Question

Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years. If a sample containing carbon-14 now has 71% of its original concentration of carbon-14, how much time has passed in years? 4.09 ~ 103 years 5.73 x 103 years 2.38 x 103 years 2.83 x 103 years 3.52 * 104 years

Answer 1

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

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  = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$