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/half life = ?
Then ln(No/N) = kt
You know k.
No = initial amount. It is given but it may be in your problem and you didn't include that in your post. If so put that in. If not, you can't answer in amounts except in %. In that case place a 100 for No.
N = amount left
t = 60 days. After you find N,
Then N/No will be the fraction left.
To determine how much phosphorus-32 will remain after 60 days, we can use the concept of radioactive decay and the formula for calculating the amount of a radioactive substance remaining after a certain time.
The formula for calculating the amount of a radioactive substance remaining after a certain time is:
N = N0 * (1/2)^(t / T)
Where:
N = amount of the radioactive substance remaining after time t
N0 = initial amount of the radioactive substance
t = time that has passed
T = half-life of the radioactive substance
In this case, the initial amount of phosphorus-32 (N0) is not specified, so we'll assume it to be 1 (you can take any value for N0 as long as you use it consistently in the calculation).
Using the given information, we know that the half-life of phosphorus-32 is 14.26 days, and we want to find the amount remaining after 60 days (t = 60 days).
Plugging the values into the formula:
N = 1 * (1/2)^(60 / 14.26)
Now let's calculate it.