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.
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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