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
How much energy is released in this reaction? (1 u = 931.5 MeV/c²)

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
How much energy is released in this reaction? (1 u = 931.5 MeV/c²)

Quizplus · Accepted Answer

The mass defect is:$$\Delta m = (m_U + m_n) - (m_Xe + m_Sr + 2m_n)$$$$= (235.043924 u + 1.008665 u) - (139.921620 u + 93.915367 u + 2(1.008665 u))$$$$= 0.199267 u$$The energy released is:$$E = \Delta m c^2 = (0.199267 u)(931.5 MeV/c^2)$$$$= 185 MeV$$

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 $^{235}_{92}$