Re: [fluka-discuss]: neutron yield

From: Ivan <gordeev_at_jinr.ru>
Date: Tue, 15 Jan 2019 23:33:52 +0300

Hello again!

The first thing which is really confuses me is that you define polar
angle from 0 to 6.28(?) which is 2pi. The thing is that in input of
USRYIELD if you choose "Polar Theta lab" as a second quantity you should
set value in *radians*, or, if you choose "Polar Theta lab deg" in
*degrees*, but, in both cases polar angle should go from 0 to *3.14(180
deg)*.

Check also FLUKA manual carefully:

> = 14 : polar angle in the lab frame    (#)

> ( = 24 : like 14, but with input data given
>                                in degrees rather than in radians ) (#)

You also can check the information from .sum output file, it says:

> 2nd variable (x2) is: Laboratory Angle (radians)
So, I think the first thing is that you should change *6.28* to *3.14*
on your input. If you want the full angle of emission.


And yes, in the previous letter I noticed that you should multiply on
angle interval in *steradians! *Because for output you must use *solid
angle*, no matter how you defined it is in input.**And that means you
need to use *steradians* for data in the output, not for input!
**

Here, info from FLUKA manual again:
**

> 6) In the case of polar angle quantities (|ie| or |ia| = 14,15,17,18,
>             24,25) the differential yield is always referred to solid
> angle in
>             steradian, although input is specified in radian or degrees.
So, taking into account what I wrote above, I think this:

> Ihave used this normalize: Yield = d2N/dx1dx2 * dOmega(=6.28) * 10^10
> = neutrons/MeV

Also should be like this:

Energy Differential Yield  = d2N/dx1dx2 * dOmega(*12.56* which is 4pi -
full solid angle) * 10^13(why 10^10? if your intensity 10^13?) =
neutrons/(*GeV**s) (be careful about energy units as well!)

and for yield:

Yield  = d2N/dx1dx2 * dOmega(*12.56* which is 4pi - full solid angle) *
dE(col2-col1) * 10^13 = number of neutrons per second


Best wishes,

Ivan Gordeev


15.01.2019 13:06, kabytayeva пишет:
>
> Ivan Gordeev, many thanks for your help!
>
> I am studying the neutron yield from an Al target hitten by a 12C beam
> with energy 6MeV/nucleon and intensity 10^13 ion/s
>  Could you please take a look at my input card and tell me if I made
> some mystake? Please find in the attached files also the output of the
> USRYIELD card I used to score the neutron yield,because neutron output
> seems to me very high. Ineed not to normalize to Nbins, is not it?
>
> Ihave used this normalize: Yield = d2N/dx1dx2 * dOmega(=6.28) * 10^10
> = neutrons/MeV
>
>
> Thank you very much for your help
>
> С уважением,Раушан Кабытаева
> Best regards, Raushan Kabytayeva
> engineer of the RIB group
>
> Flerov Laboratory of Nuclear Reactions
> Joint Institute for Nuclear Research
> Joliot-Curie str, 6. Bld 131, office 408.
> 141980 Dubna, Moscow region - Russia
>
> phone: +7(49621)62517
> e-mail:kabytayeva_at_jinr.ru


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Received on Tue Jan 15 2019 - 22:41:04 CET

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