Answer:
The half-life is ![t_h = 3.856*10^{3} minute](https://tex.z-dn.net/?f=t_h%20%3D%203.856%2A10%5E%7B3%7D%20minute)
Explanation:
From the question we are told that
The sample is 90 Y
The first activity is
per minute
The second activity is
per minute
The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006 is
![t = 14 \ days \ 1 hr \ 15 min](https://tex.z-dn.net/?f=t%20%3D%2014%20%5C%20days%20%5C%201%20hr%20%5C%2015%20min)
Converting to minutes we have
![t = (14 * 24 * 60) + (1* 60) + 15](https://tex.z-dn.net/?f=t%20%3D%20%20%2814%20%2A%2024%20%2A%2060%29%20%2B%20%20%281%2A%2060%29%20%2B%2015)
![t = 20235 \ minutes](https://tex.z-dn.net/?f=t%20%3D%2020235%20%5C%20minutes)
The first order rate constant for this disintegrations can be mathematically represented as
![ln \frac{A_2}{A_1} = - \lambda t](https://tex.z-dn.net/?f=ln%20%5Cfrac%7BA_2%7D%7BA_1%7D%20%20%3D%20%20-%20%5Clambda%20t)
Where
is the rate constant
Substituting values
![-3.6358 = - \lambda * 20235](https://tex.z-dn.net/?f=-3.6358%20%3D%20-%20%20%5Clambda%20%2A%2020235)
So
![\lambda = \frac{3.6358}{20235}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%20%5Cfrac%7B3.6358%7D%7B20235%7D)
![\lambda = 1.7968 *10^{-4} minute^{-1}](https://tex.z-dn.net/?f=%5Clambda%20%3D%201.7968%20%2A10%5E%7B-4%7D%20minute%5E%7B-1%7D)
The half life is mathematically represented as
![t_{h} = \frac{0.693}{\lambda }](https://tex.z-dn.net/?f=t_%7Bh%7D%20%3D%20%5Cfrac%7B0.693%7D%7B%5Clambda%20%7D)
So ![t_h = \frac{0.693}{1.7968 *10^{-4}}](https://tex.z-dn.net/?f=t_h%20%3D%20%5Cfrac%7B0.693%7D%7B1.7968%20%2A10%5E%7B-4%7D%7D)
![t_h = 3.856*10^{3} minute](https://tex.z-dn.net/?f=t_h%20%3D%203.856%2A10%5E%7B3%7D%20minute)