A detector is used to examine the radiation emitted from an uknown source. In on

Question

A detector is used to examine the radiation emitted from an uknown source. In one minute it records counts of two specific energies:(3640+-120) counts at (1.3+-0.1) MeV and (3850+-120) counts at (1.2+-0.1) MeV.The mass of the detector is (450+-10)g. (all uncertanties are at 95% confidence levels) Find the absorbed dose (gamma ray energy times number of gammas absorbed divided by the mass of the material), in joules per kilogram due to:
1) The (1.3+-0.1)MeV radiation
2) The (1.2+-0.1) MeV radiation
3) Both energies
Note :1 MeV=10^6 eV and 1eV=1.6*10^-19J

Explanation / Answer

dose=1.3*3640*1.6*10-13/0.450=1.68*10-9
error=(120/3640+0./1.3+10/450)*100%=13.21%=2.2*10-10

dose=1.68*10-9(+-)2.2*10-10 J/kg

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