Chapter 31, Problem 51E

(a)

To determine

Energy absorbed by child in a year.

Answer to Problem 51E

The child absorbs energy $0.617\text{J}$ in a year.

Explanation of Solution

Write the expression of number of gamma rays $N$ emitted in one year

  $N=a\left(1\text{year}\right)$        (I)

Here, $a$ is source activity.

Write the expression of energy $E$ absorbed by child

  $E=pNe$        (II)

Here, $p$ is fraction of gamma rays absorbed and $e$ is energy of gamma rays.

Conclusion:

Substitute $10\text{microcurie}$ for $a$ in expression (I).

  $\begin{array}{l}N=\left(10\text{microcurie}\left(\frac{1\text{curie}}{{10}^6\text{microcurie}}\right)\left(\frac{3.70\times {10}^{10}{\text{s}}^{-1}}{1\text{curie}}\right)\right)\left(1\text{year}\left(\frac{3.15\times {10}^7\text{s}}{1\text{year}}\right)\right)\\ =1.17\times {10}^{13}\end{array}$

Substitute $\left(1/2\right)$ for $p$, $0.66\text{MeV}$ for $e$ and $1.17\times {10}^{13}$ for $N$ in expression (II).

  $\begin{array}{c}E=\left(\frac{1}{2}\right)\left(1.17\times {10}^{13}\right)\left(0.66\text{MeV}\left(\frac{1.6\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\right)\\ =0.617\text{J}\end{array}$

Thus, the child absorbs energy $0.617\text{J}$ in a year.

(b)

To determine

Absorbed dose in a year in rads.

Answer to Problem 51E

The absorbed dose in a year is $3.08\text{rad}$.

Explanation of Solution

Mass of child is $20\text{kg}$.

Write the expression of absorbed dose $A$ in one year in rads

  $A=\frac{E}{m}\left(\frac{1\text{rad}}{0.01\text{J}\cdot {\text{kg}}^{-1}}\right)$        (III)

Here, $m$ is mass of child.

Conclusion:

Substitute $0.617\text{J}$ for $E$ and $20\text{kg}$ for $m$ in expression (III).

  $\begin{array}{c}A=\frac{\left(0.617\text{J}\right)}{\left(20\text{kg}\right)}\left(\frac{1\text{rad}}{0.01\text{J}\cdot {\text{kg}}^{-1}}\right)\\ =3.08\text{rad}\end{array}$

Thus, the absorbed dose in a year is $3.08\text{rad}$.

(c)

To determine

Ratio of biologically equivalent dose to average natural annual dose.

Answer to Problem 51E

The ratio of biologically equivalent dose to average natural annual dose is $28$.

Explanation of Solution

Write the expression of biologically equivalent dose $B$ per annum

  $B=AQ$        (IV)

Here, $Q$ is the quality factor.

Write the expression of ratio $R$ of biologically equivalent dose to natural dose per annum

  $R=\frac{B}{n}$        (V)

Here, $n$ is natural dose per year.

Conclusion:

Substitute $3.08\text{rad}$ for $A$ and $0.8$ for $Q$ in expression (IV).

  $\begin{array}{c}B=\left(3.08\text{rad}\right)\left(0.8\right)\\ =2.46\text{rem}\end{array}$

Substitute $2.46\text{rem}$ for $B$ and $88\text{mrem}$ for $n$ in expression (V).

  $\begin{array}{c}R=\frac{\left(2.46\text{rem}\right)}{\left(88\text{mrem}\left(\frac{1\text{rem}}{{10}^3\text{mrem}}\right)\right)}\\ =28\end{array}$

Thus, the ratio of biologically equivalent dose to average natural annual dose is $28$.