Q24.65 P - Chemistry: Molecular Nature Of Matter And Change

Question

A small region of a cancer patient’s brain is exposed for 24.0 min to 475 Bq of radioactivity from  60Co for treatment of a tumor. If the brain mass exposed is  $\text{1.858}$g and each  particle emitted has an energy of  ${\mathbf{\text{5.05×10}}}^{\mathbf{\text{-14}}}\mathbf{\text{ J}}$, what is the dose in rads

Step-by-Step Solution

Answer

The dose in rads is  _{${\text{1.86×10}}^{\text{-3}}\text{rad}$}

Step 1: The give data

-  _{$\text{1rad=0.01 J/kg}$}

- The mass of exposed part of the brain is ${\text{1.858 g=1.858×10}}^{\text{-3}}\text{ kg}$ 

- A brain is exposed to ${\text{475 Bq from }}^{\text{60}}\text{Co(1 Bq=1 d/s)}$ 

- The duration of exposure is$24 \text{min}$  

- Energy of each disintegration is _{$5.05\times {10}^{-14} \text{J}$}

Step 2: Calculation

We have to calculate the dose in rads.

First, let us calculate the number of disintegrations per second

 _{$\text{A =475 Bq×}\frac{\text{1 d/s}}{\text{1 Bq}}\text{=475 d/s}$}

Now, let us calculate the number of disintegrations in $24 \text{min}$ 

 _{$\begin{array}{c}\text{ No of d in 20 min=475}\frac{\text{d}}{\text{s}}\text{×}\frac{\text{60 s}}{\text{1 min}}\text{×24 min}\\ {\text{ =6.84×10}}^{\text{5}}\text{ d}\end{array}$}

Total number of  that the brain receives in 20 minutes is

 _{$\begin{array}{c}{\text{ No of J=6.84×10}}^{\text{5}}{\text{ d×5.05×10}}^{\text{-14}}\text{ J/d}\\ {\text{=3.454×10}}^{\text{-8}}\text{ J}\end{array}$}

Therefore, the dose in rads is

 _{$\begin{array}{c}\text{Dose in rads =}\frac{{\text{3.454×10}}^{\text{-8}}\text{ J}}{{\text{1.858×10}}^{\text{-3}}\text{ kg}}\text{×}\frac{\text{1rad}}{\text{0.01 J/kg}}\\ {\text{=1.86×10}}^{\text{-3}}\text{rad}\end{array}$}

Hence, the dose in rads is ${\text{1.86×10}}^{\text{-3}}\text{rad}$