half life question

[spoog74]

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found this question in achiever, and they have many such as this one;

Would this be something that we would need to be able to computate in order to get a good score on the gen section of the dat?

Determine the rate constant for an isotope disintegrating to 1/8th of its original sample in 9 years

 

[Troyvdg]

found this question in achiever, and they have many such as this one;

Would this be something that we would need to be able to computate in order to get a good score on the gen section of the dat?

Determine the rate constant for an isotope disintegrating to 1/8th of its original sample in 9 years

I think you use the equation:
T1/2 = ln2/k=.693/k

halflife is 3 years

so 3 = .693/k rearrange to get k=.693/3 = .231

it's easy points, I'd just memorize the equation lol

 

[spoog74]

I think you use the equation:
T1/2 = ln2/k=.693/k

halflife is 3 years

so 3 = .693/k rearrange to get k=.693/3 = .231

it's easy points, I'd just memorize the equation lol

how do you know half life is 3 years? and which equation exactly are you talking about? And how would you use it? ( what do i plug in, when and how?)

 

[gtk06]

the half life is 3 years because 1/8th comes from 1/2 to the 3rd power since thats from the formula. the equation is pretty common. that makes sense! hope it does

 

[spoog74]

the half life is 3 years because 1/8th comes from 1/2 to the 3rd power since thats from the formula. the equation is pretty common. that makes sense! hope it does

so you dont have to use the the fact of 9 years in this problem? it could have been 15 years and the answer would have been the same?

 

[rockclock]

so you dont have to use the the fact of 9 years in this problem? it could have been 15 years and the answer would have been the same?

you follow that after each half-life, half of the remaining sample is still there right?

so starting with a "whole" sample: 1/2 the original remains after one half-life, 1/4 the original remains after two half-lives, and 1/8 the original after three half-lives, etc.

since there's 1/8 left after 9 years, you know that three half-lives went by during those 9 years. three half-lives in a total of 9 years = 3 years per half-life.

the answer would absolutely be different if it had said 15 years...how could it not?

 

[hoylematt]

Anyone else think of Gordon Freeman when they read the title to this post?

 

[indianjatt]

I'd use the first order rate law equation.

LnAt = -kt + ln A0
Ln(At/A0)/-t=k
LN(1/8)/-9 yrs= k
k = .231/year

 

[tmirchel]

I had that question on the dat I took on the 22nd, I did not remember the equation and got confused. I new there was an equation but i figured I would not see it cause its pretty specific. my own fault, but it was there.

 

[wired202808]

so you dont have to use the the fact of 9 years in this problem? it could have been 15 years and the answer would have been the same?

spoog, think of it this way, if after 9 years, you only had 12.5% (1/8) left that means

Current Time Current %
0 100
X 50
X 25
9 12.5

Therefore, looking at this, its clear it has to be 3, because each X is decreased by half.

if it was 15, then X is 5 (5+5+5=15), because to get to 12.5 you have to have 3 half lifes. so you take the number you have like 9 and divide by 3 to get 3 as the half-life year value.

You then plug in in that value into the .693/k = t. which will get you the .263.

Hope this helps.

 

[vnicolas]

I am new to SD and I dont know how to post a thread, s u can see I only know how to reply.