Re: A "dirty bomb" question

["Syd Levine \(AnaLog\)" <AnaLog@logwell.com>]

Yep, I saw that after the first one to point it out.  It is indeed a weak

gamma, which only a fraction of can even get through a scintillation

detector housing.



----- Original Message ----- 

From: "Franz Schoenhofer" <franz.schoenhofer@chello.at>

To: "Syd Levine (AnaLog)" <AnaLog@logwell.com>; "Thomas M Lashley"

<lashleyt@DTEENERGY.COM>; "Richard L. Hess" <lists@richardhess.com>

Cc: <radsafe@list.vanderbilt.edu>

Sent: Monday, January 01, 1996 1:24 AM

Subject: AW: A "dirty bomb" question





>

>

> -----Ursprüngliche Nachricht-----

> Von: owner-radsafe@list.vanderbilt.edu

> [mailto:owner-radsafe@list.vanderbilt.edu]Im Auftrag von Syd Levine

> (AnaLog)

> Gesendet: Dienstag, 28. Oktober 2003 18:23

> An: Thomas M Lashley; Richard L. Hess

> Cc: radsafe@list.vanderbilt.edu

> Betreff: Re: A "dirty bomb" question

>

>

> True, but it also puts out a 59.5 meV gamma, and I think a 14 meV gamma as

> well.  The 59.5 meV gamma is handy for setting the discrimination circuit

in

> gross count scintillation gamma ray oil well logging tools.

> ------------------------------

>

> The units you use are wrong, it should read "keV" (kiloelectronvolts).

"meV"

> would mean "millielectronvolts" or "MeV" would mean "megaelectronvolts",

> both of which are wrong by three orders of magnitude.

>

> Best regards,

>

> Franz

>

>





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