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

N_{1} ( for Li - 7) = 0.05 / 7 * 6.02e23 where 6.02e23
is avogadro's number

N_{1} = 4.3e21

as per given note in the question

dN_{1}/ dt = ???N_{1}

Therefore, N_{2} ( for Li - 8)

dN_{2}/ dt = N_{1}??e^{-??t} -
gN_{2} ----------- (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 dN_{2} / dt = 0 , we get

t = 1/g ln( g/??)

t = 3.6257 seconds

Therefore, activity = ??*N_{1}

activity = 1.11e-13*4.3e21

activity = 4.773e8 Bq

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