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Questions
Consider the fission of `""_92^238"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 `""_58^140"Ce"` and `""_44^99"Ru"`. Calculate Q for this fission process. The relevant atomic and particle masses are
`"m"(""_92^238"U")` = 238.05079 u
`"m"(""_58^140"Ce")` = 139.90543 u
`"m"(""_44^99"Ru")` = 98.90594 u
In a fission event of `""_92^238"U"` by fast-moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are `""_58^140"Ce"` and `""_44^99"Ru"` Calculate Q for this process. Neglect the masses of electrons/positrons emitted during the intermediate steps.
Given:
`"m"(""_92^238"U")` = 238.05079 u
`"m"(""_58^140"Ce")` = 139.90543 u
`"m"(""_44^99"Ru")` = 98.90594 u
`"m"(""_0^1"n")` = 1.008665 u
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Solution
In the fission of `""_92^238"U"`, 10 β− particles decay from the parent nucleus. The nuclear reaction can be written as:
\[\ce{^238_92U + ^1_0n -> ^140_58Ce + ^99_44Ru + 10^0_{-1}e}\]
It is given that:
Mass of a nucleus `"m"(""_92^238"U")` m1 = 238.05079 u
Mass of a nucleus `"m"(""_58^140"Ce")`m2 = 139.90543 u
Mass of a nucleus`"m"(""_44^99"Ru")`, m3 = 98.90594 u
Mass of a neutron `"m"(""_0^1"n")`, m4 = 1.008665 u
Q-value of the above equation,
`"Q" = ["m'"(""_92^238"U") + "m"(""_0^1"n") - "m'"(""_28^140"Ce") - "m'"(""_44^99"Ru") - 10"m"_("e")]"c"^2`
Where,
m’ = Represents the corresponding atomic masses of the nuclei
`"m'"(""_92^238"U")` = m1 − 92me
`"m'"(""_58^140"Ce")`= m2 − 58me
`"m'"(""_44^99"Ru")` = m3 − 44me
`"m"(""_0^1"n")`= m4
`"Q" = ["m"_1 - 92"m"_"e" + "m"_4 - "m"_2 + 58"m"_"e" - "m"_3 + 44"m"_"e" - 10"m"_"e"]"c"^2`
`= ["m"_1 + "m"_4 - "m"_2 - "m"_3]"c"^2`
`= [238.05079 + 1.008665 - 139.90543 - 98.90594]"c"^2`
`= [0.248535 "c"^2]"u"`
But 1 u = `931.5 " MeV"//"c"^2`
`"Q" = 0.248535 xx 931.5 = 231.510 "MeV"`
Hence, the Q-value of the fission process is 231.510 MeV.
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