Answer :
Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%
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 =[tex]0.1s^{-1}[/tex]
t = age of sample
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process
a) for tracer concentration to drop by 50%
a - x = amount left after decay process = 50
[tex]t=\frac{2.303}{k}\log\frac{100}{50}[/tex]
[tex]t=\frac{0.693}{0.1}[/tex]
[tex]t=6.93sec[/tex]
It will take 6.93 sec for tracer concentration to drop by 50%
b) for tracer concentration to drop by 75 %
a - x = amount left after decay process = 25
[tex]t=\frac{2.303}{k}\log\frac{100}{100-75}[/tex]
[tex]t=\frac{2.303}{0.1}\times \log\frac{100}{25}[/tex]
[tex]t=13.9sec[/tex]
It will take 13.9 sec for tracer concentration to drop by 75%.