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.

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.