Question
Answer (192)
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
[tex] \frac{N}{ N_{0} } = e^{-( \frac{0.693 X T_{2} }{T_{1}}) [/tex]
3.8 / [tex]N_{0}[/tex] = [tex]e^[/tex] ((0.693 X 162.5 ) / 32.5) = 121.5
On solving we get, The initial activity as 121.5 μci
Answer : The initial activity, in microcuries, of the sample was, 120.9 μci
Explanation :
Half-life = 32.5 days
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{32.5\text{ days}}[/tex]
[tex]k=0.0213\text{ days}^{-1}[/tex]
Now we have to calculate the time passed.
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]0.0213\text{ days}^{-1}[/tex]
t = time passed by the sample = 162.5 days
a = initial amount of the reactant = ?
a - x = amount left after decay process = 3.8 μci
Now put all the given values in above equation, we get
[tex]162.5=\frac{2.303}{0.0213}\log\frac{a}{3.8\mu ci}[/tex]
[tex]a=120.9\mu ci[/tex]
Therefore, the initial activity, in microcuries, of the sample was, 120.9 μci