Question

Thermal neutrons fall normally on the surface of a thin gold foil consisting of stable `Au^(197)` nuclide. The neutron flux density is `J=1.0.10^(10) "part"//(1s,cm^(2))`. The mass of the foil is `m= 10 mg`. Then neuteron capture produces beta-active `Au^(198)` nuclei with half-life `T= 2.7 days`. The effective capture cross-section is `sigma= 98 b`. Find :
(a) the irradiation time after which the number of `Au^(197)` nuclei decreases by `eta= 1.0%`,
(b) the maximum number of `Au^(198)` nuclei that can be formed during protracted irradiation.

Answer 1

(a) Suppose `N_(0)=` no. of `Au^(197)` nuclei in the foil. Then the number of `Au^(197)` nuclei transformed in time `t` is
`N_(0).J.sigma.t`
For this to equal `etaN_(0)`, we must have
`t= eta//(J. sigma)= 323 years`
(b) Rate of formation of the `Au^(198)` nuclei is `N_(0).J.sigma` per sec and rate of decay is `lambda n,` where `n` is the number of `Au^(198)` at any instant.
Thus `(dn)/(dt)=n_(0).J.sigma-lambda n`
The maximum number of `Au^(198)` is clearly
`n_(max)=(N_(0).J.sigma)/(lambda)=(N_(0).J.sigma.T)/(In 2)`
beacuse if `n` is smaller, `(dn)/(dt) gt 0` and `n` will increase further and if `n` is larger `(dn)/(dt) lt 0` and `n` decrease. (Actually `n_(max)` is approached steadily as `trarr oo`)
Substitution gives using `N_(0)= 3.05xx10^(19), n_(max)= 1.01xx10^(13)`