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?

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 %

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