The radioactive isotope, phosphorus-32, has a half-life of 14.26 days. How much phosphorus-32 will remain after 60 days?
Please help.
k = 0.693/t1/2
Substitute k into the below equation.
ln(No/N) = kt
No = The problem doesn't say how much to start with so start with 100 grams.
N = what you end up with after 60 days.
k from above.
t = 60 days
Starting with 100 g will give you a fraction and that x 100 will give you a percentage of the amount you start with that will remain after 60 days.
3.743
To determine how much phosphorus-32 will remain after 60 days, we can use the concept of half-life. The half-life of 14.26 days tells us that after 14.26 days, half of the phosphorus-32 will decay and half will remain.
We can calculate the number of half-lives in 60 days by dividing the total time (60 days) by the half-life (14.26 days).
60 days / 14.26 days = 4.2066 (approximately)
This means that there are approximately 4.2066 half-lives in 60 days.
To find how much phosphorus-32 remains after 60 days, we can use the formula:
Remaining amount = Initial amount * (1/2)^(number of half-lives)
Since we don't have the initial amount, we can assume it to be 1 (for simplicity).
Remaining amount = 1 * (1/2)^(4.2066)
Using a calculator:
Remaining amount ≈ 0.125
So after 60 days, approximately 0.125 (or 12.5%) of the initial amount of phosphorus-32 will remain.