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The radioactive element, Beatrium, decays over time. It has a half-life of 4078 years. An archeologist samples a relic that has 52% of its original amount of Beatrium. How old is the relic?

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Final answer:

To determine the age of a relic with 52% of its original amount of the radioactive element Beatrium left, we calculate the number of half-lives that have passed and multiply it by the half-life period (4078 years). The relic is approximately 3792.54 years old.

Step-by-step explanation:

The question asks us to determine the age of a relic by using the half-life of a radioactive element it contains, known as Beatrium. Given that the relic has 52% of its original amount of Beatrium and that the half-life of Beatrium is 4078 years, we need to calculate how many half-lives have passed to reach this percentage.

The general formula for radioactive decay, which helps us find the number of half-lives (n) that have passed, is:

N = N

0

(1/2)

n

where N is the remaining amount of substance, N0 is the initial amount, and (1/2) is raised to the power of the number of half-lives (n). Since the relic has 52% of its Beatrium remaining, we can express this as N/N0 = 0.52. We can find n by solving this equation:

0.52 = (1/2)n

To find n, we take the logarithm of both sides:

log(0.52) = n log(0.5)

n = log(0.52) / log(0.5)

By inserting the appropriate values for the logs, we find that n is approximately equal to 0.93 half-lives.

We then multiply the number of half-lives by the half-life duration to find the age of the relic:

Age = n × half-life

Age ≈ 0.93 × 4078 years

Age ≈ 3792.54 years

The relic is therefore approximately 3792.54 years old.

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