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# Consider the radioactive decay formula A=Aoe^-kt where a is the amount of radium remaining at the time t. Ao is the amount present initially

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Consider the radioactive decay formula A=Aoe^-kt where a is the amount of radium remaining at the time t. Ao is the amount present initially and k is decay constant. After How many years would 10 grams of radium decay so that only 8 grams remain? The half life of radium is 1590 years

Maltesewolfman  Aug 24, 2015

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We have5 grams remaining after 1590 yeats.....so.....

5 = 10 e^[-k(1590)]     divide both sides by 10

(.5)  = e^[-k(1590]     take the ln of both sides

ln (.5)  = ln e^[-k(1590]    and  we can write

ln (.5)  = [-k(1590] *ln e      and ln e = 1....so we have

ln(.5) = -k(1590)    divide both sides by 1590

ln(.5)/1590  = -k

-0.0004359416229937 = -k   .....so.....k = 0.0004359416229937

And we need to find  t for 8 grams remaining....so.....

8 = 10 e^[-0.0004359416229937 (t)]       divide both sides by 10

.8  = e^[-0.0004359416229937 (t)]    take the ln of both sides

ln(.8)  =  ln  e^[-0.0004359416229937 (t)]     and we can write

ln(.8)  = [-0.0004359416229937 (t)] ln e     and ln e = 1   .....so we have...

ln(.8) = -0.0004359416229937 (t)    divide both sides by  -0.0004359416229937

ln(.8) / -0.0004359416229937  = t = about 511.87 years

CPhill  Aug 24, 2015
Sort:

#1
+80928
+10

We have5 grams remaining after 1590 yeats.....so.....

5 = 10 e^[-k(1590)]     divide both sides by 10

(.5)  = e^[-k(1590]     take the ln of both sides

ln (.5)  = ln e^[-k(1590]    and  we can write

ln (.5)  = [-k(1590] *ln e      and ln e = 1....so we have

ln(.5) = -k(1590)    divide both sides by 1590

ln(.5)/1590  = -k

-0.0004359416229937 = -k   .....so.....k = 0.0004359416229937

And we need to find  t for 8 grams remaining....so.....

8 = 10 e^[-0.0004359416229937 (t)]       divide both sides by 10

.8  = e^[-0.0004359416229937 (t)]    take the ln of both sides

ln(.8)  =  ln  e^[-0.0004359416229937 (t)]     and we can write

ln(.8)  = [-0.0004359416229937 (t)] ln e     and ln e = 1   .....so we have...

ln(.8) = -0.0004359416229937 (t)    divide both sides by  -0.0004359416229937

ln(.8) / -0.0004359416229937  = t = about 511.87 years

CPhill  Aug 24, 2015

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