Carbon-14 is used in radiocarbon dating. The half-life of carbon-14 is 5.73E3 years. If a sample of carbon-14 has an activity of 0.350 Ci, what is the difference in activities between this sample and the same sample after 1.719E4 years?

0.30625 Ci

Explanation:

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

Half-life (t½) = 5.73E3 years

Original activity (Nₒ) = 0.350 Ci

Time (t) = 1.719E4 years

Difference in activities =?

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

Half-life (t½) = 5.73E3 years = 5.73×10³ years

Time (t) = 1.719E4 years = 1.719×10⁴ years

Number of half-lives (n) =?

n = t / t(½)

n = 1.719×10⁴ / 5.73×10³

n = 3

Thus, 3 half-lives has elapsed.

Next, we shall determine the activity after 1.719E4 years. This can be obtained as follow:

Original activity (Nₒ) = 0.350 Ci

Number of half-lives (n) = 3

Activity remaining (N) =?

N = 1/2ⁿ × Nₒ

N = 1/2³ × 0.350

N = 1/8 × 0.350

N = 0.125 × 0.350

N = 0.04375 Ci

Finally, we shall determine the difference in activities. This can be obtained as follow:

Original activity (Nₒ) = 0.350 Ci

Activity remaining (N) = 0.04375 Ci

Difference in activities =?

Difference in activities = Nₒ – N

Difference in activities = 0.350 – 0.04375

Difference in activities = 0.30625 Ci.

answer; i believe that the correct answer is (a) potassium(k)

the answer is a. co2, this would be colder than the : )

explanation: