The percentage of an iodine-131 sample that will remain after 40.20 days is 3.125%
Half-life (t½) = 8.040 days.
Time (t) = 40.20 days
[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{40.20}{8.040}\\\\[/tex]
Thus, 5 half-lives has elapsed.
Number of half-lives (n) = 5
[tex]N = \frac{1}{2^{n} } * N_{0} \\\\N = \frac{1}{2^{5} } * N_{0} \\\\N = \frac{1}{32} * N_{0}\\\\[/tex]
Divide both side by N₀
[tex]\frac{N}{N_{0}} = \frac{1}{32} \\\\\frac{N}{N_{0}} = 0.03125\\\\[/tex]
Multiply by 100 to express in percent
[tex]\frac{N}{N_{0}} =[/tex] 0.0312 × 100
Therefore, the percentage of an iodine-131 sample that remains is 3.125%
[tex]\frac{N}{N_{0}}[/tex] => Fraction remaining
Learn more: https://brainly.com/question/14883322
