Lead-214 has a half-life of 43.5 seconds. What percent of a sample of this isotope would be left after a time equivalent to 7 half-lives?

Lead has an half life of 43.5 seconds

The radioactive decay law says

N (t) = No•2^ (-t / t½)

Hence, the percent of a sample of the Lead-214 isotope that decay after a time equivalent to 7 half-lives

N (t) = No•2^ (-t / t½)

N (t) / No = 2^ (-t / t½)

N (t) / No = 2^-7

N (t) / No = 7.81 * 10^-3

To percentage

N (t) / No = 7.81 * 10^-3 * 100

N (t) / No = 0.781 %