For a PET diagnostic imaging is used 18F with an activity of 5,2 mCi and a T1/2 of 110 min.. How much fluorine 18 is there at the beginning of the exam?
Those are the information that i know, it's a problem that i try to resolve, but i arrived to a very low quantity of 18F (like 10^-9 g i don't remember the exact number), i don't know why, but it's impossible (i found 5,2 mCi and T1/2 on the web)
To calculate the amount of fluorine-18 (18F) present at the beginning of the PET exam, you need to use the concept of radioactive decay and the decay constant, which is related to the half-life.
The decay of a radioactive isotope can be described by the equation: N(t) = N0 * e^(-λt), where N(t) is the remaining quantity at time t, N0 is the initial quantity, e is the base of the natural logarithm (approximately 2.71828), λ is the decay constant, and t is the time.
In this case, you know the activity of the 18F at the time of the exam, which is given as 5.2 mCi. Activity is defined as the rate at which the radioactive substance decays, and it is measured in units called curies (Ci). In the SI unit, 1 Ci is equal to 3.7 x 10^10 decays per second.
To convert the activity from Ci to disintegrations per second (Bq), you multiply by the conversion factor: 1 Ci = 3.7 x 10^10 Bq.
Thus, 5.2 mCi = 5.2 * (3.7 x 10^10) Bq = 1.924 x 10^11 Bq.
Given that the half-life of fluorine-18 is 110 minutes, you can calculate the decay constant (λ) using the equation: λ = 0.693 / T1/2. Plugging in the value of T1/2 = 110 min, we get:
λ = 0.693 / 110 min ≈ 0.00629 min^-1.
Now, let's solve for the initial quantity (N0):
N(t) = N0 * e^(-λt).
At the beginning of the exam, t = 0 minutes. Plugging in the known values:
N(0) = N0 * e^(-0.00629 * 0).
Since e^0 = 1, we have:
N0 = N(0) = 1.924 x 10^11 Bq.
So, there is approximately 1.924 x 10^11 Bq of fluorine-18 present at the beginning of the exam.
I hope this helps! Let me know if you have further questions.