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 [tex]1.28 * 10^9[/tex]years.
Answer:
The potassium-40 present in 80 kg is [tex]Z = 0.0288 *10^{-3}\ kg[/tex]
The effective dose absorbed per year is [tex]x = 2.06 *10^{-24}[/tex] per year
Explanation:
From the question we are told that
The mass of potassium in 1 kg of human body is [tex]m = 3g= \frac{3}{1000} = 3*10^{-3} \ kg[/tex]
The mass of the person is [tex]M = 80 \ kg[/tex]
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 [tex]E = 1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J[/tex]
Now 1 kg of human body contains [tex]3.0*10^{-3}\ kg[/tex] of Potassium
So 80 kg of human body contains k kg of Potassium
=> [tex]k = \frac{ 80 * 3*10^{-3}}{1}[/tex]
[tex]k = 0.240\ kg[/tex]
Now from the question potassium-40 is 0.012% of the total potassium so
Amount of potassium-40 present is mathematically represented as
[tex]Z = \frac{0.012}{100} * 0.240[/tex]
[tex]Z = 0.0288 *10^{-3}\ kg[/tex]
The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as
[tex]D = \frac{E}{M}[/tex]
Substituting values
[tex]D = \frac{1.7622*10^{-13}}{80}[/tex]
[tex]D = 2.2*10^{-15} J/kg[/tex]
Converting to Sieverts
We have
[tex]D_s = REB * D[/tex]
[tex]D_s = 1.2 * 2.2 *10^{-15}[/tex]
[tex]D_s = 2.64 *10^{-15}[/tex]
So
for half-life ([tex]1.28 *10^9 \ years[/tex]) the dose is [tex]2.64 *10^{-15}[/tex]
Then for 1 year the dose would be x
=> [tex]x = \frac{2.64 *10^{-15}}{1.28 * 10^9}[/tex]
[tex]x = 2.06 *10^{-24}[/tex] per year
