Question

The half-life of $^{14}C$ is 5570 years. How many years will it take 90 % of a sample to decompose?
(A) 5,570 years
(B) 17,700 years
(C) 18,510 years
(D) 50,100 years

Answer 1

Hint: Half life period of a reaction can be defined as the time in which the concentration of a reactant gets reduced to the half of the initial concentration. It is the time in which half of the reaction takes place.
$^{14}C$ is a radioactive element and the radioactive decay reactions are first order reactions. Further, the half life of first order reaction is not dependent on the initial concentration of reactant.

Formula used: Half life for first order reaction ${t_{1/2}} = \dfrac{{0.693}}{k}$
Where k = decay constant
a = Initial amount of element
a – x = amount of element after t time
Time (t) = $\dfrac{{2.303}}{k}\log \left( {\dfrac{a}{{a - x}}} \right)$

Complete step by step answer:
Given Half life for of $^{14}C({t_{1/2}}) = 5570\,years$
When sample is 90 % decomposed,
Suppose initial amount of $^{14}C$ = 100
After t time amount of $^{14}C$ = 100 - 90 = 10
Here, $k = \dfrac{{0.693}}{{{t_{1/2}}}}$
$k = \dfrac{{0.693}}{{5570}}y{r^{ - 1}}$
It is known that $t = \dfrac{{2.303}}{k}\log \dfrac{a}{{(a - x)}}$
On substituting the value of k, a and a – x
$t = \dfrac{{2.303 \times 5570}}{{0.693}}\log \dfrac{{100}}{{10}}$
$ = \dfrac{{2.303 \times 5570}}{{0.693}}\log 10$..................(because log 10 = 1)
t = 18510 years

Hence, the correct answer is (C) 18510 years.

Note: Sometime, students get confused in a, a – x and x. So
a = Initial amount of element
a – x = Amount of element at time t
x = Decomposed amount after t time
The above discussed terms denote the certain amount of reactant concentration at different time-periods. One should carefully distinguish the different concentrations occurring at different times.