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In a comparison of two radioisotopes, isotope A requires 15.0 hours for its decay rate to fall to 1/16 its initial value, while isotope B has a half-life that is 2.5 times that of A. How long does it take for the decay rate of isotope B to decrease to 1/32 of its initial value?

2007-12-08 21:35:51 · 2 answers · asked by mceb 1 in Science & Mathematics Chemistry

2 answers

We know that t(1/16)=4t(1/2)there fore t1/2=15/4 hrs=3hrs 45 min
isotope B then have a t1/2=2.5 times of A=9hrs and 22(1/2)mins
so to decrease to 1/32 value it will be 5*t1/2
=46hrs and 52(1/2)mins

2007-12-09 02:44:31 · answer #1 · answered by kartheek 2 · 0 0

for isotope A 15 hours corresponds to 4 half-lives : one half-life corresponding to half decay =1/2 and 1/16 = 1/2^4.
So the half-life of A is 15/4=3.75 hours.
The half life of B is 2.5*3.75=9.375 hours
to decay to 1/32 = 1/2^5 you need 5 hal-times
and 9.375*5=46.875 hours or 46 hours 52 minutes, 30 seconds

2007-12-09 06:24:57 · answer #2 · answered by maussy 7 · 0 0

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