The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 14.3 days. If an initial sample of phosphorus-32 has an activity of 2.36E4 Ci, what is the activity after 42.9 days have passed?

2.95E3 Ci

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 14.3 days.

Original amount (Nₒ) = 2.36E4 Ci

Time (t) = 42.9 days

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 14.3 days.

Time (t) = 42.9 days

Number of half-lives (n) =?

n = t / t½

n = 42.9 / 14.3

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the amount remaining. This can be obtained as follow:

Original amount (Nₒ) = 2.36E4 Ci = 2.36×10⁴ Ci

Number of half-lives (n) = 3

Amount remaining (N) =?

N = 1/2ⁿ × Nₒ

N = 1/2³ × 2.36×10⁴

N = 1/8 × 2.36×10⁴

N = 0.125 × 2.36×10⁴

N = 2.95×10³ Ci = 2.95E3 Ci

Thus, activity after 42.9 days have passed is 2.95E3 Ci.

law of conservation of matter