1)Fluorine-21 has a half life of approximately 5 seconds. If you start with 100g, how much would remain after 10 seconds?
What’s the answr

25 g

Explanation:

The following data were obtained from the question:

Half life (t½) = 5 s

Original amount (N₀) = 100 g

Time (t) = 10 s

Amount remaining (N) =..?

Next, we shall determine the decay constant for the disintegration. This can be obtained as follow:

Half life (t½) = 5 s

Decay constant (K) =.?

K = 0.693 / t½

K = 0.693 / 5

K = 0.1386 s¯¹

Finally, we shall determine the amount of the isotope that remained after 10 s as follow:

Original amount (N₀) = 100 g

Time (t) = 10 s

Decay (K) = 0.1386 s¯¹

Amount remaining (N) =..?

Log (N₀/N) = kt /2.303

Log (100/N) = (0.1386 × 10) / 2.303

Log (100/N) = 1.386 / 2.303

Log (100/N) = 0.6018

Take the anti log of 0.6018

100/N = antilog (0.6018)

100/N = 4

Cross multiply

100 = N × 4

Divide both side by 4

N = 100/4

N = 25 g

Therefore, the amount remaining after 10 s is 25 g

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