Question

The radioactive isotope 14C has a half-life of approximately 5715 years. Now there are 50g of 14C.(1) How much of it remains after 1600 years? (Round your answer to three decimal places.) g Tries 0/99(2) How much of it remains after 16000 years? (Round your answer to three decimal places.) g Tries 0/99

Answer 1

Given

The half life is T=5715 years

The initial amount is N=50 g

To find

1) How much of it remains after 1600 years?

(2) How much of it remains after 16000 years?

Step-by-step explanation

The amount of carbon remains after t yaers is

`N^(\prime)=N((1)/(2))^{(t)/(T)}`

1. Thus putting t=1600 years

`\begin{gathered} N^(\prime)=50((1)/(2))^{(1600)/(5715)} \\ \Rightarrow N^(\prime)=41.208\text{ g} \end{gathered}`

2.Putting t=16000

`\begin{gathered} N^(\prime)=50((1)/(2))^{(16000)/(5715)} \\ \Rightarrow N^(\prime)=7.184\text{ g} \end{gathered}`

Conclusion

1.Amount remains after 1600 year is 41.208 g

2.Amount remains after 16000 year is 7.184 g