Question

some fossil bones contain 1/8 of their original amount of carbon-14.how many half-lives have passed?how old are the bones?

Answer 1

1... 1/2...1/4...1/8. That's 3 half lives.. therefore do 5730 *3 since each half life requires 5730 years , in order to find the total : 17190 years

Answer 2

3 half-lives have passed.

The bones are 17 190 years old.

Step-by-step explanation:

The half-life of a radioactive element is the time required for the substance to lose half of its initial activity. In other words, the half-life is denoted by this symbol
`t_{(1)/(2) }`

Let's take the half-life of carbon as
`(1)/(2)`

After a certain number of years, the activity degenerates to half of the original activity. So it will be
`t_{(1)/(2) } = (1)/(2) *(1)/(2) \\ =(1)/(4)`

After some years, the activity will be
`(1)/(8)`

Thus three half-lives have passed. A simple formula to remember the pattern is to use the exponential expression :

`(1)/(2) ^(x) = (1)/(8)`

solving for x gives x = 3

hence there half-lives.

For carbon,
`t_{(1)/(2) } = 5730 years`

since there half-lives have passed, the age of the sample will be
`t_{(1)/(2) } after three years = 5730 * 3\\ = 17 190 years`

The sample will be 17 190 years.