13.8 Activation Analysis
213
where h A and h B are some unknown constants. Differentiating this equation gives
us
d N B
dt
= −h B e
−λ B t
λ B
(13.3)
substituting this value into Eq. (13.1), we have
−h B λ B e
−λ B t
=
N w A nφσ
M
− N B λ B .
(13.4)
Thus
−h B λ B e
−λ B t
=
N w A nφσ
M
− (h A + h B e
−λ B t
)λ B .
(13.5)
or
−h B λ B e
−λ B t
=
N w A nφσ
M
− h A λ B − h B λ B e
−λ B t
(13.6)
hence
N w A nφσ
M
= h A λ B
(13.7)
or
h A =
N w A nφσ
Mλ B
(13.8)
Now using Eq. (13.1)
N B =
h A + h B e
−λ B t
and solving it for time t = 0, we have N B = 0. Thus, h A = −h B . Substituting this
condition into the Eq. (13.2), we have
N B =
N w A nσ φ
Mλ B
−
N w A nσ φ
Mλ B t
e
−λ B t
(13.9)
i.e.
N B λ B =
N w A nσ φ
M
(e
−λ B t
) = activity of atom B per unit time
(13.10)
Thus, by measuring the activity of the irradiated sample (if it contains one type
of atoms) or of the sample chemically separated, the amount of atom A can be
determined from this equation, because all other terms are known.