A question about radioactive decay

AI Thread Summary
In a mixture of two radioactive species with disintegration constants of λ and λ/3, the long-term behavior reveals that species B will dominate due to its longer half-life. As time progresses, species A decays significantly faster, leading to its near-complete disappearance. The mean life of the mixture effectively becomes that of species B, as it is the only significant species remaining. The discussion emphasizes that even a small difference in decay rates can result in one species prevailing over time. Ultimately, the mean life of the mixture aligns with the characteristics of the surviving species.
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Homework Statement


Two species of radioactive atoms are mixed in equal numbers. The disintegration constant of first species is ##\lambda## and that of second species is ##\frac \lambda {3}##. After a long time, the mixture will behave as a species with mean life ________.

Homework Equations


##\lambda=\frac{0.693} {t_{1/2}}##
##t_{mean}=\frac{1}\lambda##

The Attempt at a Solution


What I did was to equate the mean life of mixture being the average of the mean lives of the individual species.
$$t_{mean}=\frac{\frac{1} \lambda + \frac{3}\lambda} {2}$$

However in this article https://en.wikipedia.org/wiki/Half-life it says that ##\frac{1} {t{eff}} = \frac{1} {t_{1}} + \frac{1} {t_{2}}## where ##t_{eff}##,##t_{1}## & ##t_{2}## are half lives of mixture, species A and B respectively.
 
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Actually the question can be answered from the equations you gave without calculating a combined mean life. After a long time, which of the two species will dominate?
 
Species dominates as in amount of nuclei ? Then should be species B
 
phyzguy said:
Actually the question can be answered from the equations you gave without calculating a combined mean life.
How so?
 
If you have two quantities, ##a(t)## and ##b(t)##, which decay exponentially:

##a(t) = Ae^{-k_a t}##, ##b(t) = Be^{-k_b t}##,

then the ratio of the quantities, ##\frac{a(t)}{b(t)} = \left(\frac{A}{B}\right)e^{(k_b - k_a )t}##, will approach either zero or infinity when ##t\rightarrow\infty##, depending on whether ##k_a > k_b## or ##k_b > k_a##.

Based on only this, you can deduce that if two mixed radioactive species have even a small difference in half-life, one of them will be the only significant species after a long enough time has passed.
 
Ashes Panigrahi said:
Species dominates as in amount of nuclei ? Then should be species B

Correct. So after a long time, species B dominates and basically all of species A has decayed away. So it isn't really a mixture any more, it is only species B. So what then is the mean life?
 
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