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AmTran_Bus

  • one year ago

Integration

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  1. AmTran_Bus
    • one year ago
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  2. AmTran_Bus
    • one year ago
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    How do you solve the above integral with this one from the table of integrals?

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  3. AmTran_Bus
    • one year ago
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    I can't see for the life of me how to do it. I've worked others just fine.

  4. phi
    • one year ago
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    the numerator's highest multiplier is 2n-1= 2 -1 = 1 so just 1 the bottom is 4 alpha

  5. phi
    • one year ago
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    and then you have sqr(pi/alpha) I would flip alpha to get 2KT/m \[ \frac{2 K T}{4m} \cdot \sqrt{\frac{2 KT\pi}{m} }\] and simplify that

  6. AmTran_Bus
    • one year ago
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    Ok, thanks Phi. I will try to go back and start from the beginning using what you said.

  7. Empty
    • one year ago
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    Essentially what you need to do is pattern matching. You need to bring your integral in line with this by finding what \(\alpha\) and \(n\) are. Don't let the constant out front of the integral distract you. So start by looking at this integral: \[\int_0^\infty x^{2n} e^{-\alpha x^2} dx \] and let's make it easier on ourselves by switching out the dummy variable \(x\) for \(\nu\) \[\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu \] Ok, so we look in the integral we're given, ignoring the constant out front, that will just multiply our final answer: \[\int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu \] So now it might be easier to see what \(n\) and \(\alpha\) have to be for this to be true: \[\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu = \int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu \] Once you have those, you just plug the values of \(n\) and \(\alpha\) for the solved form of the integral they have given you. Give it a try and see what you get.

  8. AmTran_Bus
    • one year ago
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    So are you saying that 2n =n and the alpha is just m/2pikbt?

  9. phi
    • one year ago
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    n=1, not sure what you mean by 2n=n

  10. AmTran_Bus
    • one year ago
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    Whoops, yes, that's what I mean.

  11. phi
    • one year ago
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    alpha is \[ \alpha= \frac{m}{2 K_B T} \] and \[\frac{1}{ \alpha}= \frac{2 K_B T}{m} \]

  12. phi
    • one year ago
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    I am getting \[ \frac{\sqrt{2\pi}}{2} \left( \frac{K_B T}{m}\right)^\frac{3}{2}\]

  13. Empty
    • one year ago
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    No, not quite, \[\int_0^\infty \nu^{2n} e^{-\alpha \nu^2} d \nu = \int_0^\infty \nu^{2} e^{-m \nu^2/2 k_B T} d \nu \] What I'm saying is the terms are equal here: \(\nu^{2n} = \nu^2\) So we will have \(2n=2\), so \(n=1\). Then we equate the other two terms as well: \[e^{-\alpha \nu^2} =e^{-m \nu^2/2 k_B T} \] Which will give us \[\alpha = m/2 k_B T\] Then we take these values and plug them into the closed form solution: \[\frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^{n+1}\alpha^n} \left( \frac{\pi}{\alpha} \right) ^{1/2} \] So the top part is the odd factorial of all the odd numbers multiplied up to 2n-1, which since n=1, that means 2n-1=2*1-1 = 1. Simple! \[\frac{1}{2^{1+1}( m/2 k_B T)^1} \left( \frac{\pi}{ m/2 k_B T} \right) ^{1/2} \] Now I just plug and chug in my alpha and n values.

  14. Empty
    • one year ago
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    Sorry I was responding to stuff you guys kinda already figured out I guess but w/e lol I'm a bit distracted so my replies are kinda behind a step from you haha

  15. AmTran_Bus
    • one year ago
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    Thanks for your help all! I feel much better about it. The factorial stuff up top was throwing me for a loop.

  16. Empty
    • one year ago
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    Cool fun problem! Is this Thermodynamics you're studying or Physical Chemistry?

  17. AmTran_Bus
    • one year ago
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    Quantum chem!!! Its a pain!

  18. AmTran_Bus
    • one year ago
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    So we are saying this is normalized then.

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