The annual radiation dose from \(^{14}{\rm{C}}\)in our bodies is \(0.01\,{\rm{mSv}}/{\rm{y}}\). Each \(^{14}{\rm{C}}\) decay emits a \({\beta ^ - }\) averaging \(0.0750\,{\rm{MeV}}\). Taking the fraction of \(^{14}{\rm{C}}\)to be \(1.3 \times {10^{ - 12}}\;{\rm{N}}\)of normal \(^{12}{\rm{C}}\), and assuming the body is \(13\% \) carbon, estimate the fraction of the decay energy absorbed. (The rest escapes, exposing those close to you.)

Energy is released spontaneously as energetic subatomic particles or electromagnetic waves from radioactive atoms. Radiation is the term for the emissions.

A radiation dose is a measurement of how much energy a source of radiation deposits into a material.

In this problem we are to estimate the fraction of energy decay absorbed for\(^3{\rm{He}}{ + ^3}{\rm{He}}{ \to ^4}{\rm{He}}{ + ^1}{\rm{H}}{ + ^1}{\rm{H}}.\)

To calculate the mass, first we should remind the definition of the activity:\(R = \frac{{0.693N}}{{{t_{1/2}}}}\)

Where,\(N\) is the number of the atoms, which can also be calculated via molar mass and Avogadro number:\(N = \left( {\frac{m}{M}} \right){N_A}\)
Joule of energy is delivered to each kilogram of absorbing material:\(1\,\frac{{{\rm{J/kg}}}}{{{\rm{Gy}}}} = 1\,\,{\rm{rad}}\)

And also:\({\rm{ }}Sv{\rm{ }} = {\rm{ }}Gy \times RBE,{\rm{ }}and{\rm{ }}1{\rm{ }}Sv{\rm{ }} = 100{\rm{ }}rem{\rm{ }}\)

So to get energy from this radioactive decay of we use the following relation:

\(E = Rt\Delta E\)

Where\(t\) is the time in which dose was received and\(\Delta E\)is the energy per decay. We put in the relation for activity and we get:\(E = \frac{{0.693N}}{{{t_{1/2}}}}t\Delta E\)

Because we are estimating only a fraction of energy and not whole (because the rest escapes), we can note that fraction of the energy by\(x\):\(E = \frac{{0.693N}}{{{t_{1/2}}}}t\Delta Ex\)

And since we do not know the number of particles we can use the formula with the molar mass and Avogadro number, so we write: \(E = \frac{{0.693\left( {\frac{m}{M}} \right){N_A}}}{{{t_{1/2}}}}t\Delta {E_x}\)

Consider the given information.

We need to take into account the received dose. 50 Kg is exposed into ionizing radiation over the body she absorbs 1.00J her received dose is:\((1.00\;{\rm{J}})/(50.0\;{\rm{kg}}) = 0.0200\;{\rm{J}}/{\rm{kg}} = 2.00\,{\rm{rad}}\)

So we know that expression to calculate the energy of radiation is:\(E = {\rm{ dose}} \times m\)

And this is the dose in Joules, to get it in rem, we remember that\(1\,rad\, = \,0.01\;J/kg,\,\,50\):

\(E = {\rm{ dose}} \times m \times 0.01\;{\rm{J}}/{\rm{kg}}\)

And here\(m\)is exactly the mass of the radioactive element that is producing the radiation dose.

For the naturally occurring\(^{14}{\rm{C}}\) we would like to know the radiation dose in\({\rm{J}}/{\rm{kg}}\), so we must include the\(RBE\)factor, which is for the\(\beta \)rays exactly\(1\). So the does will be:\({\rm{ dose rad = }}\frac{{{\rm{ dose rem }}}}{{{\rm{RBE}}}}\)

Putting in the values we have (remembering that (\(1\;{\rm{Sv}} - 100\)rem):

\({\rm{ }}dose{\rm{ }}rad{\rm{ }} = \frac{{0.01m\,\,rem/y}}{1}\)

To get the units we want we have:\({\rm{ }}dose{\rm{ }}rad = \frac{{1 \times {{10}^{ - 3}}1/y}}{1} \times 0.01\frac{J}{{kg}}\)

So the received does in \({\rm{J}}/{\rm{kg}}\)in one year will be:\({\rm{ }}dose{\rm{ }}rad{\rm{ }} = \frac{{1 \times {{10}^{ - 5}}}}{1}\frac{J}{{kgy}}\).

Let us calculate fraction of energy decay.

We have already written the relation that is connecting the radiation dose and delivered energy to the body:

\(E = dose \cdot {m_C}\)

We put in expression for the energy here (and we are taking only 13% of the energy so we write:

\(\frac{{0.693\left( {\frac{m}{M}} \right){N_A}}}{{{t_{1/2}}}}t\Delta Ex = {\rm{ dose}} \times {m_C}\)

And expressing only the mass of our radioactive element we have:

\(x = \frac{{{\rm{ dose }} \cdot {m_C} \cdot {t_{1/2}} \cdot M}}{{t \cdot \Delta E \cdot {N_A} \cdot 0.693 \cdot m}}\)

We need to take into account we are taking only a fraction of carbon\(14,1.3 \times {10^{ - 12}}\;{\rm{N}}\), which is\(13\% \)equals to\(1.69 \times {10^{ - 13}}\), so we got:\(x = \frac{{{\rm{ dose}} \times {m_C} \times {t_{1/2}} \times M}}{{t \times \Delta E \times {N_A} \times 0.693 \times 1.69 \times {{10}^{ - 13}}}}\)

The molar mass is\(14.003\)as found in the Appendix. We also convert\(0.075{\rm{MeV}}\)to Joules (given\(1{\rm{eV}} = 1.6 \times {10^{ - 19}}\;{\rm{J}}\)) which is\(1.2 \times {10^{ - 14}}\)Joules, while also seen from the Appendix half-life is\(1.28 \times {10^9}{\rm{y}}\)\(1.28 \times {10^9}{\rm{y}}\). Now we have all the information to calculate the mass:

\(x = \frac{{1 \times {{10}^{ - 5}}\frac{{\rm{J}}}{{kgy}} \times 5730\,{\rm{y}} \times 0.014003\,\;{\rm{kg}}/{\rm{mol}}}}{{1\,{\rm{y}} \times 1.2 \times {{10}^{ - 14}}\;{\rm{J}} \times 6.022 \times {{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{ - 1}} \times 0.693 \times 1.69 \times {{10}^{ - 13}}}}\)

Therefore, we get the result of:\(x = 0.95218 = 95\% \).