Question

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

Answer 1

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 − 92m_e

`"m'"(""_58^140"Ce")`= m₂ − 58m_e

`"m'"(""_44^99"Ru")` = m₃ − 44m_e

`"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.