Question

the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes from the 5.49 MeV alpha particle emitted by Rn-222, how much energy is deposited in your body each year from radon. Approximately how many decays does this represent.

Answer 1

Answer:

The approximate number of decays  this represent  is  $N= 23*10^{10}$  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is $I_a = 2.28 \ mSv$

     The source of the radiation is $S = 5.49 MeV \ alpha \ particle$

 Generally

            $1 \ J/kg = 1000 mSv$

   Therefore  $2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg$

Also  $1eV = 1.602 *10^{-19}J$

  Therefore  $2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg$

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        $1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x$

Cross-multiplying and making x the subject

           $x = 88.7 * 1.423*10^{16} eV$

              $x = 126.2*10^{16}eV$

Therefore the total  energy  deposited is $x = 126.2*10^{16}eV$

The approximate number of decays  this represent  is mathematically evaluated as

            N = $\frac{x}{S}$

Where n is the approximate number of decay

   Substituting values

                             $N = \frac{126 .2*10^{16}}{5.49*10^6}$  

                                  $N= 23*10^{10}$