Answer :
Answer:
The approximate number of decays this represent is [tex]N= 23*10^{10}[/tex]
Explanation:
From the question we are told that
The amount of Radiation received by an average american is [tex]I_a = 2.28 \ mSv[/tex]
The source of the radiation is [tex]S = 5.49 MeV \ alpha \ particle[/tex]
Generally
[tex]1 \ J/kg = 1000 mSv[/tex]
Therefore [tex]2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg[/tex]
Also [tex]1eV = 1.602 *10^{-19}J[/tex]
Therefore [tex]2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg[/tex]
An Average american weighs 88.7 kg
The total energy received is mathematically evaluated as
[tex]1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x[/tex]
Cross-multiplying and making x the subject
[tex]x = 88.7 * 1.423*10^{16} eV[/tex]
[tex]x = 126.2*10^{16}eV[/tex]
Therefore the total energy deposited is [tex]x = 126.2*10^{16}eV[/tex]
The approximate number of decays this represent is mathematically evaluated as
N = [tex]\frac{x}{S}[/tex]
Where n is the approximate number of decay
Substituting values
[tex]N = \frac{126 .2*10^{16}}{5.49*10^6}[/tex]
[tex]N= 23*10^{10}[/tex]