Calculating Radioactive Decay with Simultaneous Particle Emission

[erisedk]

Homework Statement

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:

Homework Equations

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?

 

[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.

 

[erisedk]

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

 

[Orodruin]

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

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

 

[gleem]

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

 

[erisedk]

I'm extremely sorry for replying after so long!

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

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 λ_iN.
So,
dN/dt = λN = λ₁N + λ₂N where λ₁, λ₂ are the decay constants of the two different daughter nuclei.
So, N cancels.
I get t_1/2 = 1620*810/1620+810 = 540 years
So, 1/4th will remain after 540*2 = 1080 years.

Which is the right answer. Thank you!