Question

How long a time t will it take for the 133 54xe to decay so that eventually its activity decreases by a factor of 1024?

Answer 1

Base on my research the radioactive isotope X-133 has a half-life of 5 days. So base on the given it decreases by a factor of 1024, it represents 10 halvings. To get how long will it take for the 133 54xe to decay, just multiply 5 days with 10 halvings. The answer is 50 days for the 133 54xe to decay. 

Answer 2

This is an incomplete question, here is a complete question.

The radioactive isotope $^{133}_{54}Xe$ 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 $^{133}_{54}Xe$ 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 :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5\text{ days}}$

$k=0.1386\text{ days}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $0.1386\text{ years}^{-1}$

t = time passed by the sample  = ?

a = initial amount of the reactant  = 1

a - x = amount left after decay process = $\frac{1}{1024}$

Now put all the given values in above equation, we get

$t=\frac{2.303}{0.1386}\log\frac{1}{(\frac{1}{1024})}$

$t=50.02\text{ days}\approx 50\text{ days}$

Therefore, the time passed in days is, 50 days