Question

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

Answer 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