The situation is illustrated in the figure. N0 is the total amount of isotope in the bat at t = 0. Its value is the answer to part (b). It is divided into 4 equal parts of which 2 are drunk by the bats. This amount (2N0/4) decays in bats with  leb for 40 minutes, becoming the amount;

The amount N40 is taken in by the owl and decays with the rate leb for 360 min (6 hr) becoming;

The problem states that this whole body count is 200 per minute.
Eliminating N40 between these equations:

But  leb = 0.693/80 = 8.66X10-3 min-1
 leo = 0.693/240 = 2.89X10-3 min-1

 Therefore:

(N0/2)e-0.364 e-1.04 = 200
(N0/2)(0.695)(0.353) = 200
N0 = 1600 count/min {Answer to part (b)}
All this time the isotope remaining in the pot has been decaying with the rate  lp and also has a count rate of 200 counts per minute.

- lp 400 = ln ½ = -0.693
 lp = 1.73X10-3
In bats  le lp lb
 lb le lp = 8.66X10-3 - 1.73X10-3 = 6.93X10-3
Tbb = 0.693/6.93X10-3 = 100 min
This is the answer to part (a)

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