Question

An ancient club is found that contains 160 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assuming that in living trees the ratio of ( 14 C 12 C ) atoms is about 1.40 × 10 − 12 . Note that the half life of carbon-14 is 5700 years and the Avogadro number is 6.02 × 10 23.

Answer in years

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 = 1.40*10^-12

N_12 = number of atoms of 12C carbon

Given that club contains 160 gm of pure carbon (which is 12C)

N_12 = [160 gm/(12 gm/mol)]*(6.023*10^23 atmos/mol)

N_12 = 8.03067*10^24 atoms

Now

N_14 = N_12*1.40*10^-12

N_14 = 8.03067*10^24*1.40*10^-12 = 1.1243*10^13 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 = 6.5 decays/sec

= 3.853*10^-12 per sec, So

N(t) = |(-1/(3.853*10^-12))*6.5|

N(t) = 1.687*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 (1.1243*10^13/(1.687*10^12))

t = 4.9229*10^11 sec

t = (4.9229*10^11 sec)*(1 yr/(3.156*10^7 sec))

t = 15598.54 yrs = 15600 yrs

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