# 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

Khimraj
3007 Points
4 years ago
Let the two particle are a and b. and as decay constant $\lambda$ are given by
$\lambda$a = ln2/(t1/2)a = ln2/1620
$\lambda$b = ln2/(t1/2)b = ln2/810
Total decay constant  $\lambda$ =$\lambda$a+$\lambda$b
$\lambda$ = 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.