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

Step-by-step 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 =$A_0-\frac{0.1}{100}\times A_0=0.999A_0$

$0.999A_0=A_0e^{k\times 1}$

$0.999=e^k$

$k=-0.001year^{-1}$

for half life : $t=t_\frac{1}{2}$

$A_t=\frac{1}{2}A_o$

Putting in the values , we get

$\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}$

$t_\frac{1}{2}=693 years$

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