Derivation for Time to Max Radioactivity Transient Equilibrium

[matthewt]

Hi,

I'm having a bit of difficulty deriving the time to max activity for the case of transient equilibrium for a parent-daughter.

This is where I want to get to , tm = (1 / (λ1-λ2)) * 1n(λ1/λ2)

I believe there is an alternative equation for tm as well expressed in terms of half-life.

I would be gratefeul if someone could walk me through a derivation,

BW,
Matt

 

[voko]

Do you mean that you have some parent substance, with a given half-life time, which produces some daughter substance, with another given half-life time (which then decays into an inactive substance), and you need to find out the time it gets for the mix to attain maximum activity?

 

[matthewt]

that's right. For the case of a Mo-99 and Tc-99m generator, at t=0, let the activity of Tc-99m be zero. The activity of Tc99m will increase until it reaches a maximum value, but will then start to decline as per transient equilibrium. I think using the bateman equations you can derive an expression for Tmax, the time it takes for the Tc-99m to reach it's max activity (which is ~ 24 hours). A derivation of that equation is what I'm after.

thanks,
matt

 

[voko]

Do you understand how the single-step decay equation is obtained and solved?

 

[paranormal]

Hello Matt,

Sure, I would be happy to walk you through a derivation for the time to maximum radioactivity in transient equilibrium for a parent-daughter system. First, let's define some variables:

- A: Activity of the parent isotope
- B: Activity of the daughter isotope
- λ1: Decay constant of the parent isotope
- λ2: Decay constant of the daughter isotope
- t: Time

In transient equilibrium, the rate of decay of the parent is equal to the rate of production of the daughter, so we can write the following equation:

dA/dt = -λ1*A = λ2*B

Solving for B, we get:

B = (-λ1/λ2)*A

To find the time at which the activity of the daughter is at its maximum, we can take the derivative of B with respect to time and set it equal to zero:

dB/dt = (-λ1/λ2)*dA/dt = 0

This means that dA/dt = 0, and since we know that dA/dt = -λ1*A, we can solve for A:

-λ1*A = 0

A = 0

This tells us that the activity of the parent isotope must be zero at the time of maximum daughter activity. Now, we can plug this value for A into our equation for B to find the time at which B is at its maximum:

B = (-λ1/λ2)*0 = 0

So, at the time of maximum daughter activity, the activity of the daughter isotope is also zero. This means that the time to maximum daughter activity, tm, is when both A and B are equal to zero. We can express this as:

tm = (1/λ1)*ln(A0/Af)

Where A0 is the initial activity of the parent isotope and Af is the final activity of the parent isotope (which is equal to zero in this case).

Now, we can substitute our equation for A into this expression to get:

tm = (1/λ1)*ln(A0/((-λ1/λ2)*A0))

Simplifying, we get:

tm = (1/(λ1-λ2))*ln(λ2/λ1)

This is the same equation you mentioned in your question and does not involve the half-life of the parent isotope. However, if