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 > 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
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 > 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
Related questions
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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
