Q24.45 P - Chemistry: Molecular Nature Of Matter And Change

Question

Due to decay of ${}^{\mathbf{40}}\mathit{K}$ , cow’s milk has a specific activity of about ${\mathbf{\text{6x10}}}^{\mathbf{\text{-11}}}$ mCi per milliliter. How many disintegrations of ${}^{\mathbf{40}}\mathit{K}$  nuclei are there per minute in an $\mathbf{\text{8.0}}$ -oz glass of milk?

Step-by-Step Solution

Answer

The number of disintegrations per minute is  _{$31.5 \text{d}/\text{min}$}

Step 1: Specific activity

Specific activity of a radionuclide is defined as the activity per its quantity.

Step 2: Calculation

- The specific activity of cow's milk is  _{$6\times {10}^{-11}\text{mCi}/\text{mL}$$6\times {10}^{-11}\text{mCi}/\text{mL}$}

-  _{${\text{1Ci=3.70×10}}^{\text{10}}\text{ d/s}$} (disintegrations per second)

-  _{$1\text{\hspace{0.17em}mCi}=3.70\times {10}^7 \text{d}/\text{s}$}

We have to find the number of disintegrations of ${}^{40} \text{K}$ nuclei per min, in $8.0\text{\hspace{0.17em}\hspace{0.17em}oz}$ glass of milk.

First, let us convert the volume of the glass of milk, from oz to   _{data-custom-editor="chemistry" $\text{mL}$}

_{$\begin{array}{c}\text{V=8.0oz×}\frac{\text{29.5735 mL}}{\text{1oz}}\\ \text{=236.59 mL}\end{array}$}

Calculate the number of disintegrations:

 _{$\begin{array}{c}\text{Activity = Specific activity ×V}\\ {\text{ =6×10}}^{\text{-11}}\text{mCi}\times \text{236.59 mL}\\ {\text{ =1.419×10}}^{\text{-8}}\text{mCi}\\ {\text{ =1.419×10}}^{\text{-8}}\text{mCi×}\frac{{\text{3.70×10}}^{\text{7}}\text{ d/s}}{\text{1mCi}}\end{array}$}

On further simplification,

 _{$\begin{array}{l}\text{Activity=0.525 d/s}\\ \text{ =0.525}\frac{\text{d}}{\text{s}}\text{×}\frac{\text{60 s}}{\text{1 min}}\\ \text{ =31.5 d/min}\end{array}$}

Hence, the number of disintegrations per minute is $31.5 \text{d}/\text{min}$