Asked by Jannet
I don't get how to get N, please show me cause I don't understand
Answers
Answered by
DrBob222
You follow the math.
k = 0.693/t1/2
Then
ln(No/N) = kt
No = 120 ug
N = solve for this
k = from above
t = 120 sec.
<b>First, you substitute the half life of this material which is listed as 20 seconds and that allows you to solve for k. You need k to solve the equation below it in the solution above.
Then you substitute 120 ug for No (No is what you have at the beginning). N is what you have at the end and that is what you're solving for. N is the only unknown in the equation. Then substitute k from the first calculation above. t in the problem is 2 minutes but since the half life was given in seconds this 2 min need to be convert to seconds which is 120 s. So the final equation looks like this
ln(120/N) = k(120). I don't remember k but that is what you obtained from above. Solve for N.</b>
k = 0.693/t1/2
Then
ln(No/N) = kt
No = 120 ug
N = solve for this
k = from above
t = 120 sec.
<b>First, you substitute the half life of this material which is listed as 20 seconds and that allows you to solve for k. You need k to solve the equation below it in the solution above.
Then you substitute 120 ug for No (No is what you have at the beginning). N is what you have at the end and that is what you're solving for. N is the only unknown in the equation. Then substitute k from the first calculation above. t in the problem is 2 minutes but since the half life was given in seconds this 2 min need to be convert to seconds which is 120 s. So the final equation looks like this
ln(120/N) = k(120). I don't remember k but that is what you obtained from above. Solve for N.</b>
Answered by
Jannet
Um so is it like this :
0.693 x 120 = 83.16
120/n=83.16
=9979.2?
0.693 x 120 = 83.16
120/n=83.16
=9979.2?
Answered by
DrBob222
No, you aren't following the math.
The half life is 20 seconds.
k = 0.693/t<sub>1/2</sub> = 0.693/20 = 0.03465.
Then ln(120 ug/N) = 0.03465*120
Solve for N.
By the way, that ln (which you don't have anywhere in your work) in front of the (No/N) is not for looks. That means to take the natural log (that's the log base e) of No/N.
The half life is 20 seconds.
k = 0.693/t<sub>1/2</sub> = 0.693/20 = 0.03465.
Then ln(120 ug/N) = 0.03465*120
Solve for N.
By the way, that ln (which you don't have anywhere in your work) in front of the (No/N) is not for looks. That means to take the natural log (that's the log base e) of No/N.
Answered by
Jannet
Alright, my final answer is 166.32
is this correct?
is this correct?
Answered by
DrBob222
No. If the sample starts out at 125 ug and it loses half of the sample every 20 seconds, how can it have more than it started with at the end of 2 minutes. It MUST be less than 125 ug.
Answered by
Jannet
Yah that's wut I thought so I tried again and I know I sound dumb but is it .0115 ?
Answered by
Jannet
K this was my equation 120in/n=1.386
How do I solve for n?
How do I solve for n?
Answered by
DrBob222
Much closer but not quite.
I think part of the problem is the 1.386. I think that should be 4.158.
k = 0.693/20 = 0.03465
ln(125/N) = 0.3465*120
ln(125/N) = 4.158
Take the antilog of both sides.
antilog ln(125/N) = 125/N
antilog 4.158 = 63.94
So 125/N = 63.94
and N = 125/63.94 = 1.95 ug
I think part of the problem is the 1.386. I think that should be 4.158.
k = 0.693/20 = 0.03465
ln(125/N) = 0.3465*120
ln(125/N) = 4.158
Take the antilog of both sides.
antilog ln(125/N) = 125/N
antilog 4.158 = 63.94
So 125/N = 63.94
and N = 125/63.94 = 1.95 ug
Answered by
Jannet
Oh now I get it! Thank u so much for ur help:)
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