Question

Be sure to answer all parts. A freshly isolated sample of 90Y was found to have an activity of 2.2 × 105 disintegrations per minute at 1:00 p.m. on December 3, 2006. At 2:15 p.m. on December 17, 2006, its activity was measured again and found to be 5.8 × 103 disintegrations per minute. Calculate the half-life of 90Y. Enter your answer in scientific notation.

Answer 1

Answer:

The half-life is  $t_h = 3.856*10^{3} minute$

Explanation:

From the question we are told that

     The sample is  90 Y

      The first  activity is  $A_1 = 2.2 *10^5$  per minute

       The second  activity is  $A_2 = 5.8 *10^3$  per minute

        The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006   is

               $t = 14 \ days \ 1 hr \ 15 min$

Converting to minutes we have  

               $t = (14 * 24 * 60) + (1* 60) + 15$

               $t = 20235 \ minutes$

The first order rate constant for this disintegrations can be mathematically represented  as

               $ln \frac{A_2}{A_1} = - \lambda t$

 Where $\lambda$ is the rate constant

    Substituting values

                     $ln [\frac{5.8 * 10^{3}}{2.2 *10^{5}} ] = - \lambda * 20235$

                    $-3.6358 = - \lambda * 20235$

So

                $\lambda = \frac{3.6358}{20235}$

                    $\lambda = 1.7968 *10^{-4} minute^{-1}$

The half life is mathematically represented as

               $t_{h} = \frac{0.693}{\lambda }$

So           $t_h = \frac{0.693}{1.7968 *10^{-4}}$

              $t_h = 3.856*10^{3} minute$