The level of thorium in a sample decreases by a factor of one-half every 4.2 million years. A meteorite is discovered to have only 7.6% of its original thorium remaining. How old is the meteorite?
Let N be the number of half-lives in its age
(1/2)^N = 0.076
Solve for N and multiply N by 4.2 million years.
N = log 0.076 / log 0.5 = 3.72
Age = 15.6 million years
To determine the age of the meteorite, we can use the concept of radioactive decay and the given information. The level of thorium in the sample decreases by a factor of one-half every 4.2 million years. We are told that the meteorite has only 7.6% of its original thorium remaining.
Let's denote the initial amount of thorium in the meteorite as "A" and the current amount as "B." From the information given, we know that B = 0.076A, since 7.6% is equivalent to 0.076.
The number of half-lives elapsed can be found using the formula B = A * (1/2)^(t/h), where "t" is the age in years and "h" is the half-life. Rearranging the formula, we get t = h * log2(B/A).
Plugging in the values, we have h = 4.2 million years, A = 100% = 1, and B = 0.076.
t = 4.2 million * log2(0.076/1)
= 4.2 million * log2(0.076)
Now, we can use a calculator to find the logarithm and calculate the age of the meteorite. Using base 10 logarithm as an example:
t ≈ 4.2 million * (-2.12)
≈ -8.924 million years
Since time cannot be negative, we take the absolute value of -8.924 million years to find the age of the meteorite.
Therefore, the meteorite is approximately 8.924 million years old.