I've got this question and i cant figure out my teacher's comments to my answer.
This is the question: "Radio-isotopes of different elements have different half-lives. Magnesium-27 has a half-life of 9.45 minutes. What is the decay constant for Magnesium-27? Round to five decimal places."
This is my answer: e^(-kt) = 0.5
e^(-9.45k) = 0.5
log(base e) e^(-9.45k) = log(base e) 0.5
-9.45k = -0.6931471806
k = (-0.6931471806) ÷ (-9.45)
k = 0.07335 (rounded to 5 decimal places)
Therefore, the decay constant for Magnesium-27 is 0.07335.
But my teacher said that the answer should be -0.07335...so i'm guessing that he's saying the decay constant formula should be e^kt = 0.5. I don't get this!! i can't understand it...can any1 help pleeeeeeeeeez???????
2007-05-07 19:48:05 · 4 answers · asked by ♥ Victory ♥ 3 in Science & Mathematics ➔ Mathematics
The idea of using a constant like k is not to have any preconceived notions about it. You know it is a decay equation so you put in -k. You should keep it as k and when you solve and get a negative number, you know it is decaying. Well, that and the .5 which indicates you have less than what you started with! Decay constants should have a negative number. This formula is almost identical to the one for continuously compounded interest and in that case, your k is a positive number indicating you are making money.
2007-05-07 20:00:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The "error" is in the very first step, then. Your teacher wants you to start the problem using e^(kt) = 0.5, instead of using the negative sign in the exponent. Otherwise, your logic is good. I think you answered your own question at the bottom there.
2007-05-07 19:56:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Go along with your teacher on this one, for now, at least. He is indeed using the more generalized form of the exponential, where you don't use minus signs, and using this form the constant is negative. It's kinda like generalizing a quadratic as ax^2 + bx + c. I favor your view, but that won't help your grade any.
2007-05-07 20:01:37 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
1. 3x+ (n+2)x² =5 Since we want to find the value of n, we need to get it to one side. 3x+ nx² +2x² =5 Remove parenthesis nx² =5-3x-2x² Isolate the n nx²/x²=(5-3x-2x²)/x² undo the operation of multiplication n = (5-3x-2x²)/x² 2. isolate p 3p ≥ -x² + x +2 3p/3 ≥( -x² + x +2)/3 divide both sides by 3 p ≥( -x² + x +2)/3 This is fun
2016-05-18 00:31:02 · answer #4 · answered by Anonymous · 0⤊ 0⤋