24.99 CP - Chemistry: Molecular Nature Of Matter And Change

Question

How many rads does a 65-kg  human receive each year from the approximately  10^-8 g of ${}_6{}^{14}C$  naturally present in her body (t = 5730 yrs;   each disintegration releases 0.156MeV) ?

Step-by-Step Solution

Answer

Number of Rads  (Dose) received by 65kg  human each year is $2\times {10}^{-3}rad$ .

Step 1: Definition of Concept

Nuclear reaction: A nuclear reaction occurs when a lighter nucleus fuses together to form new stable nuclei, or when a heavier nucleus splits into stable daughter nuclei, releasing a large amount of energy.

Step 2: Given information

Considering the given information:

 

Mass of human  = 65 Kg

Energy released in each disintegration  =0.156MeV

Amount of  ¹⁴C receiving in each yr=10^-8 g

Half-life time, t_1/2   of  ¹⁴C=5730 yr

Step 3: Calculating the decay constant from half-life

The half-life equation will be used to determine the value of k.

 $t_{1/2}=\frac{\ln 2}{k}$

After that, rearrange the equation.

 $$\begin{gathered} k=\frac{\ln 2}{t_{1/2}} \\ =\frac{\ln 2}{5730yr} \\ k=1.209\times {10}^{-4} y{r}^{-1} \end{gathered}$$

Step 4: Calculating the number of 14 C carbon atoms

Number of moles of  ¹⁴C present in   ${10}^{-8} g =\frac{{10}^{-8}g}{14 g/mol}=\frac{{10}^{-8}}{14} mol$

  $$\begin{gathered} \text{No\hspace{0.05em}of\hspace{0.05em}atoms\hspace{0.05em}in\hspace{0.05em}}\frac{{10}^{-8}}{14} \text{moles\hspace{0.05em}}= 6.023\times {10}^{23}\times \frac{{10}^{-8}}{14} \\ = 4.3014\times {10}^{14} {}^{14}C \text{atoms} \end{gathered}$$

 Let us calculate the activity.

  $$\begin{gathered} A=kN \\ A=\left.\left(1.209\right.x{10}^{-4}y{r}^{-}\right)\left(4.3014x{10}^{14}\right.atom\left.s^{14}C\right)\left(1\frac{di\sin t.}{atom}\right) \\ A=5.2033x{10}^{10}\frac{di\sin t.}{atom} \end{gathered}$$

 

Step 5: Calculating the number of rads the human body receives each year

We can then calculate the dose value using conversion.

 $\frac{5.2033x{10}^{10}\frac{di\sin t.}{year}}{65 kg}\left(1year\right)\left(\frac{0.156MeV}{di\sin t.}\right)\left(\frac{1.602x{10}^{-13}}{1MeV}\right)\left(\frac{1rad}{0.01\frac{J}{kg}}\right)=2.001\times {10}^{-3}rad$

Therefore, The required number of  Rads (Dose) received by 65 kg  human each year is  $2\times {10}^{-3}rad$.