The half-life of a substance is the amount of time it takes for half of the substance to decay.
If an initial population of size P has a half-life of d years (or any other unit of time), then the formula to find the final number A in t years is given by
A = P(½)^{ᵗ⁄d}
Problem 1 :
The half-life of carbon-14 is approximately 6000 years. How much of 800 g of this substance will remain after 30,000 years?
Solution :
Half-Life Decay Formula :
A = P(½)^{ᵗ⁄d}
Substitute.
P = 800
t = 30000
d = 6000
Then, we have
A = 800(½)^{³⁰⁰⁰⁰⁄₆₀₀₀}
= 800(0.5)^{5}
= 800(0.03125)
= 25
So, 25 g of carbon-14 will remain after 30,000 years.
Problem 2 :
A certain radioactive substance has a half-life of 12 days. This means that every 12 days, half of the original amount of the substance decays. If there are 128 milligrams of the radioactive substance today, how many milligrams will be left after 48 days?
Solution :
Half-Life Decay Formula :
A = P(½)^{ᵗ⁄d}
Substitute.
P = 128
t = 48
d = 12
Then, we have
A = 128(½)^{⁴⁸⁄₁₂}
= 128(0.5)^{4}
= 128(0.0625)
= 8
So, 8 milligrams of radio active substance will be left after 48 days.
Problem 3 :
Cafine has a half-life of 2 hours, meaning the amount of caffeine in the body decreases by 50% every 2 hours. Write a general formula, C(t), to represent the amount of caffeine (in mg) after t hours. Use C_{o} to represent the initial amount. State the decay factor and the decay rate.
Solution :
Half-life Decay Formula :
A = P(½)^{ᵗ⁄d}
Substitute.
A = C(t)
P = C_{o}
d = 2 (half-life time)
Then, we have
C(t) = C_{o}(½)^{ᵗ⁄2}
C(t) = C_{o}[(½)^{1⁄2}]^{ᵗ}
C(t) = C_{o}(0.707)^{ᵗ}
Growth factor = 0.707
Growth rate = 1 - 0.707
= 0.293
= 29.3% per hour
Problem 4 :
The function M models the mass of the radioactive isotope carbom-14 in a sample, in picograms, t years after the initial measurement. How much time, in years, does it take for the mass of carbon-14 in the sample to decrease to 5 picograms?
Solution :
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