The radioactive nuclide 18F decays by first order kinetics with a half life of 1.83 hr.
How many hours will it take for 68.5% of a give quantity of 18F to decay?
k for the decay = 0.693/1.83 = approx 0.4 but you need a better answer than that. Then,
ln(No/N) = kt
You can pick any number you like for No, then N is that number x (100-68.5) but I suggest you make it easy and do this.
No = 100
N = 31.5
k is from above
t = ? solve for this.
Remember to calculate k more accurately.
To find out how many hours it will take for 68.5% of a given quantity of 18F to decay, we can use the concept of half-life and the decay equation.
First, let's understand the concept of half-life. The half-life of a radioactive substance is the time it takes for half of the initial quantity of the substance to decay. In this case, we are given that the half-life of 18F is 1.83 hours.
Now, if we have a given quantity of 18F, and we want to find out how much will decay after a certain time, we use the decay equation:
N = N0 * e^(-kt)
where N is the final quantity, N0 is the initial quantity, k is the rate constant, and t is the time.
In our case, we are given that 68.5% of the substance will decay. This means that only 31.5% will remain.
Now, let's substitute the given values into the equation:
0.315 * N0 = N0 * e^(-kt)
We can simplify this equation as:
e^(-kt) = 0.315
Now, we need to solve for t, the time it will take for 68.5% of the substance to decay.
To do this, we take the natural logarithm of both sides of the equation:
ln(e^(-kt)) = ln(0.315)
This simplifies to:
-kt = ln(0.315)
Finally, we solve for t by dividing both sides of the equation by (-k):
t = ln(0.315) / (-k)
Now, we know that the half-life of 18F is 1.83 hours. The rate constant (k) can be determined using the relationship between half-life and rate constant:
k = ln(2) / half-life
Substituting the given values:
k = ln(2) / 1.83
With this value of k, we can calculate the time (t) it will take for 68.5% of the substance to decay:
t = ln(0.315) / (-ln(2) / 1.83)
Evaluating this expression will give us the answer in hours.