The half-life of a certain radioactive element is 800 years. How old is an object if only 12.5% of the radioactive atoms in it remain?

a
3200 years
b
6400 years
c
1600 years

b) 6400 years

To find out how old the object is, we can use the formula for exponential decay:

N = N0 * (1/2)^(t / T)

Where:
N = Remaining amount of radioactive atoms
N0 = Initial amount of radioactive atoms
t = Time passed
T = Half-life of the radioactive element

In this case, we know that only 12.5% of the radioactive atoms remain, which means N = 0.125 * N0.

0.125 * N0 = N0 * (1/2)^(t / 800)

We can cancel out N0 from both sides of the equation:

0.125 = (1/2)^(t / 800)

To solve for t, we can take the logarithm of both sides:

log(0.125) = log((1/2)^(t / 800))

Using the logarithmic property log(a^b) = b * log(a):

log(0.125) = (t / 800) * log(1/2)

Dividing both sides by log(1/2):

log(0.125) / log(1/2) = t / 800

Now, we can solve for t:

t = 800 * (log(0.125) / log(1/2))

Using a calculator, we find that t ≈ 6400 years.

Therefore, the correct answer is b) 6400 years.

To determine the age of the object based on the remaining radioactive atoms, you need to use the concept of half-life. The half-life of a radioactive element is the amount of time it takes for half of the atoms in a sample to decay.

In this case, the half-life of the radioactive element is given as 800 years. This means that after 800 years, half of the radioactive atoms in the object will have decayed, leaving the other half remaining.

Now, if only 12.5% of the radioactive atoms remain, we need to find out how many half-lives have passed. To do this, we need to determine how many times we need to divide 100 (representing 100% of atoms) by 12.5 to get the number of half-lives.

100 / 12.5 = 8

So, 8 half-lives have passed. Since each half-life is 800 years, we can multiply the number of half-lives by the half-life time to find the total time that has passed.

8 x 800 = 6400 years

Therefore, the object is 6400 years old. The correct answer is option b: 6400 years.