Chapter 31, Problem 48E

(a)

To determine

Reason why exposure inversely varies with square of distance from source.

Answer to Problem 48E

The exposure varies inversely with distance from source because the flux of radiation is inversely proportional to the distance.

Explanation of Solution

Exposure to radiation is defined by extent of dose that is received by matter. The radiation energy is continuously distributed in all direction at a certain distance from radiating source.

The flux of radiation is defined as amount of radiation per unit area per unit time. If a spherical region of radius same as distance from source is imagined, then the flux of radiation inversely varies with distance because area of sphere is directly proportional to square of its radius.

Thus, the exposure which is a measure of flux also varies inversely with square of distance.

Conclusion:

Thus, exposure varies inversely with distance from source because the flux of radiation is inversely proportional to the distance.

(b)

To determine

Time required to produce dose of $100\text{rems}$ at distance $0.01\text{m}$.

Answer to Problem 48E

The time required to produce dose of $100\text{rems}$ at distance $0.01\text{m}$ is $758.25\text{s}$.

Explanation of Solution

A dose of $8.2\text{rads}$ is produced by $1\text{mg}$ of radium per hour at distance $0.01\text{m}$. For gamma rays of radium the quality factor is $0.965$.

Write the expression of biologically equivalent dose $D$ of $1\text{mg}$ radium source

  $D=AQ$        (I)

Here, $A$ is absorbed dose in rads and $Q$ is quality factor.

Write the expression of required time for second radium source

  $T_1=\left(\frac{D_1}{D}\right)\left(\frac{M}{M_1}\right)\left(\frac{R_1{}^2}{R^2}\right)T$        (II)

Here, $D_1$ is dose produced by second source of mass $M_1$ at distance $R_1$ and $M$ is mass of first radium source that produces dose $D$ at distance $R$ over time $T$.

Conclusion:

Substitute $8.2\text{rads}$ for $A$ and $0.965$ for $Q$ in expression (I)

  $\begin{array}{c}D=\left(8.2\text{rads}\right)\left(0.965\right)\\ =7.913\text{rems}\end{array}$

Substitute $7.913\text{rems}$ for $D$, $100\text{rems}$ for $D_1$, $0.01\text{m}$ for $R_1$ and $R$, $1\text{hour}$ for $T$, $1\text{mg}$ for $M$ and $60\text{mg}$ for $M_1$ in expression (II)

  $\begin{array}{c}{T}_1=\left(\frac{100\text{rems}}{7.913\text{rems}}\right)\left(\frac{1\text{mg}}{60\text{mg}}\right)\left(\frac{{\left(0.01\text{m}\right)}^2}{{\left(0.01\text{m}\right)}^2}\right)\left(1\text{hour}\left(\frac{3600\text{s}}{1\text{hour}}\right)\right)\\ =758.25\text{s}\end{array}$

Thus, the time required to produce dose of $100\text{rems}$ at distance $0.01\text{m}$ is $758.25\text{s}$.

(c)

To determine

Time required to produce dose of $100\text{rems}$ at distance $0.05\text{m}$.

Answer to Problem 48E

The time required to produce dose of $100\text{rems}$ at distance $0.05\text{m}$ is $18956.15\text{s}$.

Explanation of Solution

Write the expression of required time for third radium source

  $T_2=\left(\frac{D_2}{D}\right)\left(\frac{M}{M_2}\right)\left(\frac{R_2{}^2}{R^2}\right)T$        (III)

Here, $D_2$ is dose produced by third source of mass $M_2$ at distance $R_2$.

Conclusion:

Substitute $7.913\text{rems}$ for $D$, $100\text{rems}$ for $D_2$, $0.01\text{m}$ for $R$, $0.05\text{m}$ for $R_2$, $1\text{hour}$ for $T$, $1\text{mg}$ for $M$ and $60\text{mg}$ for $M_2$ in expression (III)

  $\begin{array}{c}{T}_2=\left(\frac{100\text{rems}}{7.913\text{rems}}\right)\left(\frac{1\text{mg}}{60\text{mg}}\right)\left(\frac{{\left(0.05\text{m}\right)}^2}{{\left(0.01\text{m}\right)}^2}\right)\left(1\text{hour}\left(\frac{3600\text{s}}{1\text{hour}}\right)\right)\\ =18956.15\text{s}\end{array}$

Thus, the time required to produce dose of $100\text{rems}$ at distance $0.05\text{m}$ is $18956.15\text{s}$.