Problem 24E

In Example 2 the decay constant for isotope RA1 was 10/sec, which expresses itself in the exponent of the rate term 50e-10t kg/sec. When the decay constant for RA2 is k = 2/sec, we see that in formula (14) for y the term (185/4)e-2t eventually dominates (has greater magnitude for t large).

(a) Redo Example 2 taking k = 20/sec. Now which term in the solution eventually dominates?

(b) Redo Example 2 taking k = 10/sec.

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.