Answer: [tex]6.54\times 10^5years[/tex]
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{2.44\times 10^5}=0.284\times 10^{-5}yr^{-1}[/tex]
b) for [tex]6.40\times 10^2[/tex] g of the isotope to decay to [tex]1.00\times 10^2[/tex]
[tex]t=\frac{2.303}{0.284\times 10^{-5}}\log\frac{6.40\times 10^2}{1.00\times 10^2}[/tex]
[tex]t=6.54\times 10^5years[/tex]
The time for [tex]6.40\times 10^2[/tex] g of the isotope to decay to [tex]1.00\times 10^2[/tex] is [tex]6.54\times 10^5years[/tex]
