The radioisotope radon-222 has a half-life of 3.8 days. How much of a 98.4 g sample of radon -222 would be left after approximately 19 days?
98.4 * (1/2)^(19/3.8) = 3.075 g
Cf = Ci(e^-kt)
Ci = 98.4 gms
Cf = ?
k = (0.693/3.8)days^-1 = 0.8663 da^-1
t = 19 days
Cf = Ci e^-kt
Cf = 94.4gms e^-[(0.8663da^-1)(19da)]
Cf = (94.4)(7.11 x 10^-8)gms
Cf = 6.7 x 10^-6 gm remains
two errors, DrRebel:
(0.693/3.8) ≠ 0.8663
it was 98.4, not 94.4
To determine how much of the 98.4 g sample of radon-222 would be left after approximately 19 days, we can use the concept of half-life.
The half-life of a radioactive substance is the time it takes for half of the sample to decay. In this case, the half-life of radon-222 is 3.8 days.
To calculate the amount of radon-222 remaining after a certain number of half-lives, we can use the formula:
Amount remaining = Initial amount × (1/2)^(number of half-lives)
First, let's find out how many half-lives occur in 19 days:
Number of half-lives = 19 days ÷ 3.8 days per half-life
Number of half-lives = 5
Now we can use the formula to find the remaining amount of radon-222:
Amount remaining = 98.4 g × (1/2)^5
Amount remaining = 98.4 g × 1/32
Amount remaining ≈ 3.075 g
Therefore, approximately 3.075 grams of the 98.4 g sample of radon-222 would be left after approximately 19 days.