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A sample originally contains 24 grams of a radioactive isotope. After 18 days, 18.8 grams of the isotope remains in the sample.

What is the half-life of the isotope? 

 Jan 13, 2020
 #1
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We have that

 

18.8  = 24 b^18         divide both sides  by  24

 

18.8 / 24  = b^18       take the 18th root of both sides

 

(18.8/24)^(1/18)  = b

 

So....we  want to find this   [the half-life is the tme when 12 g   remain ]

 

12  = 24 [ (18.8.24)^(1/18)]^(t)      where t is the number of days we seek

 

12 = 24 [ 18/8/24]^(t/18)

 

Divide both sides by 24

 

(1/2)  = (18.8/24)^(t/18)      take the log of both sides

 

log (1/2)  = log (18.8/24)^(t/18)     and we can write

 

log (1/2)  = (t/18)  log (18.8/24)    isolate t

 

18 log (1/2)  / log (18.8 / 24)  =  t  ≈   51.1 days

 

 

cool cool cool

 Jan 13, 2020

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