In each of the following questions, calculate theamount remaining from a 100 mg sample of thegiven radioisotope after the specified time.a 105Rh has a half-life of 35 h. Calculate themass remaining after 70 h.b 209Po has a half-life of 105 y. Calculate themass remaining after 420 y.c 219Rn has a half-life of 3.96 s. Calculate themass remaining after 39.6 s.
Question
In each of the following questions, calculate theamount remaining from a 100 mg sample of thegiven radioisotope after the specified time.a 105Rh has a half-life of 35 h. Calculate themass remaining after 70 h.b 209Po has a half-life of 105 y. Calculate themass remaining after 420 y.c 219Rn has a half-life of 3.96 s. Calculate themass remaining after 39.6 s.
Solution
a) The half-life of 105Rh is 35 hours. This means that after 35 hours, half of the original amount will remain. After 70 hours, which is two half-lives, the amount will have halved twice.
First half-life: 100 mg / 2 = 50 mg Second half-life: 50 mg / 2 = 25 mg
So, after 70 hours, 25 mg of 105Rh will remain.
b) The half-life of 209Po is 105 years. After 420 years, which is four half-lives, the amount will have halved four times.
First half-life: 100 mg / 2 = 50 mg Second half-life: 50 mg / 2 = 25 mg Third half-life: 25 mg / 2 = 12.5 mg Fourth half-life: 12.5 mg / 2 = 6.25 mg
So, after 420 years, 6.25 mg of 209Po will remain.
c) The half-life of 219Rn is 3.96 seconds. After 39.6 seconds, which is ten half-lives, the amount will have halved ten times.
After ten half-lives, the remaining amount can be calculated as: 100 mg / (2^10) = 0.09765625 mg
So, after 39.6 seconds, approximately 0.098 mg of 219Rn will remain.
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