Question

What is the total number of grams of a 32-gram sample of 32P remaining after 71.5 days of decay?

Answer 1

Answer:

1 g

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 32 g

Time (t) = 71.5 days

Half-life of phosphorus (t½) = 14.3 days

Amount remaining (N) =?

Next, we shall determine the rate of disintegration. This can be obtained as follow:

Half-life of phosphorus (t½) = 14.3 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 14.3

K = 0.0485 /day

Finally, we shall determine the amount remaining as illustrated below:

Original amount (N₀) = 32 g

Time (t) = 71.5 days

Decay constant (K) = 0.0485 /day

Amount remaining (N) =?

Log (N₀/N) = Kt/2.303

Log (32/N) = 0.0485 × 71.5 / 2.303

Log (32/N) = 3.46775 / 2.303

Log (32/N) = 1.5058

Take the antilog of 1.5058

32/N = antilog (1.5058)

32/N = 32.05

Cross multiply

32 = N × 32.05

Divide both side by 32.05

N = 32 / 32.05

N = 0.998 ≈ 1 g

Thus, the amount remaining after 71.5 days is approximately 1 g.