The half life of ^{14}C is 5730 years. What fraction of the ^{14}C nuclei in a sample are left after 7060 years?

the formula is A = Ao (1/2)^(t/h)

where, A is the final amountAo is the initial amountt = timeh= half lifeso, A = Ao (1/2)^(7060/5730)or, (A/Ao) = (1/2)^ 1.232 = 0.4257so, the fraction of ^{14}C left = 0.4257 of initial amount.Please best my answer if helpful. Happy New Year :)

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up