Question

A. The half-life of iron-61 is 6 minutes. If a sample originally contains 8 g of the isotope, how much remains after 12 minutes? Write 2 - 3 sentences explaining your reasoning.

b. The ratio of carbon-14 to nitrogen-14 in an artifact is 1:7. In other words 1/8 of the original carbon-14 remains. Given the half-life of carbon-14 is 5730 years, how old is the artifact? Write 2 - 3 sentences explaining your reasoning.

Answer 1

a) in 6 minutes 8 grams will turn to 4 grams of pure iron-61. Then in the next 6 minutes these 4 grams will turn to 2 grams.

b) Carbon-14 turns to Nitrogen-14, i.e. in the very beginning of the artifact's life there were 8 parts of carbon and no nitrogen-14.

8 parts existed until 5730

4 parts existed until 5730*2

2 parts existed till 5730*3

1 part exists since the year 5730*3=17190 , so it's 17190 y.o.

Answer 2

PART A)

One half life means the times after which half of the radioactive substance will decay. So here we have 8 g of isotope present in the sample

Half life of the sample = 6 minutes

Now we need to find the amount after 12 minutes

So it will complete the cycle of two half life

So it will decay to half two times in given time

So final amount will be m = 2 gram

PART B)

After every half life of the cycle the original amount will be reduced to half of initial

So here if the proportion of carbon-14 is 1/8 times of initial

So we can say it is
`(1)/(8) = ((1)/(2))^3`

so it will complete half life 3 times

so it is 5730 yrs + 5730 yrs + 5730 yrs

so total life is T = 17190 years