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A radioactive material decays by simultaneous emission of two particles with respective half life 1620 and 810 years . The time in years after which one fourth of material decays a) 1080 b)2340 c) 3240 d)4860

A radioactive material decays by simultaneous emission of two particles with respective half life 1620 and 810 years . The time in years after which one fourth of material decays a) 1080 b)2340 c) 3240 d)4860

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
Dear Samrat,
 
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.
So,
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.
we get t1/2 = 1620*810/1620+810 = 540 years
So, 1/4th will remain after 540*2 = 1080 years.
This should be the answer i think yeah!
 
Option a)
 

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