Question

Transcribed Image Text:The rate at which radon is being created is the rate at which radium is decaying,
namely A1 N1 or A1 Noe-Ait. But the radon is also decaying at the rate A,N2. Hence,
we have
dN2
= \1N1 – A2N2,
dt
or
dN2
+ A2N2 = A1N1 = 11 Noe¬A1t_
dt
This equation is of the form (3.1), and we solve it as follows:
I =
/ A2 dt = Azt,
(3.10)
N2e^2t
A, Noe-Aite^2t dt + c
=
e(A2-A1)t dt + c=
d1 No
(A2-A1)t + c,
A2 - A1
12, see Problem 19.) Since N2 = 0 at t = 0 (we
if A1 + d2. (For the case A1
assumed pure Ra at t = 0), we must have
d1 No
A1 No
0 =
d2 – di
or
c =
Substituting this value of c into (3.10) and solving for N2, we get
d, No
(e-
d2 - A1
- e-Aat).
Ait
N2

Transcribed Image Text:> Example 2. Radium decays to radon which decays to polonium. If at t = 0, a sample is
pure radium, how much radon does it contain at time t?
Let
No = number of radium atoms at t = 0,
N1 = number of radium atoms at time t,
N2 = number of radon atoms at time t,
A1 and A2 = decay constants for Ra and Rn.
As in Section 2, we have for radium
dN1
-A1 N1,
N1 = Noe-^t.
dt
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