The half-life of a radioactive substance is 14 days. After 15 days a sample of the substance has been reduced to a mass of 7 mg. What is the initial mass? When will there only be 0.2 left?

Answer 1

7 = a₀(0.5)^15/14

ln 7 = ln (a₀(0.5)^15/14)

ln 7 = ln a₀ + (15/14)(ln 0.5)

1.9459 ≈ ln a₀ + (15/14)(-0.6931)

1.9459 ≈ ln a_0¡ + -0.7426

2.6885 ≈ ln a₀

14.710 ≈ a₀

0.2 = 14.710(0.5)^t/7

0.0136 ≈ (0.5)^t/7

ln 0.136 = (t/7)(ln 0.5)

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