Как выбрать гостиницу для кошек
14 декабря, 2021
Consider the general radioactive decay chain
ATj —— ►ЛГ,—— >N3——- ►——————— ►———- >N,——- ► • • •
with Ni atoms of the first member at time zero and none of the other members present at that time. The differential equations are
(2.140s) |
|
^ = x. Af, — 2N2 |
(2.140b) |
(2.140/) |
|
The boundary conditions at t = 0 are |
|
a? її —© |
(2.141s) |
о II II s? II II a? и £ |
(2.141b) |
The system of differential equations (2.140) may be transformed to a system of linear equations by taking the Laplace transform and using Eq. (2.138) for the Laplace transform of the first derivatives: |
|
—N° + sNi = ~iNi |
(2.142s) |
sN2 = 1Nl — 2N2 |
(2.142b) |
■a? i T T II |
(2.1420 |
where N is the transform of N. These equations may be solved successively for the Ns: |
|
r< II ■as |
(2.143s) |
r, _ x.^1 _ Х, л? 2 2 + s (Xj + s)(X2 + s) |
(2.143b) |
_ П W? Y _ Xf-rty-t _ *=1 1 i + S і П (X* + s) Jt= і |
(2.1430 |
Nt may be found by taking the inverse transform of Eq. (2.143s): |
Nt =N? e~Kt (2.144)
|”I 0* + — s) |
To find the inverse transform of Eqs. (2.143b) to (2.143/) it is necessary to express the denominator as a sum of partial fractions. For Eq. (2.143/) this would be
To find a specific coefficient Aj, multiply each side of Eq. (2.145) by (X;- + s):
І
—— ————— — Aj + (X,- + s) £
її (x* + *) кФІ
k*i
and let s approach —X;. When s = — X,-,
(2.147)
(2.148)
Because the inverse transform of 1 /(X/ + s) is e V,
i — X — f
Ni = ZV? X,X2 • • • X,_, 2 —’—————————- O’ > 1) (2.149)
/=i ri(x*-w
k=
k*i
which is the Bateman equation (2.17). The product term ї’кФІ (X* — X,) has no meaning when the / species is the initial member of the chain, so Eq. (2.149)necessarily applies only to the daughter species, i. e., />1.