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Physics question 2
- To: KellySt@aol.com, hous0042@maroon.tc.umn.edu, T.L.G.vanderLinden@student.utwente.nl, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, rddesign@wolfenet.com, David@interworld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com, sl0c8@cc.usu.edu, MLEN3097@Mercury.GC.PeachNet.EDU, 101765.2200@compuserve.com
- Subject: Physics question 2
- From: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)
- Date: Wed, 29 May 1996 13:57:31 -0500
In case your woundering, the procedure I used to conver MeV to Watts/kg was:
Power is given in Million electron volts (MeV). An electron volt is equal
to 1.60219 E-19 Joule, and a Joule /
sec is equal to a Watt. Mass in this case is the number of Protons and
neutrons involved in the reaction. They
each weigh about 1.673 E-27 kg. So since p + 11B has 12 P's and N's per
8.68 MeV reaction. (Note I'm
ignoring electrons. Life's to short, they're too light).
8.7 MeV
= 8.7 E6 eV per 12 * 1.673 E-27 kg
= 4.324 E32 eV per / kg
= 6.926 E13 watts / kg (1 ev/second = 1.60219 E-19
Joule / sec)
1 j/s = Watt
Kelly
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Kelly Starks Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)
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