A meteorite has landed at UOIT! Chemical analysis shows that it contains 1
gram of potassium and 1E-5 grams of argon, assumed to be present entirely
from the decay of K-40. Assuming that no argon has escaped the meteorite,
estimate the age of the meteorite.
Data given:
Natural potassium has an atomic weight of 39.089 and contains 0.0118 atomic
percent K-40, which has 2 decay modes:
40
K →40
Ca + β +anti-neutrino
40
K + e
-
→40
Ar* + neutrino →40
Ar + γ
On average there are 12 gamma-rays emitted to every 100 beta-particles
emitted.
To estimate the age of the meteorite, we need to use the concept of radioactive decay and half-life. The key information we have is that the argon (Ar) present in the meteorite is derived entirely from the decay of potassium-40 (K-40).
First, let's calculate the number of atoms of K-40 present in 1 gram of potassium. To do this, we need to determine the number of moles of potassium.
Given:
Atomic weight of potassium (K) = 39.089 g/mol
Atomic percent of K-40 = 0.0118%
To calculate the number of moles:
1 gram of K / (Atomic weight of K) = 1 g / 39.089 g/mol ≈ 0.02558 mol
Next, we need to calculate the number of moles of K-40:
0.02558 mol * 0.0118% = 3.0152 x 10^-6 mol
Now, we can convert the number of moles of K-40 to the number of K-40 atoms:
Number of K-40 atoms = (Number of moles of K-40) * Avogadro's number
Number of K-40 atoms = (3.0152 x 10^-6 mol) * 6.022 x 10^23 mol^-1 ≈ 1.8156 x 10^18 atoms
Since there is an equal number of argon atoms produced from the decay of K-40, we have 1.8156 x 10^18 atoms of Ar.
Given that the ratio of beta-particles (β) emitted to gamma-rays (γ) emitted is 100:12, we can calculate the number of beta-particles emitted as follows:
Number of beta-particles emitted = (Number of gamma-rays emitted) * (100 / 12)
Number of beta-particles emitted = (Number of gamma-rays emitted) * 8.333
We know that for each beta-particle emitted, one argon atom is produced. Therefore, the number of argon atoms is equal to the number of beta-particles emitted:
Number of Ar atoms = Number of beta-particles emitted ≈ 1.8156 x 10^18
Now, we can calculate the fraction of K-40 atoms that have decayed to form Ar:
Fraction of K-40 atoms decayed = (Number of Ar atoms) / (Total number of K-40 atoms)
Fraction of K-40 atoms decayed = (1.8156 x 10^18) / (1.8156 x 10^18) ≈ 1
Since we assume that no argon has escaped the meteorite, we can say that the fraction of K-40 atoms that have decayed is equal to the fraction of K-40 atoms remaining:
Fraction of K-40 atoms remaining = 1 - (Fraction of K-40 atoms decayed) ≈ 1 - 1 = 0
The half-life of K-40 is approximately 1.28 billion years. The half-life is the time it takes for half of the K-40 atoms to decay. Since we have found that all the K-40 atoms have decayed, we can estimate the time as follows:
Time = (Half-life of K-40) * (Number of half-lives)
Time = 1.28 billion years * (Number of half-lives)
Since the fraction of K-40 atoms remaining is 0, we can estimate the number of half-lives as follows:
Number of half-lives = log (Fraction of K-40 atoms remaining) / log (0.5)
Using logarithms:
Number of half-lives = log10(0) / log10(0.5) = undefined
Therefore, since all the K-40 atoms have decayed, we can estimate that the age of the meteorite is greater than the half-life of K-40, which is approximately 1.28 billion years.