Re: BCAL and KLOE

[Alex Dzierba <dzierba@indiana.edu>]

Yes - of course - thanks much Elton

At 5:27 PM -0500 11/15/07, Elton Smith wrote:
>HI Alex,
>
>I assume their number of 1p.e./MeV is relative to incident photon energy.
>Therefore, our number should be 25 p.e. for mip/3.8 cm = 0.5 (scint
>frac)*3.8 *2 MeV/cm = 3.8 MeV in scintillator, but this is equivalent to
>3.8MeV/sampling fraction = 3.8/0.12 = 32 MeV / 25 p.e. or 0.8 p.e./MeV of
>equivalent energy.
>
>Cheers, Elton.
>


-- 
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Alex R. Dzierba
Chancellor's Professor of Physics (Emeritus)
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