A foil of 7Li of mass 0.05gram is irradiated with
thermal neutrons (capture cross section σ = 37milllibars) and forms
8Li, which decays by β-decay with a half-life of 0.85 sec.
Find the equilibrium activity (number of β-decays per second) when the foil is exposed to a steady neutron ux of φ =3×1012neutrons/sec · cm2.
Note: Use dN1 dt =−σφN1 1
N1 ( for Li - 7) = 0.05 / 7 * 6.02e23 where 6.02e23 is avogadro's number
N1 = 4.3e21
as per given note in the question
dN1/ dt = ???N1
Therefore, N2 ( for Li - 8)
dN2/ dt = N1??e-??t - gN2 ----------- (1)
where g - decay constant and is equal to g = ln 2 / half life = ln 2 / 0.85 = 0.815
?? = 3e12*3.7e-26 = 1.11e-13
Integrating (1) and setting dN2 / dt = 0 , we get
t = 1/g ln( g/??)
t = 3.6257 seconds
Therefore, activity = ??*N1
activity = 1.11e-13*4.3e21
activity = 4.773e8 Bq
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