Answer
See the detailed answer below.
Work Step by Step
a)
First, we need to find the activity in decay per day.
$$R=\rm \dfrac{2000\times 10^{-12}\;\color{red}{\bf\not}Ci}{1\;\color{red}{\bf\not}L}\times \dfrac{\frac{1}{2}\;\color{red}{\bf\not}L}{1\;d}\times12\;\color{red}{\bf\not}h\times \dfrac{ 60^2\;\color{red}{\bf\not}s}{1\;\color{red}{\bf\not}h}\times \dfrac{3.7\times 10^{10}\;decay}{1\;\color{red}{\bf\not}s\cdot \color{red}{\bf\not}Ci}$$
$$R= \bf 1.598\times10^6\;\rm decay/d $$
Now we need to find the energy absorbed in one your;
$$\dfrac{E}{\Delta t}=0.1\times \rm \dfrac{1.598\times10^6\;\rm \color{red}{\bf\not}decay}{\color{red}{\bf\not}day}\times\dfrac{365.25\;\color{red}{\bf\not}day}{1\;year}\times \dfrac{1.5\times 10^6\;\color{red}{\bf\not}eV}{1\;\color{red}{\bf\not}decay}\times \dfrac{1.6\times 10^{-19}\;J}{1\;\color{red}{\bf\not}eV}$$
$$\dfrac{E}{\Delta t}=\bf 1.4\times 10^{-5}\;\rm J/year$$
Now we can find the effective dose for the adult (60 kg), and recall that the
potassium decays by $\gamma$ and $\beta^-$ where $\gamma$ and $\beta^-$ have an RBE of 1.
Thus, the number of Sv is the same as the number of Gy, and hence the number of rem is the same as that of rad.
$${\rm Effective \;dose }=\rm \dfrac{1.4\times 10^{-5}\;\color{red}{\bf\not}J}{1\;year}\times \dfrac{1}{60\;\color{red}{\bf\not}kg}\times \dfrac{1\;Sv}{1\;\color{red}{\bf\not}Gy}\times \dfrac{1\;\color{red}{\bf\not}Gy}{1\;\frac{\color{red}{\bf\not}J}{\color{red}{\bf\not}kg}}\times \dfrac{10^5\;mrem}{1\;\color{red}{\bf\not}Sv}$$
$${\rm Effective \;dose }=\color{red}{\bf 2.33\times 10^{-2}}\;\rm mrem/year$$
which is less than the allowed dose by about 4300 times.
$$\rm \dfrac{Allowed\;dose}{Effective\;dose}=\dfrac{100\;mrem/year}{0.0233\;mrem/year}=\color{red}{\bf4283}$$
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b) For the 6-kg baby;
$${\rm Effective \;dose }=\rm \dfrac{1.4\times 10^{-5}\;\color{red}{\bf\not}J}{1\;year}\times \dfrac{1}{6 \;\color{red}{\bf\not}kg}\times \dfrac{1\;Sv}{1\;\color{red}{\bf\not}Gy}\times \dfrac{1\;\color{red}{\bf\not}Gy}{1\;\frac{\color{red}{\bf\not}J}{\color{red}{\bf\not}kg}}\times \dfrac{10^5\;mrem}{1\;\color{red}{\bf\not}Sv}$$
$${\rm Effective \;dose }=\color{red}{\bf 0.233}\;\rm mrem/year$$
which is less than the allowed dose by about 430 times.
$$\rm \dfrac{Allowed\;dose}{Effective\;dose}=\dfrac{100\;mrem/year}{0.0233\;mrem/year}=\color{red}{\bf 429}$$
