Solution Found!
In Example 2 the decay constant for isotope RA1 was
Chapter 2, Problem 24E(choose chapter or problem)
In Example 2 the decay constant for isotope \(R A_{1}\) was \(10 / s e c \), which expresses itself in the exponent of the rate term \(50e^{-10t}\mathrm{\ kg}/\mathrm{\sec}\). When the decay constant for \(R A_{2}\) is \(k=2 / \mathrm{sec}\), we see that in formula (14) for y the term \((185 / 4) e^{-2 t}\) eventually dominates (has greater magnitude for t large).
(a) Redo Example 2 taking \(k=20 / \mathrm{sec}\). Now which term in the solution eventually dominates?
(b) Redo Example 2 taking \(k=20 / \mathrm{sec}\).
Equation Transcription:
Text Transcription:
R A_{1}
10/sec
50 e^{-10t} kg/sec
R A_{2}
k=2 / sec
(185/4) e^{-2 t}
k=20/sec
k=10/sec
Questions & Answers
QUESTION:
In Example 2 the decay constant for isotope \(R A_{1}\) was \(10 / s e c \), which expresses itself in the exponent of the rate term \(50e^{-10t}\mathrm{\ kg}/\mathrm{\sec}\). When the decay constant for \(R A_{2}\) is \(k=2 / \mathrm{sec}\), we see that in formula (14) for y the term \((185 / 4) e^{-2 t}\) eventually dominates (has greater magnitude for t large).
(a) Redo Example 2 taking \(k=20 / \mathrm{sec}\). Now which term in the solution eventually dominates?
(b) Redo Example 2 taking \(k=20 / \mathrm{sec}\).
Equation Transcription:
Text Transcription:
R A_{1}
10/sec
50 e^{-10t} kg/sec
R A_{2}
k=2 / sec
(185/4) e^{-2 t}
k=20/sec
k=10/sec
ANSWER:
Solution:
Step 1:
Accept that the rate at which RA1 decays into RA2 is 50e-10t kg/sec. Since the rate of rot of RA2 is relative to the mass y(t) of RA2 introduce, the rate of progress of RA2 is
50e-10t - ky ……..(i)
Where k > 0 is the decay steady.