1. An excited U * nucleus undergoes fission into two fragments., as described by
ID: 1398159 • Letter: 1
Question
1.
An excited U* nucleus undergoes fission into two fragments., as described by the following equation:
U* Ba + Kr
The following atomic masses are known:
Kr 91.926270 amu
Ba 143.922845 amu
U* 236.045563 amu
What is the reaction energy? Let 1 amu = 931.5 MeV.
150 MeV
170 MeV
190 MeV
160 MeV
180 MeV
2.
The absorbed dose of X-rays is 70 mrad over a mass of 6.0 kg. What is the total energy absorbed?
2.6 x 1016 eV
7.0 x 104 eV
6.7 x 104 eV
4.2 x 1016 eV
3.
A 70 kg researcher absorbs 4.5 × 108 neutrons in a work day. The energy of the neutrons is 1.2 MeV. The relative biological efficiency (RBE) for fast neutrons is 10. The equivalent dosage of the radiation exposure, in mrem, is closest to which of the following values?
3.7
12
1.2
0.77
0.39
1.
An excited U* nucleus undergoes fission into two fragments., as described by the following equation:
U* Ba + Kr
The following atomic masses are known:
Kr 91.926270 amu
Ba 143.922845 amu
U* 236.045563 amu
What is the reaction energy? Let 1 amu = 931.5 MeV.
150 MeV
170 MeV
190 MeV
160 MeV
180 MeV
2.
The absorbed dose of X-rays is 70 mrad over a mass of 6.0 kg. What is the total energy absorbed?
2.6 x 1016 eV
7.0 x 104 eV
6.7 x 104 eV
4.2 x 1016 eV
Explanation / Answer
energy of the reactant = E1 = 236.045563*931.5 = 219876.44193450001 MeV
energy of the product = (91.926270+143.922845)*931.5 = 219693.45062250001 MeV
reaction energy = 219876.44193450001 - 219693.45062250001
reaction energy = 182.991312 MeV
180 MeV <<<----------answer
+++++++++++++
absorbed dose = 70 mrad = 70*10^-2 my = 0.7*10^-3 J/kg
energy absorbed = 0.7*10^-3*6 = 4.2*10^-3 J
E = (4.2*10^-3)/(1.6*10^-19) = 2.6*10^16 eV
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