Question

In the 1986 disaster at the Chernobyl reactor in eastern Europe, about 18 of the 137Cs present in the reactor was released.

The isotope 137Cs has a half-life of 30.07 y for b decay, with the emission of a total of 1.17 MeV of energy per decay. Of this, 0.51 MeV goes to the emitted electron; the remaining 0.66 MeV goes to a g ray. The radioactive 137Cs is absorbed by plants, which are eaten by livestock and humans. How many 137Cs atoms would need to be present in each kilogram of body tissue if an equivalent dose for one week is 3.5 Sv? Assume that all of the energy from the decay is deposited in 1.0 kg of tissue and that the RBE of the electrons is 1.5.

Short answer

N₀= 3.67 x 10¹⁶atom.

Step-by-step solution

Given

The half-life of\(137\)is\({T_{1/2}} = 30.07yr\)and the time of the radiation is t=1.00 week.

The total energy released from its decay is Q=1.17 MeV/decay, the energy carried by the electron is\({K_e}\)= 0.510 MeV and the energy carried by the y-ray is\({E_\gamma }\)=0.66 MeV

The equivalent dose of one week is Equivalent dose =3.70 Sv/week, the mass of the tissue absorbed in the radiation is m= 1.00 kg, and RBE= 1.50 for electrons.

Calculate the absorbed dose.

First, we convert the energies of the electron and the\(\gamma \)-ray to joules

We know that the absorbed dose is the total energy delivered to the tissue per unit mass

So the absorbed dose delivered to 1.0 kg of tissue by the electron is:

\(Absorbed - dos{e_e} = \frac{{{E_e}}}{m} = \frac{{8.16 \times {{10}^{ - 14}}{\rm{J}}}}{{1.00{\rm{kg}}}} = 8.16 \times {10^{ - 14}}{\rm{Gy}}/{\rm{\;electron\;}}\)

And the absorbed dose delivered to 1.0 kg of tissue by the y-ray is:

\(Absorbed - dos{e_\gamma } = \frac{{{E_\gamma }}}{m} = \frac{{1.06 \times {{10}^{ - 13}}{\rm{J}}}}{{1.00{\rm{kg}}}} = 1.06 \times {10^{ - 13}}{\rm{Gy}}/{\rm{\;electron\;}}\)

Calculate the total number of decays.

The equivalent dose is given by:

Equivalent dose (Sv) = RBE (Sv/Gy) x Absorbed dose (Cy)

And, RBE = 1.00 for\(\gamma \)-rays;

So, the equivalent dose delivered to the tissue for each type of radiation per decay is:

\(\begin{array}{l}{\rm{Equivalent }}dos{e_e} = \left( {1.50Sv/Gy} \right) \times \left( {8.16 \times {{10}^{ - 14}}{\rm{Gy}}/{\rm{\;electron\;}}} \right)\\{\rm{Equivalent }}dos{e_e} = 1.22 \times {10^{ - 13}}Sv/{\rm{\;electron}}\\{\rm{Equivalent }}dos{e_\gamma } = \left( {1.00Sv/Gy} \right)\left( {1.06 \times {{10}^{ - 13}}{\rm{Gy}}/{\rm{\;electron}}} \right)\\{\rm{Equivalent }}dos{e_\gamma } = 1.06 \times {10^{ - 13}}{\rm{Sv}}/{\rm{\;electron}}\end{array}\)

Thus, the total equivalent dose delivered to 1.00 kg of tissue per decay is:

\(\begin{array}{l}{\rm{Equivalent }}dos{e_{decay}} = 1.22 \times {10^{ - 13}}Sv + 1.06 \times {10^{ - 13}}Sv\\{\rm{Equivalent }}dos{e_{decay}} = 2.28 \times {10^{ - 13}}Sv\end{array}\)

Thus, for a total equivalent dose of 3.70 Sv, we need a total number of decays of:

\({N_{decay}} = \frac{{{\rm{Equivalent }}dose}}{{{\rm{Equivalent }}dos{e_{decay}}}} = \frac{{3.70Sv}}{{2.28 \times {{10}^{ - 13}}Sv}} = 1.62 \times {10^{13}}Sv\)

Calculate the total number of atoms.

the number of remaining nuclei at t in a sample of a radioactive element is given by:

\(N = {N_0}{e^{ - \lambda t}}\)

Where\({N_0}\)is the number of nuclei at t = 0.

\({N_0}\)is what we need in this problem.

Since\({N_0}\)is the initial number and N is remaining, so the number of decayed nuclei is\({N_{decay}} = {N_0} - N\).

we get,

\({N_{decay}} = {N_0} - {N_0}{e^{ - \lambda t}} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right)\)

Solving for\({N_0}\), we get:

\({N_0} = \frac{{{N_{decay}}}}{{1 - {e^{ - \lambda t}}}}\) (1)

the decay constant of\(^{137}Cs\)is:

Finally, we plug our values for\({N_{decay}}\),\(\lambda \)and t into equation (1), so we get: