Question

What is the age of an ancient wooden piece if it is known that the specific activity of C¹⁴ nuclide in it amounts to $\frac{3}{5}$ of that in freshly felled trees ? Given : the half-life of C¹⁴ nuclide is 5570 years

• A. 1000 years
• B. 2000 years
• C. 3000 years
• D. 4000 years

Answer 1

Answer: (D)

4000 years

Answer 2

$\frac { \mathrm { A } } { \mathrm { A } _ { 0 } } = \frac { 3 } { 5 } = \left( \frac { 1 } { 2 } \right) ^ { \frac { \mathrm { t } } { \mathrm { T } } }$

ln $\left( \frac { 3 } { 5 } \right)$ = ln $\left( \frac { 1 } { 2 } \right)$t = $\frac { \mathrm { T } ( \ln 5 - \ln 3 ) } { \ln 2 } \cong$ 4000 yrs. or, $\frac { 3 } { 5 }$

= 60% ; 50% decay takes 5570 yrs. So 40% decay takes a

little less number of years