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

The half-life of a radioactive substance refers to the time taken for the substance to Decay to half of its initial size. Hence, the mass of the substance left after 71.5 days is 1 gram.

After 14.3 days ;

After 14.3 × 2 = 28.6 days :

After 14.3 × 3 days = 42.9 days :

After 14.3 × 4 days = 57.2 days :

After 14.3 × 5 days = 71.5 days :

Hence, the mass of the sample left after 71.5 days would be 1 gram.

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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.

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