Question

Question: The positron emission emitted by ${\mathbf{11}}_{\mathbf{C}}$ have a maximum kinetic energy of$\mathbf{0}\mathbf{.}\mathbf{99}\mathbf{MeV}$ , and those emitted byrole="math" localid="1663390945192" ${}^{15}\mathbf{O}$ have a maximum kinetic energy of $\mathbf{1}\mathbf{.}\mathbf{72}\mathbf{MeV}$. Calculate the ratio of number of millisieverts of radiation exposure caused by ingesting a given fixed chemical amount(equal number of atoms) of each of these radioisotopes.

Short answer

The ratio of number of millisieverts of radiatiation exposure is$1.72:099$.

Step-by-step solution

Conversion of MeV to Sievert

1 Sievert is defined as the amount of radiation absorbed which deposits 1 Joule of energy in 1 kilogram of tissue.

$1MiliSievert=1000Sievert=1000J/kg$

$1\hspace{0.17em}MeV=1.602\times {10}^{-13}J$

Ratio

Equal number of atoms of and are ingested. Let the number of atoms present in per kilogram of tissue be N.

Then the amount of energy absorbed due to decay of $=N\times 0.99MeV$

Then the amount of energy absorbed due to decay of$=N\times 1.72MeV$

Therefore,

$\frac{{}^{15}O}{{}^{11}C}=\frac{1.72NMeV}{0.99NMeV}=\frac{1.72}{0.99}$