help with decay question (1 Viewer)

hatty

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the half life of radium is 1600 years.
a) find the % of radium that will be decayed after 500 years.
b) find the # of years that it will take for 75% of the radium to decay

this is straight of the of Margaret Grove book

thanks
 

Grey Council

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hehe, here goes:
1/2 = e^(1600k)
(log (1/2)) / 1600 = k

then let t=0 and find out initial radium.

then:
place the radium after 500 years (let t=500) over total radium times one hundred. This will give you the percentage. Thats teh first questions.

The second one is similar. I'll let you figure it out, it shouldn't be too hard.

And before anyone else gives him the answer, DONT! Let him work, otherwise he won't learn. Post your answer here, and we can verify it.

and if you don't understand something, let me know. :)
 
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Grey Council

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umm, a bit more detailed explanation:
hehe, here goes:
1/2 = e^(1600k)
(ln (1/2)) / 1600 = k
let G be initial radium mass.

then G/2 = Ge^(k1600)
bring G over:
G/2G = e^(1600k)
G's cancel, so
1/2 = e^(1600k)
use log laws, you get:
ln(1/2) = 1600k
divide both sides by 1600
(ln(1/2))/1600 = k
now you know k, everything else just falls into place. Try it.

notice that to find any proportion of the original mass doesn't require you to know the original mass. Its like half of decay questions will require you to know that. :)
 

kimmeh

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blah :p to save the shame.. i'll delete it :D

ahahaha man.. i dont know how the hell i came up with that :p i think it was because i was doing an english essay, then i saw this and decided to post. ahaha i guess i'll never do that again.. :p so kids the moral is: switch your brain onto maths mode before doing any maths DO NOT leave it on english mode :rolleyes: :p
 

Grey Council

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come on, hatty. Say thank you.

You went after Keypad for not saying thank you cause you complimented him. Now I help you and I think you should say thanks. ^__^
 

Grey Council

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humph

blah to you hatty. ok, i'll solve it now
and you had better say thank you. and not quote some random cicero bs quote. :(

hehe, here goes:
1/2 = e^(1600k)
(log (1/2)) / 1600 = k

then let t=0 and find out initial radium.

the reason i'm not writing values is that in exponential questions, a slight rounding off may lead to maybe thousands of years difference. Its called exponential for a reason. anyway, now you know k. Everytime i say k, sub in that equation for k. Let A be initial radium (which you know, look above)

"place the radium after 500 years (let t=500) over total radium times one hundred. This will give you the percentage. Thats teh first questions." watch:

((Ae^(500k)) / (A)) x 100
= 100e^(500k)
that gives you the EXACT answer for question one.
 

Grey Council

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for the second, here goes:

3/4 = e^(kt)
you know k, so:
ln (3/4) = kt
(ln(3/4)) / k = t
so there you go

I'm waiting hatty. ;)
 
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freaking_out

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Originally posted by kimmeh
blah :p to save the shame.. i'll delete it :D

ahahaha man.. i dont know how the hell i came up with that :p i think it was because i was doing an english essay, then i saw this and decided to post. ahaha i guess i'll never do that again.. :p so kids the moral is: switch your brain onto maths mode before doing any maths DO NOT leave it on english mode :rolleyes: :p
stop making up excuses! :p
 

Grey Council

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lol, keypad, its me and hatty. :) I don't ask anyone else, but hatty is different. you don't understand, thats all. He refused to say a word of thanks (in pm) and sent me this crap about how it was a work well done, but not finished. He even quoted some Cicero guy.

anyway, i love the way Hatty talks. Best thing EVER. :D

and np hatty.
 

Grey Council

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im guardian. Read my signiture. I changed my nickname.

LOL, "And who MIGHT you be?" not "who MAY you be". :p
tsk tsk tsk
 

Grey Council

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umph, i have my reasons. :)

ANd i still haven't typed up those economics questions. sorry :(
 

Xayma

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Originally posted by Grey Council
im guardian. Read my signiture. I changed my nickname.
And this signature would be where?
 

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