation

N(t) = N0*exp(-*t)

dN/dt = -N0**exp(-*t)

dN/dt = -*N(t)

N(t) = |(-1/)*(dN/dt)|

Here we know that

dN/dt = 5.5 decays/sec

= 3.853*10^-12 per sec, So

N(t) = |(-1/(3.853*10^-12))*5.5|

N(t) = 1.4275*10^12 atoms

Now Using above values:

N(t) = N0*exp(-*t)

t = (1/)*ln (N0/N(t))

t = (1/(3.853*10^-12))*ln (7.5287*10^12/(1.4275*10^12))

t = 4.3156*10^11 sec

t = (4.3156*10^11 sec)*(1 yr/(3.156*10^7 sec))

t = 13675.6 yrs

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