When a neutron collides with a uranium-235 nucleus it can induce a variety of fission reactions. One such reaction is $^{235}_{92}$ U + $^{1}_{0}$ n → $^{140}_{54}$ Xe + $^{94}_{38}$ Sr + 2 $^{1}_{0}$ n The following mass values are known: $^{140}_{54}$ Xe: 139.921620 u $^{94}_{38}$ Sr: 93.915367 u $^{235}_{92}$ U: 235.043924 u $^{1}_{0}$ n: 1.008665 u What mass of uranium is needed to produce a 10-kiloton yield?
(1 u = 931.5 MeV/c² = 1.66054 × 10^-27 kg, 1-kiloton yield = 5.0 × 10¹² J)

Question

When a neutron collides with a uranium-235 nucleus it can induce a variety of fission reactions. One such reaction is

^{235}_{92}

U +

^{1}_{0}

n →

^{140}_{54}

Xe +

^{94}_{38}

Sr + 2

^{1}_{0}

n The following mass values are known:

^{140}_{54}

Xe: 139.921620 u

^{94}_{38}

Sr: 93.915367 u

^{235}_{92}

U: 235.043924 u

^{1}_{0}

n: 1.008665 u What mass of uranium is needed to produce a 10-kiloton yield?
(1 u = 931.5 MeV/c² = 1.66054 × 10^-27 kg, 1-kiloton yield = 5.0 × 10¹² J)

Quizplus · Accepted Answer

First, we need to calculate the energy released in the fission reaction:
Mass of reactants = 235.043924 u + 1.008665 u = 236.052589 u
Mass of products = 139.921620 u + 93.915367 u + 2(1.008665 u) = 236.854317 u
Mass defect = (236.052589 u - 236.854317 u) = -0.801728 u
Energy released = (0.801728 u)(931.5 MeV/c²u) = 746.2 MeV

Next, we need to calculate the amount of uranium needed to produce a 10-kiloton yield, which is 5.0 x 10¹² J:
Number of fissions required = (5.0 x 10¹² J) / (746.2 MeV/fission x 1.6 x 10⁻¹³ J/MeV) = 4.7 x 10²⁰ fissions
Mass of uranium required = (4.7 x 10²⁰ fissions) x (235.043924 u/1 fission) / (1.66054 x 10⁻²⁷ kg/u) = 0.66 kg

Therefore, the best choice is B (0.66 kg).

When a Neutron Collides with a Uranium-235 Nucleus It Can 92235^{235}_{92}92235​

When a Neutron Collides with a Uranium-235 Nucleus It Can $^{235}_{92}$