Question
Carbon-14 is a radioactive substance produced in the Earth's atmosphere and then absorbed by plants and animals on the surface of the earth. It has a half-life (the time it takes for half the amount of a sample to decay) of approximately 5730 years. Using this known piece of information, scientists can date objects such as the Dead Sea Scrolls. The function N = N0e-λt represents the exponential decay of a radioactive substance. N is the amount remaining after time t in years, N0 is the initial amount of the substance and λ is the decay constant.
1. Find the rate of change of an initial amount of 1 gm of carbon-14 found in the scrolls, if the decay constant is given as λ = 1.21 x 10-4.
2. If the percentage of carbon-14 atoms remaining in a sample is 79%, how old is the sample?
1. Find the rate of change of an initial amount of 1 gm of carbon-14 found in the scrolls, if the decay constant is given as λ = 1.21 x 10-4.
2. If the percentage of carbon-14 atoms remaining in a sample is 79%, how old is the sample?
Answers
MathMate
1.
dN/dt = -λN
so the initial rate of change
= -1.21*10^4 × 1 g/year
= -0.000121 g/year
2.
N = N<sub>0</sub>e<sup>-λt</sup>
e-λt = N/N<sub>0</sub>
taking logs on both sides
-λt = ln(N/N<sub>0</sub>)
t = -ln(N/N<sub>0</sub>)/λ (before present, year 1950)
= -ln(0.79)/(1.21*10<sup>4</sup>) (BP)
= 1948 (BP)
dN/dt = -λN
so the initial rate of change
= -1.21*10^4 × 1 g/year
= -0.000121 g/year
2.
N = N<sub>0</sub>e<sup>-λt</sup>
e-λt = N/N<sub>0</sub>
taking logs on both sides
-λt = ln(N/N<sub>0</sub>)
t = -ln(N/N<sub>0</sub>)/λ (before present, year 1950)
= -ln(0.79)/(1.21*10<sup>4</sup>) (BP)
= 1948 (BP)