Question

A sample of charred animal bones found in a cave has a carbon-14 decay rate of 5.2 disintegrations per minute per gram of carbon (5.2 dis/min-g C). Living organisms have a decay rate of 15.3 dis/min-g C. The half-life of carbon-14 is 5715 yr. How old is the bone sample? (All radioactive elements decay according to first order kinetics.)

a)2.8 x 10 3  yrs

b)3.9 x 10 3  yrs

c)1.9 x 10 3  yrs

d)8.9 x 10 3  yrs

Answer 1

SOLUTION :

Radioactive carbon - 14  has half-life of 5715 years.

Decimal fraction of C-14 remaining = 5.2 / 15.3 = 0.3399 

So,

A / A0 = e^(- lamda * t)

=> 1/2 = e^(- lamda * 5715)

Taking ln :

=> ln(1/2) = - lamda * 5715

=> lamda = - ln(1/2) / 5715 = 1.21286 * 10^(-4) 

So,

A(now) / A0 = e^(- 1.21286*10^(-4)  * t)

=> 0.33399 =  e^(- 1.21286*10^(-4)  * t)

Takin ln :

=> ln(0.3399) = - 1.21286*10^(-4) * t

=> t = ln(0.3399) / (- 1.21286*10^(-4)

=> t =  8897.18  = 8.9 * 10^3 years approx.

Hence, the bones are 8.9 * 10^3  years (approx.) old :  OPTION d (ANSWER)