The situation is illustrated in the figure. N_{0}
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 (2N_{0}/4) decays in bats with
l_{eb} for 40 minutes, becoming the amount; |

The amount N_{40} is taken in by the owl and decays with the
rate l_{eb} for 360 min (6 hr) becoming;

The problem states that this whole body count is 200 per minute.

Eliminating N_{40} between these equations:

But l_{eb} = 0.693/80 = 8.66X10^{-3}
min^{-1}

l_{eo} = 0.693/240 = 2.89X10^{-3}
min^{-1}

Therefore:

(N_{0}/2)e^{-0.364} e^{-1.04} = 200

(N_{0}/2)(0.695)(0.353) = 200

N_{0} = 1600 count/min {Answer to part (b)}

All this time the isotope remaining in the pot has been decaying with
the rate l_{p} and also has a
count rate of 200 counts per minute.

- l_{p} 400 = ln ½ = -0.693

l_{p} = 1.73X10^{-3}

In bats l_{e} =
l_{p} + l_{b}

l_{b} =
l_{e} - l_{p} =
8.66X10^{-3} - 1.73X10^{-3} = 6.93X10^{-3}

T_{bb} = 0.693/6.93X10^{-3} = 100 min

This is the answer to part (a)