CGPET 2021 Solved Paper

(b) The half-life t

1

2

of C14 is 5760 years.
Initial amount of C14 is 1%
Final amount of mathrm‌C14 is 0.25%
k=

2303

t

‌log‌

a

a−x

.........(i)
where, k= rate constant, a= initial amount of reactant,
a−x= amount left/final amount of reactant leftand t= time taken.
Also, k=

0.6932

t 1 2

.........(ii)
From (i) and (ii),

0.6932

t 1 2

=

2.303

t

‌log‌

a

a−x

0.6932

5760

=

2303

t

‌log‌

1

0.25

⇒ t=

2303

0.6932 5760

‌log(4)=11520yr