Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 x 10°years.

Answer 1

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is $1.28 * 10^9$years.

Answer:

The potassium-40 present in 80 kg is  $Z = 0.0288 *10^{-3}\ kg$

The effective dose absorbed per year is  $x = 2.06 *10^{-24}$ per year

Explanation:

From the question we are told that

      The mass of potassium in 1 kg of human body is $m = 3g= \frac{3}{1000} = 3*10^{-3} \ kg$

      The mass of the person is $M = 80 \ kg$

       The abundance of Potassium-39 is   93.26%

        The abundance of Potassium-40 is   0.012%

         The abundance of Potassium-41 is   6.78 %

         The energy absorbed is  $E = 1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J$

Now  1 kg of human body contains       $3.0*10^{-3}\ kg$ of  Potassium

So      80 kg of human body contains      k kg of  Potassium

=>   $k = \frac{ 80 * 3*10^{-3}}{1}$

     $k = 0.240\ kg$

Now from the question potassium-40 is  0.012% of the total  potassium so

     Amount of potassium-40  present is mathematically represented as

            $Z = \frac{0.012}{100} * 0.240$

            $Z = 0.0288 *10^{-3}\ kg$

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

           $D = \frac{E}{M}$

Substituting values

          $D = \frac{1.7622*10^{-13}}{80}$

            $D = 2.2*10^{-15} J/kg$

Converting to Sieverts

We have

           $D_s = REB * D$

           $D_s = 1.2 * 2.2 *10^{-15}$

           $D_s = 2.64 *10^{-15}$

So

     for half-life ($1.28 *10^9 \ years$)  the dose is  $2.64 *10^{-15}$

     Then for 1  year the dose would be  x

=>         $x = \frac{2.64 *10^{-15}}{1.28 * 10^9}$

             $x = 2.06 *10^{-24}$ per year