Answer:
1
The mass of the Potassium-40 is [tex]m_{40}} = 2.88*10^{-6} kg[/tex]
2
The Dose per year in Sieverts is [tex]Dose_s = 26.4 *10^{-10}[/tex]
Explanation:
From the question we are told that
The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41
Their abundance is 93.26%, 0.012% and 6.728%
The mass of potassium contained in human body is [tex]m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg[/tex] per kg of the body
The mass of the first body is [tex]m_1 = 80 \ kg[/tex]
Now the mass of potassium in this body is mathematically evaluated as
[tex]m_p = m * m_1[/tex]
substituting value
[tex]m_p = 80 * 0.0003[/tex]
[tex]m_p =0.024 kg[/tex]
The amount of Potassium-40 present is mathematically evaluated as
[tex]m_{40}} =[/tex]0.012% * 0.024
[tex]m_{40}} = \frac{0.012}{100} * 0.024[/tex]
[tex]m_{40}} = 2.88*10^{-6} kg[/tex]
The dose of energy absorbed per year is mathematically represented as
[tex]Dose = \frac{E}{m_1}[/tex]
Where E is the energy absorbed which is given as [tex]E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}[/tex]
Substituting value
[tex]Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}[/tex]
[tex]Dose = 22*10^{-10} J/kg[/tex]
The Dose in Sieverts is evaluated as
[tex]Dose_s = REB * Dose[/tex]
[tex]Dose_s = 1.2 * 22*10^{-10}[/tex]
[tex]Dose_s = 26.4 *10^{-10}[/tex]
