What is the age in years of a bone in which the C14C12ratio is measured to be1.65×10-13?

Short Answer

Expert verified

The age of bone is17,058,783Years

Step by step solution

01

Given Calculations:

Ratio of C14to C12 in living organisms is,

localid="1649227741643" N0C14N0C12=1.3×10-12

Ratio of C14to C12 in bone is,

localid="1649224990460" NC14NC12=1.65×10-13

As, localid="1649226770511" N0C12=N0C12

localid="1649227753217" NC14N0C14=1.65×10-131.3×10-12

localid="1649226377005" =1.27×10-1

=0.127

Therefore, N(14C)=0.127N0(14C)

02

Given Calculations:

Decay equation in terms of half-life is,

N=N012tt1

12tt1=NN0

Here, Nis the present number of C14atoms, N0represent the initial number of atoms of C14atoms and localid="1649226999766" t12 represents the half-life ofC14 .
Substitute 5730 years for t12and 0.127for NN0,

localid="1649227250557" 12t5730Years=0.127

localid="1649227227925" 0.5t5730Years=0.127

Take natural log on both sides,

localid="1649227331822" t5730Yearsln0.5=ln0.127

t=ln0.127ln0.55730Years

t=(2.9771)(5730years)

t=17,058,783

Hence, the age of bone is localid="1649227458973" 17,058,783Years.

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