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A radioactive metal decay by simaltenuous emission of two particles with two half live 1620 and 810 respectively. the time in year after one-forth of substances remain?1080234032404860

A radioactive metal decay by simaltenuous emission of two particles with two half live 1620 and 810 respectively. the time in year after one-forth of substances remain?1080234032404860

Grade:12th pass

1 Answers

Khimraj
3007 Points
6 years ago
Let the two particle are a and b. and as decay constant \lambda are given by
\lambdaa = ln2/(t1/2)a = ln2/1620
\lambdab = ln2/(t1/2)b = ln2/810
Total decay constant  \lambda =\lambdaa+\lambdab
\lambda = ln2/1620 + ln2/810 = 3*ln2/1620
when (1/4)th of metal remains
N0/4 = N0e-λt
λt = ln4
t = 2ln2/λ = 2*ln2/λ = (2/3)*1620 = 1080 years.
Hope it clears. If you like answer then please approve it.

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