If the physical half-life of sulfur-35 is 87.1 days, and the biologic half-life in the testicle is 632 days, what is its effective half-life in this organ?
I know i have to find the time and then the half-life.
But im having trouble finding the time
A(t)=Ao x e ^ (-0.693) (t) / T1/2 [half=life]
I'm thinking to just divide 632 days by 87.1 days to determine the half-life
25
To find the effective half-life in the testicles, you need to determine the time it takes for the activity of sulfur-35 to decrease by half in this organ.
The formula you have provided:
A(t) = Ao * e^(-0.693 * (t) / T1/2)
Let's break down the formula:
A(t) represents the activity of sulfur-35 at time t.
Ao is the initial activity of sulfur-35.
e is the base of the natural logarithm (approximately 2.71828).
-0.693 is a constant derived from the decay constant of radioactive substances.
(t) represents the time that has elapsed.
T1/2 is the half-life of sulfur-35.
To solve for t in the formula, we can rearrange it as follows:
t = (ln (A(t) / Ao)) * (-T1/2 / 0.693)
From the given information, we know the physical half-life of sulfur-35 is 87.1 days and the biological half-life in the testicles is 632 days.
Plugging these values into the formula:
t = (ln (1/2)) * (-632 / 87.1)
ln (1/2) is equal to -0.693 (since ln (1/2) = -0.693).
t = -0.693 * (-632 / 87.1)
Calculating this expression:
t ≈ 5.029
Therefore, the effective half-life of sulfur-35 in the testicles is approximately 5.029 days.