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.