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Half Life Problem

  1. Dec 7, 2004 #1
    Half Life Problem help PLEASE PLEASE

    :yuck: The plutonium isotope Pu-239 is a radioactive material. The half-life of plutonium isotope Pu-239 is 24360 years. If 10 grams of Plutonium were released in the nuclear accident, how long will it take for the 10 gram to decay to 1 gram? Round off your answer to the nearest year.

    I attempted to solve this half-life problem by using log which I havent even been taught yet. The equation I was given for these types of problems was P=Po(1/2)^(t/H) where P=amount left after time, Po=is the original amount, t is the time, and h is the half-life.
    I subbed in the numbers and got 1=10(1/2)^(t/24360)
    my final answer was t=80,922 years. Im not sure if this is correct can someone please verify? o:)
     
    Last edited: Dec 7, 2004
  2. jcsd
  3. Dec 7, 2004 #2

    dextercioby

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    Wacky coincidence,ua nd the "dick"+"jerk" (no "off" :rolleyes: ) meet again... :devil: This time,the individual says u did just fine.
    Did u use a calculator to devide those logarithms??

    Daniel.

    EDIT:You did just fine... :tongue2: You know u managed to break two of this forums rules,the one about conduct (language and personal offences) plus posing the same question on 2 different threads ?? :surprised
     
    Last edited: Dec 7, 2004
  4. Dec 7, 2004 #3
    Yes I used a calculator, is the answer correct?
     
  5. Dec 7, 2004 #4
    Ok, it took me a while to remember how solve these problems, but I eventually got it and I got the same answer you did : 80,922 yrs. Therefore I would conclude that you solved it correctly, unless others prove us wrong.
     
  6. Dec 8, 2004 #5
    Aisha, are you in grade 11 math?
     
  7. Dec 8, 2004 #6
    DEXTERCIOBY :tongue2: I ONLY POSTED THIS QUESTION ONCE :frown: What are u talking about?

    By the way I asked my teacher if I could use this log method on my assignment questions and she said only to check, so now what? I think I have to find the answer by using a graphing calculator, can someone please help me out? Every time I put the equation into the calculator I get a completely horizontal line, something I am inputting is wrong. HELP PLEASE!
     
  8. Dec 8, 2004 #7

    dextercioby

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    It one of the equations which you posted yesterday,that was on 2 different threads,but one of the moderators saw that and deleted the thread as a whole.
    On the other hand,which eq are u puting in the computer??Generally computer sofware is not wrong (unless it has bugs (virus and s***)),but the humans which manipulate it.The exception is Windows,which really sucks!!! :tongue2:

    Daniel.
     
  9. Dec 8, 2004 #8
    I am using a graphing calculator and I think I am inputing the equation wrong because its always a horizontal line that has no zeros. What am I doing wrong? Im inputing 10(1/2)^(alpha t/24360)-1.0 in y= I tried adding more brackets but nothing happens. :cry:
     
  10. Dec 8, 2004 #9

    Diane_

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    Aisha - You can use a graphing calculator to get an approximate answer. Plug it in, the trace the graph until you get to a y-coordinate of 1. Your answer will be the x-coordinate.

    Oh, and by the way - if you're putting in the equation correctly, there should be no zeros. Exponential decay never reaches zero mathematically.

    Without knowing what calculator you're using, there's no way I can tell you how to put it in. Since you're probably using a TI, I probably couldn't tell you how to put it in even if I knew what kind of calculator you're using. I will suggest this: there are two ways that students regularly end up with horizontal lines on graphing calculators. One is that they don't use what the calculator recognizes as the "proper" x-coordinate. Make sure you've plugged in the equation correctly. The other is that they don't really have a horizontal line - it just looks like it because the scale is off. With a half-life of 24000 years, it's going to take awhile before the line starts to dip on most calculators.
     
  11. Dec 8, 2004 #10
    Im using a T1 83 graphing calculator I finally changed the window and got an x-intercept, y did u say that there would be a horizontal line and it wouldnt have zeros, if I inputed the equation right?

    Also can u tell me a way of looking at the equation that can help me in settin the window on the graphing calculator. :smile:
     
  12. Dec 8, 2004 #11

    Diane_

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    I was assuming (badly, as it turns out) that you were putting in the equation as an exponential-decay equation. After reading some of your other questions, I realized that your teacher has you approaching it from a different direction. What your teacher is doing is probably easier, so the upshot is: just ignore me there.

    As far as the other question, I wish I could give you more than a generic answer, but I don't use that calculator. Most graphing calculators have an "Options" section where you can select the range of x-values to be displayed. That's the thing you'd want to change. If you have the manual, check the index for the word "range" and see what it says.

    By the way - I find your perseverence in these matters very impressive. A lot of people give up way too easily! You go, girl! :)
     
  13. Dec 8, 2004 #12
    Since you know which y value you want, i think your idea is a good one. You want x when y = 1, so plug in y = 10*((0.5)^(X/24360)) - 1. Then just use the zero function (2nd --> trace --> number 2) to solve for x. Put in the equation exactly as i said it.

    I see what youre problem is now. When you were graphing it you did alpha t, instead of X. This doesn't work because the letter t is NOT a variable, it is a constant. So regardless, it will only input one value of t. That is why you get the straight line. The y value would not change. Whenever you are graphing anythnig, you have to use the X as your variable. It is the only one that is really a variable.
     
  14. Dec 8, 2004 #13
    :rofl: THANKS SOOOOO MUCH !! :rolleyes:
     
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