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Chapter 5: Q.5.39 (page 158)

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?

Short Answer

Expert verified

a.3168Ga鈫�3068Zn++10ea.

b.16mcg sample is active

Step by step solution

01

Part (a) Step 1: Given Information

Gallium-68is taken up by tumours.

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

02

Part (a) Step 2: Explanation

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

3168Ga鈫�3068Zn++10e

03

Part (b) Step 1: Given Information

Gallium-68is taken up by tumours.

The half-life is = 68min

Calculating the sample active after136 min

04

Part (b) Step 2: Explanation

The reaction is, 3168Ga鈫�3068Zn++10e

Half-life = 68min

In 136min = 13668=2half-life

The activity initially = 64mcg

Hence after two half-lives =6422=16mcg

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