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Chemistry 20 Online
OpenStudy (anonymous):

really need help understanding a concept! Please! I will give medal

OpenStudy (anonymous):

I need help with half life!

OpenStudy (amistre64):

lol, midlife crises?

OpenStudy (anonymous):

hahaha no not midlife crisis! A radioisptope has a half life of 4 days. How much of a 20gram sample of this radioisotope remains at the end of 4 days

OpenStudy (amistre64):

if it takes 4 years for something to reduce by half then in 8 years it reduces by 1/2 + (1/2)/2 then in 12 years it reduces by 1/2 + 1/4 + (1/4)/2 etc .... it produces an exponential decay

OpenStudy (amistre64):

if its half life is 4 days, then after 4 days there is half of it left

OpenStudy (anonymous):

I don't get it for some reason ;/ so what would the equation look like?

OpenStudy (amistre64):

20/2

OpenStudy (amistre64):

if it takes you 5 hours to walk halfway home, then how far from home are you after you walk for 5 hours?

OpenStudy (anonymous):

5 hours

OpenStudy (amistre64):

well, halfway of course ... if your started 16 miles away then you have 8 miles left to go

OpenStudy (amistre64):

if you start with 20 grams of something, and half of it is gone after 4 hours then how much of the stuff is left after 4 hours?

OpenStudy (anonymous):

ohhh 10 grams!

OpenStudy (amistre64):

yep lets say you measure it at 4 hours and its 10 grams, after another 4 hours its reduced by half again .... how much is left after that extra 4 hours?

OpenStudy (anonymous):

5

OpenStudy (amistre64):

correct

OpenStudy (amistre64):

so we go from: 20 at 0 hours 10 at 4 hours 5 at 8 hours 2.5 at 12 hours 1.25 at 16 hours every 4 hours it reduces by half

OpenStudy (amistre64):

how many 3.8 days are in 23 days?

OpenStudy (anonymous):

this one seems so much harder ;/

OpenStudy (amistre64):

then we can brute it out 6 times k at 0 periods k/2 at 1 period k/4 at 2 periods k/8 at 3 periods ... seems to be k/2^n for n periods

OpenStudy (amistre64):

73.9/2^6 ... gives us what?

OpenStudy (amistre64):

yep, but if we want to get real fancy we can simply wait till the end to approximate giving us:\[\large \frac{73.9}{2^{23/3.8}}=1.11332....\]

OpenStudy (amistre64):

so at the end of 6 periods we are at 1.15 ... a little more gets us to 1.11 since its just a tad more than 6 periods

OpenStudy (anonymous):

ohhh okay! gosh that simplifies it so much more! Thank you thank you! for your time

OpenStudy (amistre64):

youre welcome

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