Fission is often achieved by bombarding a U-235 nucleus with a neutron, temporar

Question

Fission is often achieved by bombarding a U-235 nucleus with a neutron, temporarily creating a highly unstable U-236 nucleus, which then can decay in a number of different ways. One such way is shown here (omitting the temporary U-236 nucleus). Fill in the missing product, including subscripts and superscripts, in the nuclear fission reaction shown here. Use the notation *X, where X is the atomic symbol, Z is the atomic number, and A is the atomic mass number. (b) How much energy is released during this reaction? The masses involved are as follows: neutron: 1.008 665 u U-235: 235.043 923 u Sr-93: 92.914 026 u unknown daughter nucleus: 140.926 650 u

Explanation / Answer

a) The missing product is 141Xe54

b) Total mass of reactants is 1.008665 u + 235.043923 u = 236.052588 u

Total mass of products is 140.926650+92.914026 + 2*1.008665 = 235.858006 u

Hence mass defect = 236.052588 u - 235.858006 u = 0.194582 u

This mass defect is converted into energy.

0.194582 u *931.5MeV/u = 181.25 MeV

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