Two radioactive nuclei a and b are present in equal numbers to begin with. three days later, there are 4.04 times as many a nuclei as there are b nuclei. the half-life of species b is 1.37 days. find the half-life of species a (in days).

The half-life of A is 17.1 days.

Explanation:

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The half-life of B is 1.73 days.

Let´s write the elapsed time (3 days) in terms of half-lives of B:

1.37 days = 1 half-life B

3 days = (3 days · 1 half-life B / 1.37 days) = 2.19 half-lives B.

After 3 days, the amount of A in terms of B is the following:

A = 4.04 B

The amount of B after 3 days can be expressed in terms of the initial amount of B (B0) and the number of half-lives (n):

B after n half-lives = B0 / 2ⁿ

Then after 2.19 half-lives:

B = B0 /2^(2.19)

In the same way, the amount of A can also be expressed in terms of the initial amount and the number of half-lives:

A = A0 / 2ⁿ

Replacing A and B in the equation:

A = 4.04 B

A0 / 2ⁿ = 4.04 · B0 / 2^(2.19)

Since A0 = B0

A0 / 2ⁿ = 4.04 · A0 / 2^(2.19)

Dividing by A0:

1/2ⁿ = 4.04 / 2^(2.19)

Multipliying by 2ⁿ and dividing by  4.04 / 2^(2.19):

2^(2.19) / 4.04 = 2ⁿ

Apply ln to both sides of the equation:

ln( 2^(2.19) / 4.04) = n ln(2)

n = ln( 2^(2.19) / 4.04) / ln(2)

n = 0.1756

Then, if 3 days is 0.1756 half-lives of A, 1 half-life of A will be:

1 half-life ·(3 days / 0.1756 half-lives) = 17.1 days

The half-life of A is 17.1 days.

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