Question

In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t and k is a negative constant determined by the nature of the material. A certain radioactive isotope decays at a rate of 0.1% annually. Determine the half-life of this isotope, to the nearest year.

Answer 1

Answer: 693 years

Explanation:

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

`A(t) =A_0e^(kt)`

k = rate constant

t = time taken for decomposition = 1

`A_0` = Initial amount of the reactant

`A_t` = amount of the reactant left =
`A_0-(0.1)/(100)* A_0=0.999A_0`

`0.999A_0=A_0e^(k* 1)`

`0.999=e^k`

`k=-0.001year^(-1)`

for half life :
`t=t_(1)/(2)`

`A_t=(1)/(2)A_o`

Putting in the values , we get

`(1)/(2)A_0=A_0e^{-0.001* t_(1)/(2)`

`t_(1)/(2)=693 years`

Thus half life of this isotope, to the nearest year is 693.