Chapter 31, Problem 55E

(a)

To determine

Number of excess thyroid cancers induced by explosions.

Answer to Problem 55E

The number of excess thyroid cancers induced by explosions was $4.30$.

Explanation of Solution

Nuclear explosions in atmosphere caused average thyroid dose of $3.1\times {10}^{-4}\text{rad}$ to $2.2\times {10}^8\text{persons}$.

Write the expression of collective dose $D$ received

  $D=pd$        (I)

Here, $p$ is total population and $d$ is average thyroid dose.

Since 63 cancers are induced per ${10}^6\text{person rads}$ of thyroid dose therefore the number of excess cancers is:

  $n=\frac{D\left(63\right)}{\left({10}^6\text{person rads}\right)}$        (II)

Conclusion:

Substitute $3.1\times {10}^{-4}\text{rad}$ for $d$ and $2.2\times {10}^8\text{persons}$ for $p$ in expression (I)

  $\begin{array}{c}D=\left(2.2\times {10}^8\text{persons}\right)\left(3.1\times {10}^{-4}\text{rad}\right)\\ =6.82\times {10}^4\text{person rads}\end{array}$

Substitute $6.82\times {10}^4\text{person rads}$ for $D$ in expression (II)

  $\begin{array}{c}n=\frac{\left(6.82\times {10}^4\text{person rads}\right)\left(63\right)}{\left({10}^6\text{person rads}\right)}\\ =4.30\end{array}$

Thus, the number of excess thyroid cancers induced by explosions was $4.30$.

(b)

To determine

Fractional increase in present rate of thyroid cancers.

Answer to Problem 55E

The fractional increase in present rate of thyroid cancers is $1.14\times {10}^{-5}$.

Explanation of Solution

Write the expression of number of thyroid cancers in 45 years due to present rate

  $N=r\left(45\text{year}\right)$        (III)

Here, $r$ is present rate of thyroid cancer per year.

Write the expression of fractional increase in present rate

  $f=\frac{n}{N}$        (IV)

Here, $f$ is the fractional increase in percent and $N$ is the number of thyroid cancer cases.

Conclusion:

Substitute $8400{\text{year}}^{-1}$ for $r$ in expression (III).

  $\begin{array}{c}N=\left(\frac{8400}{\text{year}}\right)\left(45\text{year}\right)\\ =378000\end{array}$

Substitute $378000$ for $N$ and $4.30$ for $n$ in expression (IV).

  $\begin{array}{c}f=\frac{\left(4.30\right)}{\left(378000\right)}\\ =1.14\times {10}^{-5}\end{array}$

Thus, the fractional increase in present rate of thyroid cancers is $1.14\times {10}^{-5}$.

(c)

To determine

Chance that an infant develops thyroid cancer.

Answer to Problem 55E

The chance that an infant develops thyroid cancer is $1.26\times {10}^{-6}$.

Explanation of Solution

Thyroid doses received by infants are of $0.02\text{rad}$.

Since 63 cancers are induced per ${10}^6\text{person rads}$ of thyroid dose therefore the chances $P$ of an infant developing cancer is:

  $P=\frac{i\left(63\right)}{\left({10}^6\text{person rads }\right)}$        (V)

Here, $i$ is collective dose received by infants.

Conclusion:

Since risk for infants is same as for whole populations therefore the value of $i$ for an individual infant is:

  $\begin{array}{c}i=\left(0.02\text{rad}\right)\left(1\text{person}\right)\\ =0.02\text{person rad}\end{array}$

Substitute $0.02\text{person rad}$ for $i$ in expression (V)

  $\begin{array}{c}P=\frac{\left(0.02\text{person rad}\right)\left(63\right)}{\left({10}^6\text{person rads }\right)}\\ =1.26\times {10}^{-6}\end{array}$

Thus, an infant has chance $1.26\times {10}^{-6}$ of developing thyroid cancer.