Radioactive Decay: Proving Effective Half-Life of Nucleus

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Homework Help Overview

The discussion revolves around the effective half-life of a radioactive nucleus that can decay through two different processes, each characterized by distinct half-lives, t₁ and t₂. Participants are exploring how to derive the effective half-life formula based on these two decay processes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between decay constants and half-lives, with one attempting to use the equation λ = λ₁ + λ₂. Others express confusion regarding the derivation and the underlying reasoning.

Discussion Status

The conversation is ongoing, with some participants offering insights into the mathematical relationships involved, while others indicate a lack of understanding or clarity on the topic. There is no explicit consensus yet, but various interpretations and approaches are being explored.

Contextual Notes

Some participants are grappling with the definitions and implications of the decay equations, and there is mention of initial conditions and expected outcomes based on different decay processes.

roshan2004
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Homework Statement


A radioactive nucleus can decay by two different processes. The half life for the first processes is t_{1} and that for the second is t_{2}. Show that the effective half life t of the nucleus is given by
\frac{1}{t}=\frac{1}{t_{1}}+\frac{1}{t_{2}}

Homework Equations


t=\frac{0.693}{\lambda }

The Attempt at a Solution


Tried to use \lambda =\lambda _{1}+\lambda _{2}, and got the answer but don't know why, how and is it the correct way to prove this?
 
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Last edited by a moderator:
@tiny-tim Now you made me totally lost
 
roshan2004 said:
@tiny-tim Now you made me totally lost

Suppose the two half-lives are τ1 and τ2. Let the initial amount of substance be A.

After time t, due to the first process you expect to see remaining:

A*2^-(t/ τ1)

But the second process has a go at the other stuff that didn't go by the first process. So the remaining amount becomes:

A*2^(t/ τ1)* 2^-(t/ τ2)

Now, n^a * n^b = n^(a + b). So do the obvious with the above.
 
oooh, sorry :blushing:

the decay equation is A = A0e-λt,

which is the same as dA/dt = -λA :wink:
 

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