Question

Given that carbon c14 decays at a constant rate in such a way that it reduces to 50% in 5568 years. find the age of an old wooden piece in which the carbon is only 12.5% of the original

Answer 1

The solution is attached

Thanks

Answer 2

Answer: 16704 years

Explanation: Radioactive decay follows first order kinetics.

Half-life of C-14= Time taken to reduce the original amount to 50%= 5568 years

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5568}= 1.24\times 10^{-4}year^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process

a = initial amount of the reactant = 100 g

a - x = amount left after decay process = 12.5 g

Putting values in above equation, we get:

$t=\frac{2.303}{1.24\times 10^{-4}year^{-1}}\log\frac{100g}{12.5g}$

$t=16704years$