The radioactive bismuth isotope ²¹⁰Bi has a half-life of 5 days. If there is 100 milligrams of ²¹⁰Bi present at t = 0, then the amount f(t) remaining after t days is given by: $f ( t ) = 100 ( 2 ) ^ { - \frac { t } { 5 } }$ How much ²¹⁰Bi remains after 25 days? A) 3.125 milligrams B) 12.125 milligrams C) 5.125 milligrams D) 8.125 milligrams E) 2.125 milligrams

Question

Quizplus · Accepted Answer

After 25 days, which is 5 half-lives (25/5), the amount of ²¹⁰Bi remaining is $100 \times (2)^{-5} = 100 \times \frac{1}{32} = 3.125$ milligrams.

The Radioactive Bismuth Isotope 210 Bi Has a Half-Life f(t)=100(2)−t5f ( t ) = 100 ( 2 ) ^ { - \frac { t } { 5 } }f(t)=100(2)−5t​

The Radioactive Bismuth Isotope ²¹⁰Bi Has a Half-Life $f ( t ) = 100 ( 2 ) ^ { - \frac { t } { 5 } }$