Question

The $ _{19}{K^{40}} $ consist of $ 0.012\% $ potassium in nature. The human body contains $ 0.35\% $ potassium by weight. Calculate the total radioactivity resulting from $ _{19}{K^{40}} $ decay in a $ 75{\text{ }}kg $ human body. Half life of $ _{19}{K^{40}} $ is $ 1.3{\text{ }} \times {\text{ }}{10^9}years. $

Answer 1

Hint: Radioactive decay reactions follow first order kinetics.
The rate constant for first order reactions is calculated by following formula:- $ K = \dfrac{{0.693}}{{{t^{\dfrac{1}{2}}}}} $
Where $ K $ is the rate constant
 $ {t^{\dfrac{1}{2}}} $ is the half life period of the substance.

Complete Step By Step Answer:
Given the half life of $ _{19}{K^{40}} = 1.3 \times {10^9}years $
Also $ K = \dfrac{{0.693}}{{{t^{\dfrac{1}{2}}}}} $
Therefore rate constant $ K = \dfrac{{0.693}}{{1.3 \times {{10}^9}}} $ $ = 0.53 \times {10^9}yr{s^{ - 1}} $
 Now we have to calculate the amount of potassium present in a human body weighing $ 75{\text{ }}kgs $
Amount of K in nature is $ 0.35\% $
Amount of K in human body by weight is $ 0.012\% $
So, the amount of K in $ 75{\text{ }}kg $ or $ 75 \times {10^3}{\text{ }}g $ human body will be:
 $ K = \dfrac{{0.35}}{{100}} \times \dfrac{{0.012}}{{100}} \times 75 \times {10^3} = 0.0315g $
To calculate total radioactivity, we will use the formula of rate decay i.e. $ R = \lambda \times N $
Here, R is total radioactivity, $ \lambda $ is the decay constant and N is the no. of molecules
To calculate R we first need to calculate N
We calculate N by using formula:
 $ No.\;of\;molecules = \;\dfrac{{given\;mass}}{{molar\;mass\;}} \times avogadro's\;no $
Given mass $ = {\text{ }}0.0315{\text{ }}g $ (it is amount of potassium present in human body weighing $ 75{\text{ }}kg $ )
Molar mass of potassium $ = {\text{ }}40 $
Avogadro’s no. $ = 6.022 \times {10^{23}} $
Therefore no. of molecules $ = \dfrac{{0.0315}}{{40}} \times 6.022 \times {10^{23}} $ $ = 0.47 \times {10^{21}}molecules $
Total radioactivity R = $ \lambda \times N $
 $ \lambda $ is the rate constant already calculated above
Therefore $ \lambda = 5.3 \times {10^{ - 8}}yrs $
 $
  R = \lambda \times N = 5.3 \times {10^{ - 8}} \times 0.47 \times {10^{21}}yrs \\
    \\
  $
 $ R = 2.49 \times {10^{13}}decay{\text{ }}per{\text{ year}} $

Note:
Radioactive reactions always proceed via a first order mechanism therefore the first order rate constant is equal to the decay constant.
Units should always be written along with the answer as the same question may be asked in exam but the answer may be in different units, but if we write the units like here we may easily convert the year into hours, minutes or even seconds.