The half-life of Nobelium-259 is 58 minutes. After 3 hours a sample still contains 10 mg. What was the initial mass of the sample, and how much will remain after another 8 hours?

Question

The half-life of Nobelium-259 is 58 minutes. After 3 hours a sample still contains 10 mg. What was the initial mass of the sample, and how much will remain after another 8 hours?

Answer 1

This problem will use the equation for continuous exponential decay: P=P_oe^rt

Given the information about the half life, you get the following

1/2Po=P_oe^r(58) P_o cancels out

1/2=e^r(58) Take the ln of both sides

ln (1/2) = ln (e^r(58)) The exponent can move out front

ln (1/2) = 58r (ln e) Simplify, lne = 1

ln (1/2) = 58r divide both sides of the equation by 58

r=-.012

What is the initial mass of the sample? Use the second piece of information given to find the value for P_o. There are 10 grams left after 3 hours. You have to convert hours to minutes so the units are the same.

10=P_oe^{(-0.012)(180)} Divide both sides of the equation by e^{(-0.012)(180)}

P_o = 86.71 mg

How much will remain after 8 hours? We will need to solve for P using the same equation. Convert t to minutes. Evaluate using the calculator

P=86.71e^{r(-.012)(480)}

P=0.27 mg

Answer 2

10 = a₀(0.5)^3(60)/58

Can you calculate a₀ and answer?

Answer 3

10 = P (.5)^(180/58)

10 = P (.5)^3.103 = 0.116A

P = initial mass =10/.116 = 86 mg

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A = 86 (.5)^(480/58) = 0.28 mg remaining after 8 hours