1. Feb 13, 2016

### erisedk

1. The problem statement, all variables and given/known data
A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is:

2. Relevant equations

3. The attempt at a solution
I'm confused whether the half lives mentioned are of the original radioactive material itself, or of the the particles which are formed. In any case, I'm really not sure of how to deal with two particle emission. I've only ever done single decays and basic half life calculations. Please help?

2. Feb 13, 2016

### gleem

I believe that you have a material that can decay through two different channels, one with a H.L. 1620 years and one with a H.L. of 810 years. The term simultaneous does not refer to the emission of two types at the same time.

3. Feb 14, 2016

### erisedk

Yeah that's what I was thinking too. But what do I do after that?

4. Feb 14, 2016

### Orodruin

Staff Emeritus
You need to find an expression which describes how much of tthe substance will remain after a given time.

5. Feb 14, 2016

### gleem

How do you think the rate of decay of the parent related to the rate of formation of the daughters?

6. Feb 23, 2016

### erisedk

I'm extremely sorry for replying after so long!

I suppose the rate of decay of the parent is equal to the rate of formation of the daughter. So, I'm guessing the rate of decay would be equal to the sum of rates of formation of the two nuclei. Here's what I did.
N = N0e-λt where N is the number of nuclei left after time t and the decay constant is λ.
Rate of formation of both the nuclei would be λiN.
So,
dN/dt = λN = λ1N + λ2N where λ1, λ2 are the decay constants of the two different daughter nuclei.
So, N cancels.
I get t1/2 = 1620*810/1620+810 = 540 years
So, 1/4th will remain after 540*2 = 1080 years.

Which is the right answer. Thank you!!