You start with 420 mg of isotope which has half life of 20 days. How many mg will be left after 40 days

105 mg

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 420 mg

Half life (t½) =20 days

Time (t) = 40 days

Amount remaining (N) =..?

Next, we shall determine the rate constant K, for the disintegration.

This can be obtained as follow:

Half life (t½) =20 days

Rate constant (K) =?

K = 0.693/ t½

K = 0.693/20

K = 0.03465 /day

Finally, we shall determine the amount remaining after 40 days as follow:

Original amount (N₀) = 420 mg

Rate constant (K) = 0.03465 /day

Time (t) = 40 days

Amount remaining (N) =..?

Log(N₀/N) = kt/2.3

Log (420/N) = (0.03465 x 40)/2.3

Log (420/N) = 1.386/2.3

Log (420/N) = 0.6026

420/N = antilog (0.6026)

420/N = 4

Cross multiply

420 = N x 4

Divide both side by 4

N = 420/4

N = 105 mg

Therefore, the amount remaining after 40 days is 105 mg.

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