Question

calculate the half lives of the following samples.
1. A sample of iodine-123 whose activity falls from 1000 Bq to 250 Bq in 14.4 hours.
2.A sample of technetium -99m whose activity falls from 200 Bq to 25 Bq in 18 hours.
3. A sample of strontium-90 whose activity falls 500 Bq to 62.5 Bq in 86.4 years.

Answer 1

Question #63529, Physics / Atomic and Nuclear Physics

calculate the half lives of the following samples.

1. A sample of iodine-123 whose activity falls from 1000 Bq to 250 Bq in 14.4 hours.

2. A sample of technetium -99m whose activity falls from 200 Bq to 25 Bq in 18 hours.

3. A sample of strontium-90 whose activity falls 500 Bq to 62.5 Bq in 86.4 years.

Solution

$$A = A _ {0} 2 ^ {- \frac {I}{T _ {U 2}}}$$

1. $250 = 1000 \times 2^{-\frac{14.4}{T_{U2}}}$ ;

$$\begin{array}{l} 2 ^ {- \frac {1 4 . 4}{T _ {U 2}}} = \frac {1}{4}; \\ - \frac {1 4 . 4}{T _ {U 2}} = - 2; \\ T _ {U 2} = 7. 2 \text { hours} \\ \end{array}$$

2. $25 = 200 \times 2^{-\frac{18}{T_{U2}}}$ ;

$$\begin{array}{l} 2 ^ {- \frac {1 8}{T _ {U 2}}} = \frac {1}{8}; \\ - \frac {1 8}{T _ {U 2}} = - 3; \\ T _ {U 2} = 6 \text { hours} \\ \end{array}$$

3. $62.5 = 500 \times 2^{-\frac{86.4}{T_{U2}}}$ ;

$$\begin{array}{l} 2 ^ {- \frac {8 6 . 4}{T _ {U 2}}} = \frac {1}{8}; \\ - \frac {8 6 . 4}{T _ {U 2}} = - 3; \\ T _ {U 2} = 2 8. 8 \text { years} \\ \end{array}$$

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