Consider the reaction 235 92 U + 1 0 n 148 57 La + 87 35 Br + 1 0 n. Element Ato

Question

Consider the reaction 23592U +10n 14857La +8735Br + 10n. Element      Atomic Mass (u) 23592U 235.043923 10n 1.008665 14857La 147.932236 8735Br 86.92071119 (a) Write the conservation of relativisticenergy equation symbolically in terms of the rest energy and thekinetic energy, setting the initial total energy to the final totalenergy. (Do this on paper. Your instructor may ask you to turn inthis work.)

(b) Using values given above, find the total mass of the initialparticles.
u

(c) Using the values given above, find the total mass of theparticles after the reaction takes place.
u

(d) Subtract the final particle mass from the initial particlemass.
u

(e) Convert the answer to part (d) to MeV, obtaining the kineticenergy of the daughter particles, neglecting the kinetic energy ofthe reactants.
Mev Element      Atomic Mass (u) 23592U 235.043923 10n 1.008665 14857La 147.932236 8735Br 86.92071119 (a) Write the conservation of relativisticenergy equation symbolically in terms of the rest energy and thekinetic energy, setting the initial total energy to the final totalenergy. (Do this on paper. Your instructor may ask you to turn inthis work.)

(b) Using values given above, find the total mass of the initialparticles.
u

(c) Using the values given above, find the total mass of theparticles after the reaction takes place.
u

(d) Subtract the final particle mass from the initial particlemass.
u

(e) Convert the answer to part (d) to MeV, obtaining the kineticenergy of the daughter particles, neglecting the kinetic energy ofthe reactants.
Mev Element      Atomic Mass (u) 23592U 235.043923 10n 1.008665 14857La 147.932236 8735Br 86.92071119

Explanation / Answer

a) the total mass of the initial particles is = 235.043923u +1.008665u = 236.052588u b) the total mass of the particles after the reaction takes place= 147.932236 u + 86.92071119 u +  1.008665u                                                                                         = 235.86161219u c) Now mass defect is m = 236.052588u - 235.86161219u                                     = 0.19097581u d) Binding energy is E =(m)(931.5MeV/u)                                 = (0.19097581u)(931.5MeV/u)                                 = 177.893967015MeV d) Binding energy is E =(m)(931.5MeV/u)                                 = (0.19097581u)(931.5MeV/u)                                 = 177.893967015MeV

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