Answer:
Explanation:
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 %
