q value for a fission reaction - Consider the fission of ^{238}_{92} U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ^{140}_{58} Ce and ^{99}_{44} Ru . Calculate Q for this fission process.

Consider the fission of ^{238}_{92} U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ^{140}_{58} Ce and ^{99}_{44} Ru . Calculate Q for this fission process.

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Anonym0us (Student, UG1 Grade)

The relevant atomic and particle masses are

m(^{238}_{92} U) = 238.05079 u

m(^{140}_{58} Ce) = 139.90543 u

m(^{99}_{44} Ru) = 98.90594 u

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Ludvig Holberg, detail of an oil painting after a portrait (destroyed) attributed to Roselius, c. ... the world's first self-sustaining nuclear chain reaction, in the Chicago Pile No. 1 ... product formation, and neutron emission in uranium nuclear fission ... 1.0 https://www.britannica.com/science/number-mathematics 2021-03-29 ...

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Nuclei

Einstein gave the famous mass-energy equivalence relation. E = mc2. (13.6) ... of a nucleus and its constituents, ∆M, is called the mass defect, and is given by ... Physics. 444 in the process. The energy Eb is called the binding energy of the nucleus. ... The energy released (the Q value ) in the fission reaction of nuclei like .

For more information, see Nuclei

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Nuclear fission product - Wikipedia

Nuclear fission products are the atomic fragments left after a large atomic nucleus undergoes nuclear fission. Typically, a large nucleus like that of uranium ...

For more information, see Nuclear fission product - Wikipedia

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Veda (last edited 6 months ago) (Student, UG2 Grade)

The amount of energy consumed or emitted during a nuclear reaction is known as the Q value. I can be determined from the difference between the masses of the initial reactants and the masses of the final products of a nuclear reaction. (MeV is a unit of energy).

Given that

Nuclear reaction of Uranium - 238

^{238}_{92} U + ^1_0 n \rightarrow ^{140}_{58} Ce + ^{99}_{44} Ru

m(^{238}_{92} U) = 238.05079 u

m (^{140}_{58} Ce) = 139.90543 u

m(^{99}_{44} Ru) = 98.90594 u

m(^1_0 n) = 1.00867

Step 1: Calculating the Q-value of the Nuclear reaction

Q -value = \text{ mass of } ^{238}_{92} U + \text{ mass of } ^1_0 n - \text{ mass of } ^{140}_{58} Ce - \text{ mass of } ^{99}_{44} Ru

Q-value [math] = [238.05079 + 1.00867 - 139.90543 u - 98.90594]c^2 [/math]

Q-value = 0.24809 c^2 * 931.5 MeV/c^2 u

Q-value = 231.0 MeV

Hence, Q-value of the Nuclear reaction = 231.0 MeV