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Grogs
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I've been asked to find the most probable and average energy of a neutron produced in a fission reaction. The neutron energy spectrum is represented by this function:
[tex]
\chi (E) = 0.453 e^{-1.036E} \sinh(\sqrt{2.29E})
[/tex]
The first part, finding the most probable energy, seems fairly straightforward: take the derivative of the function and find the zeros - one of these zeros will be the peak of the curve, and thus the most probable energy:
[tex]\frac{d}{dE} \chi (E)[/tex]
I ended up with:
[tex]\frac{d}{dE} \chi (E) = 0.453[-1.036 e^{-1.036E} \sinh(\sqrt{2.29E}) + \frac{E^{-0.5}}{2} \sqrt{2.29} cosh(\sqrt{2.29E}) e^{-1.036E}][/tex]
which, by setting it equal to zero and manipulating, I end up with:
[tex]\tanh(1.513 \sqrt{E}) = \frac{0.7303}{\sqrt{E}}[/tex]
I can solve for E numerically by trying different values of E, and indeed I come up with the correct answer (~0.7) I'm just curious if it's possible to solve for E directly in this case.
The 2nd part of the problem, finding the average energy, is giving me more trouble. The average energy should be:
[tex]\overline{E} = \int \limits_{0} ^ {\infty} E \chi (E) \ dE[/tex]
or
[tex]\overline{E} = 0.453 \int \limits_{0} ^ {\infty} E e^{-1.036E} \sinh(\sqrt{2.29E}) \ dE[/tex]
My problem is that I can't figure out how to integrate this equation. When I try to integrate by parts, any value I choose for u and dv yields another integral that needs to be integrated by parts, which yields another etc. Can anyone give me advice on how to solve this?
[tex]
\chi (E) = 0.453 e^{-1.036E} \sinh(\sqrt{2.29E})
[/tex]
The first part, finding the most probable energy, seems fairly straightforward: take the derivative of the function and find the zeros - one of these zeros will be the peak of the curve, and thus the most probable energy:
[tex]\frac{d}{dE} \chi (E)[/tex]
I ended up with:
[tex]\frac{d}{dE} \chi (E) = 0.453[-1.036 e^{-1.036E} \sinh(\sqrt{2.29E}) + \frac{E^{-0.5}}{2} \sqrt{2.29} cosh(\sqrt{2.29E}) e^{-1.036E}][/tex]
which, by setting it equal to zero and manipulating, I end up with:
[tex]\tanh(1.513 \sqrt{E}) = \frac{0.7303}{\sqrt{E}}[/tex]
I can solve for E numerically by trying different values of E, and indeed I come up with the correct answer (~0.7) I'm just curious if it's possible to solve for E directly in this case.
The 2nd part of the problem, finding the average energy, is giving me more trouble. The average energy should be:
[tex]\overline{E} = \int \limits_{0} ^ {\infty} E \chi (E) \ dE[/tex]
or
[tex]\overline{E} = 0.453 \int \limits_{0} ^ {\infty} E e^{-1.036E} \sinh(\sqrt{2.29E}) \ dE[/tex]
My problem is that I can't figure out how to integrate this equation. When I try to integrate by parts, any value I choose for u and dv yields another integral that needs to be integrated by parts, which yields another etc. Can anyone give me advice on how to solve this?
Last edited: