Asked by lulu
If I have a sample from doing a half lfe lab with m&ms, not timed, just pulling out decayed or m&ms with rhe m on top, how do I calculate the half life using t1/2=(.693)/k formula. Do I have to do this 9 times as it took 9 trys to use up all the m&ms. I am not really sure what numbers to feed in. Could you please help?
Thank you. We made a graph and a chart with the numbers from the "decayed m&ms".
Thank you. We made a graph and a chart with the numbers from the "decayed m&ms".
Answers
Answered by
DrRebel
You need to make a plot of 'undecayed' M&M's vs Throw number. Undecayed M&M's = (UD) =>
UD(1) = (Total M&M's Before 1st throw) - (M's-up after 1st throw)
UD(2) = [UD(1) - (M's up after 2nd throw)]
UD(3) = [UD(2) - (M's up after 3rd throw)]
UD(4) = [UD(3) - (M's up after 4th throw)]
So on... through UD(9)...
=> Then plot UD(#) vs Throw(#) If you know how to use Excel, this works great... If not, just trace best line through points. Then ...
=> On the UD(# axis) find point of (Total M&M's)/2.
=> Draw line parallel to Throw(#) axis until you intersect the trend curve.
=> Draw line down to Throw(#) axis => this is the Half-Life Value.
=> Substitute the Half-Life Value into [0.693/t(1/2)] => Decay constant in units of (Throws)⁻¹
UD(1) = (Total M&M's Before 1st throw) - (M's-up after 1st throw)
UD(2) = [UD(1) - (M's up after 2nd throw)]
UD(3) = [UD(2) - (M's up after 3rd throw)]
UD(4) = [UD(3) - (M's up after 4th throw)]
So on... through UD(9)...
=> Then plot UD(#) vs Throw(#) If you know how to use Excel, this works great... If not, just trace best line through points. Then ...
=> On the UD(# axis) find point of (Total M&M's)/2.
=> Draw line parallel to Throw(#) axis until you intersect the trend curve.
=> Draw line down to Throw(#) axis => this is the Half-Life Value.
=> Substitute the Half-Life Value into [0.693/t(1/2)] => Decay constant in units of (Throws)⁻¹
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.