Potassium-40 has a half-life of approximately 1.25 billion years. Approximately how many years will pass before a sample of potassium-40 contains one-sixteenth the original amount of parent isotope?

2.5 billion

To solve this problem, we need to use the concept of half-life.

The half-life of potassium-40 is approximately 1.25 billion years, which means that in that time period, half of the potassium-40 will decay into another element (argon-40 in this case).

We need to find out how many half-lives it will take for the sample to contain one-sixteenth (1/16) of the original amount of parent isotope.

To do this, we can apply the formula:

Number of Half-lives = (Log(Ratio of Final Amount to Initial Amount)) / (Log(1/2))

In this case, the ratio of final amount to the initial amount is 1/16.

Number of Half-lives = (Log(1/16)) / (Log(1/2))

Calculating this using base 10 logarithms, we get:

Number of Half-lives = (-1.2041) / (-0.3010)
Number of Half-lives ≈ 4

Hence, it will take approximately 4 half-lives for a sample of potassium-40 to contain one-sixteenth the original amount of the parent isotope.

To calculate the number of years, we can multiply the number of half-lives by the half-life of potassium-40:

Number of Years = Number of Half-lives * Half-life of Potassium-40

Number of Years ≈ 4 * 1.25 billion years
Number of Years ≈ 5 billion years

Therefore, approximately 5 billion years will pass before a sample of potassium-40 contains one-sixteenth the original amount of the parent isotope.

Since (1/2)^4 = 1/16, that would require 4 half-lives, which is 5 billion years.

5 billion

5bil