This is an incomplete question, here is a complete question.
The radioactive isotope [tex]^{133}_{54}Xe[/tex] is used in pulmonary respiratory studies to image the blood flow and the air reaching the lungs. The half-life of this isotope is 5 days .
How long a time t will it take for the [tex]^{133}_{54}Xe[/tex] to decay so that eventually its activity decreases by a factor of 1024?
Express your answer numerically in days.
Answer : The time passed in days is, 50 days
Explanation :
Half-life = 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}{5\text{ days}}[/tex]
[tex]k=0.1386\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]0.1386\text{ years}^{-1}[/tex]
t = time passed by the sample = ?
a = initial amount of the reactant = 1
a - x = amount left after decay process = [tex]\frac{1}{1024}[/tex]
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{0.1386}\log\frac{1}{(\frac{1}{1024})}[/tex]
[tex]t=50.02\text{ days}\approx 50\text{ days}[/tex]
Therefore, the time passed in days is, 50 days
