When Does Alpha Emission Peak in the Bi-210 to Po-210 to Pb-208 Decay Chain?

[ghetom]

Hi I'm trying to do a question on nuclear decay chains. The question is:

Bi-210 decays to Po-210 by beta decay (half life = 7.2 days), and this decays by alpha decay to Pb-208 (half life = 200 decays). If A substance is initially pure Bi-210, when does the alpha emmision peak?

So far I've got
Bi(0)=1, Po(0)=Pb(0)=o

Bi(t) = Bi(0)exp(-Lt) (L being the decay constant, t_1/2 = ln(2)/L)

and dPo(t)/dt = L(Bi(0)exp(-Lt)) -L'Po(t)

Obviously I just need to find when Po(t) is at a maximum, to find the corresponding maximum in alpha emmision, but I can't solve for Po(t)can anyone help?

 

[Astronuc]

Well, think on mean Bi-210 => to Po-210 => Pb-206. Half-lives are 7.2 and 200 days respectively, although reliable sources indicate half-life of Po-210 is 140 d (Hyperphysics) or 138.376 d (BNL NNDC).
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/radser.html#c3

This is at the bottom end of the U-238 decay series.


Also - apparently Bi-210 can decay to Tl-206, which beta decays to Pb-206, but it's almost 100% alpha.

To solve the problem, make sure one has the correct half-lives or decay constants.

One has to develop two differential equations. One deals with the change (decay) in amount of Bi-210, which is also the production rate of Po-210. The other equation deals with the production and decay of Po-210 (production of Pb-206). The peak alpha emission from Po-210 occurs when the amount of Po-210, or when the rate of change is zero (dP/dt = 0), or when production rate = decay rate.

This is called transient equilibrium.

 

[sir-pinski]

Sure, I can help! Let's break down the problem step by step.

First, we need to understand the decay processes involved. Bi-210 decays to Po-210 through beta decay, which means it emits a beta particle (an electron) and becomes Po-210. This process has a half-life of 7.2 days, which means that after 7.2 days, half of the original Bi-210 atoms will have decayed into Po-210.

Next, Po-210 decays to Pb-208 through alpha decay, which means it emits an alpha particle (two protons and two neutrons) and becomes Pb-208. This process has a half-life of 200 days, which means that after 200 days, half of the original Po-210 atoms will have decayed into Pb-208.

Now, let's look at the equation you've already started:

Bi(t) = Bi(0)exp(-Lt)

This equation represents the number of Bi-210 atoms at a given time t. Bi(0) represents the initial number of Bi-210 atoms, which in this case is 1 (since we have a pure sample of Bi-210). L is the decay constant, which is related to the half-life by L = ln(2)/t_1/2.

So, we can rewrite the equation as:

Bi(t) = exp(-0.693/t_1/2)t

Next, we need to find the number of Po-210 atoms at a given time t. Since Po-210 is produced from the decay of Bi-210, the number of Po-210 atoms at any time t is equal to the number of Bi-210 atoms that have decayed up to that time t. So, we can write the equation for Po(t) as:

Po(t) = Bi(0)(1 - exp(-0.693/t_1/2)t)

Now, we can differentiate this equation to find the rate of change of Po(t) with respect to time t:

dPo(t)/dt = Bi(0)(0.693/t_1/2)exp(-0.693/t_1/2)t

To find the maximum value of Po(t), we need to find the time t at which dPo(t)/dt = 0. We can solve for this time by setting dPo(t)/dt = 0 and solving for t:

0 = Bi