Chapter 31, Problem 63GP

(a)

To determine

The number of decays per seconds for an adult body.

Answer to Problem 63GP

  $3700\text{ decays/s}$

Explanation of Solution

Given:

Energy of beta particle, $E=1.4\text{ MeV}$

Mass of person, $m_p=50\text{ kg}$

Quality factor ( $\beta \text{-}$ particle), $QF=1$

The adult body contained ${}_{19}^{40}K$ is $0.10\mu \text{Ci}$

Formula used:

  $1\text{ Ci}=3.7\times {10}^{10}\text{ decays/s}$

Calculation:

The number of decays per second is calculated as,

  $\begin{array}{l}N=\left(0.10\times {10}^{-6}\right)\left(3.7\times {10}^{10}\right)\\ N=3700\text{ decays/s}\end{array}$

Conclusion:

The number of decays per second is $3700$ .

(b)

To determine

Dose per year in sieverts for $50\text{ kg}$ .

To Compare: The obtained value with $\text{3}\text{.6 mSv/year}$ .

Answer to Problem 63GP

Dose per year in sieverts for $50\text{ kg}$ is $5.23\times {10}^{-4}\text{ Sv/year}$ .

The obtained value is $15\%$ greater than the obtained value or background rate.

Explanation of Solution

Given:

Energy of beta particle, $E=1.4\text{ MeV}$

Mass of person, $m_p=50\text{ kg}$

Quality factor ( $\beta \text{-}$ particle), $QF=1$

The adult body contained ${}_{19}^{40}K$ is $0.10\mu \text{Ci}$

Formula used:

The of deposition of energy, $E\text{'}=\left(N\right)\left(E\right)$

The formula of effective does, ${\text{(S}}_v)=\text{ Dose }(D)\times QF$

Calculation:

The conversion of energy from MeV to Joule is,

  $\begin{array}{l}E=\left(1.4\times {10}^6\right)\left(1.6\times {10}^{-19}\right)\\ E=2.24\times {10}^{-13}\text{ J}\end{array}$

The rate of energy deposition is,

  $\begin{array}{l}E\text{'}=\left(N\right)\left(E\right)\\ E\text{'}=\left(3700\right)\left(2.24\times {10}^{-13}\right)\\ E\text{'}=8.288\times {10}^{-10}\text{ J/s}\end{array}$

The dose rate per second for $50\text{ kg}$ person is,

  $\begin{array}{l}D=\left(\frac{8.288\times {10}^{-10}\text{ J}\text{.s}}{50\text{ kg}}\right)\\ D=1.65\times {10}^{-11}\text{ Gy/s}\end{array}$

The dose rate per year is calculated as,

  $\begin{array}{l}D\text{'}=\left(D\right)\times \left(\frac{3.156\times {10}^7\text{ s}}{1\text{ year}}\right)\\ D\text{'}=\left(1.65\times {10}^{-11}\right)\times \left(\frac{3.156\times {10}^7\text{ s}}{1\text{ year}}\right)\\ D\text{'}=5.23\times {10}^{-4}\text{Gy/year}\end{array}$

The dose in Sv/year is,

  $\begin{array}{l}\text{dose}\text{(Sv/year)}=D\text{'}\times (QF)\\ \text{dose}\text{(Sv/year)}=\left(5.23\times {10}^{-4}\times \text{1}\right)\text{ Sv/year}\\ \text{dose}\text{(Sv/year)}=5.23\times {10}^{-4}\text{ Sv/year}\end{array}$

Now, comparing the obtained value with $\text{3}\text{.6 mSv/year}$ ,

  $\frac{5.23\times {10}^{-4}\text{ Sv/year}}{\text{3}\text{.6 mSv/year}}=0.15$

Converting in percentage,

  $\begin{array}{l}=\left(0.15\times 100\right)\%\\ =15\%\end{array}$

The obtained value is $15\%$ greater than the obtained value or background rate.

Conclusion:

The obtained value is $15\%$ greater than the obtained value or background rate.