Q.5.39 - An Introduction to General, Organic, and Biological Chemistry

Question

Gallium-68 is taken up by tumours; the emission of positrons allows the tumours to be located.

a. Write an equation for the positron emission of Ga-68.

b. If the half-life is 68 min, how much of a 64-mcg sample is active after 136 min?

Step-by-Step Solution

Answer

a.${}_{31}^{68}\mathrm{Ga}{\to }_{30}^{68}\mathrm{Zn}{+}_{+1}^0\mathrm{e}$a.

b.$16$mcg sample is active  

Part (a) Step 1: Given Information

Gallium-$68$ is taken up by tumours.

Finding the equation for the positron emission of Ga-$68$. 

Part (a) Step 2: Explanation

The equation for the positron emission of Ga-$68$ is,

 ${}_{31}^{68}\mathrm{Ga}{\to }_{30}^{68}\mathrm{Zn}{+}_{+1}^0\mathrm{e}$

Part (b) Step 1: Given Information

Gallium-$68$ is taken up by tumours.

The half-life is = $68$ min

Calculating the sample active after $136$ min 

Part (b) Step 2: Explanation

The reaction is, ${}_{31}^{68}\mathrm{Ga}{\to }_{30}^{68}\mathrm{Zn}{+}_{+1}^0\mathrm{e}$

Half-life = $68$min

In $136$min = $\frac{136}{68}=2$ half-life

The activity initially = $64$ mcg

Hence after two half-lives = $\frac{64}{2^2}=16$ mcg