hey please helpp with a chemistry half life problem?

Question

here it is... Which radioactive isotope will retain only (1/8) its original radioactive atoms after approximately 43 days?

the formulas used ffor this are as follows: fraction remaining= (1/2) t/ T..
t= total time elapsed. T= half life.
and number of half life periods: t/T .. these formulas are off a referance table.
the answer to this question is phosphorus - 32 which has a half life of 14.3 days. .. the explanation in the back is terrible; all it says is ...
(1/2) t/T = 1/8....... t/T= three half lives.... i just dont understand how to arrive at threee half lives. if i would have gotten that then i would have been able to do the rest.. i was looking at this question aimlessly ffor like ten minutes..helpp'thankss =D

Answer 1

1/8 = (1/2)^3, so it is three half-lives.

If 43 days = 3 half-lives,
then 43/3 days is the half-life (14.3).

Or you can use their original equation, substituting 1/8 for the fraction and 43 days:

(1/2)^(t/ T) = fraction_remaining
(1/2)^(43/ T) = 1/8
lln( (1/2)^(43/ T) ) = ln(1/8)
(43/T) * ln(1/2) = ln(1/8)
(43/T) = ln(1/8) / ln(1/2)
(43/T) = -2.07944 / -0.69315
(43/T) = 3
T = 43/3
T = 14.3

That's the procedure you'd have to use if the remaining fraction wasn't a power of 1/2 (e.g., "after 5 days 2/3 remains"), but 1/8 you can do in your head.

Answer 2

If there is 1/8 left, then (1/2)(1/2)(1/2) = 1/8 half lifes must have occured. This is 3 half lifes. So, divide 43 by 3:

14.333

Look for something with a half life near 14 days.

Answer 3

Yes 3 half lives have elapsed since 2^3 = 8

So, 43/3 = 14.33 days.

Any isotope having this half life will show this behaviour.

Answer 4

half life means the time in which a radioactive substance decays to half of its value.
if initial value is N &half life is T then after T time N/2 substance will exist again after T time (N/2)/2 =N/4 will exist again after T time (N/4)/2=N/8 will exist.So after 3T time i.e. , three half life substance will remain 1/8 of its original value .
Thank you.

Answer 5

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Answer 6

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