Q24.63 P - Chemistry: Molecular Nature Of Matter And Change

Question

A laboratory rat weighs  $\mathbf{\text{265}}$ g and absorbs data-custom-editor="chemistry" ${\mathbf{\text{1.77×10}}}^{\mathbf{\text{10}}}{\mathbf{\text{β}}}^{\mathbf{\text{-}}}$ particles, each with an energy of data-custom-editor="chemistry" ${\mathbf{\text{2.20×10}}}^{\mathbf{\text{-13}}}$ J. 

(a) How many rads does the animal receive? 

(b) What is this dose in Gy? 

(c) If the RBE is  $\mathbf{\text{0.75}}$, what is the equivalent dose in Sv?

Step-by-Step Solution

Answer

$\begin{array}{l}\text{a)1.47rad}\\ \text{b)0.0147Gy}\\ \text{c)0.011sv}\end{array}$

Step 1: Relation between various radiation units

-  _{$\text{1rad=0.01 J/kg=0.01 Gy}$}

- no. of rems = no. of rads $\text{×RBE}$

- 1 rem  _{$\text{=0.01 Sv}$}

- The mass of a rat is  _{$\text{265 g=0.265 kg}$}

- Rat absorbed ${\text{1.77×10}}^{\text{10}}{\text{β}}^{\text{-}}$ particles

- The energy of beta particle is $2.20\cdot {10}^{-13} \text{J}$ 

Therefore, the energy that rat received is

Energy  _{${\text{=1.77×10}}^{\text{10}}{\text{β}}^{\text{-}}\text{ particles ×}\frac{{\text{2.20×10}}^{\text{-13}}\text{ J}}{{\text{1β}}^{\text{-}}}{\text{=3.894×10}}^{\text{-3}}\text{ J}$}

Step 2: Calculation

(a)

We have to find the number of rads that rat receives

 _{$\text{ No of rads =}\frac{{\text{3.894×10}}^{\text{-3}}\text{ J}}{\text{0.265 kg}}\text{×}\frac{\text{1rad}}{\text{0.01 J/kg}}\text{=1.47rad}$}

(b)

The amount of Gy that rat received is

No of  _{$\text{Gy=1.47rad×}\frac{\text{0.01 Gy}}{\text{1rad}}\text{=0.0147 Gy}$}

Step 3: Subpart (c)

We have to calculate the dose in Sv, if RBE  _$=0.75$

First, let us calculate the number of rems

_{$\begin{array}{c}\text{No or rems = No of rads ×RBE}\\ \text{ =1.47×0.75}\\ \text{ =1.1025rem}\end{array}$} 

All we have to do now is to convert rem into Sv, and get

 _{$\text{ Dose in Sv=1.1025rem×}\frac{\text{0.01 Sv}}{\text{1 rem }}\text{=0.011 Sv}$}