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* 10^3\text{ years}`

Explanation :

Half-life = 5730 years

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

`k=(0.693)/(t_(1/2))`

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

`k=1.21* 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=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`1.21* 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\% * (x)=(71)/(100)* (X)=0.71Xg`

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

`t=(2.303)/(1.21* 10^(-4))\log(X)/(0.71X)`

`t=2831.00\text{ years}=2.83* 10^3\text{ years}`

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