Modern Physics: Nuclear Physics -- Mass-Energy Calculate the energy released (in

Question

Modern Physics: Nuclear Physics -- Mass-Energy

Calculate the energy released (in MeV, rounded to the nearest MeV) in the following fission reaction:

Course Contents . Homework for Topic 8. Nuclear Physics Problem 1 (mass-energy) Calculate the energy released (in Mev, rounded to the nearest MeV) in the following fission reaction: 1 235 0 92 Masses: J- IPRh +'Ag + neutrons 115 45 47 n: 1.008665 u 235 92 115 45 U: 235.043923 u Rh: 114.92012 u 117Ag: 116.91168 u 47 1 u 931.494 Mev Submit Answer Tries 0/3

Explanation / Answer

First of all we will find the number of neutrons towards the right hand side of the given equation using the conservation of mass

Mass towards left hand side of equation is = 1 + 235 = 236

Mass towards right hand side of the equation = 115 + 117 + m (mass of neutrons) = 232 + m

equating both sides 232 + m = 236 hence m = 4

It denotes that number of neutrons right hand side of the equation= 4 neutrons

Lets now find mass defect in the given fission reaction

mass defect dm = 1.008665 + 235.043923 - 114.92012 - 116.91168 - 4 x 1.008665 = 0.186128 u

Energy released = 0.186128 X 931.494 = 173.377115232 MeV

= 173.38 MeV

Answer

Energy released is equal to 173.38 MeV

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