Question

Consider the fission of ${\displaystyle {}_{92}^{238}\text{U}}$ 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 ${\displaystyle {}_{58}^{140}\text{Ce}}$ and ${\displaystyle {}_{44}^{99}\text{Ru}}$. Calculate Q for this fission process. The relevant atomic and particle masses are

${\displaystyle \text{m}\left({}_{92}^{238}\text{U}\right)}$ = 238.05079 u

${\displaystyle \text{m}\left({}_{58}^{140}\text{Ce}\right)}$ = 139.90543 u

${\displaystyle \text{m}\left({}_{44}^{99}\text{Ru}\right)}$ = 98.90594 u

In a fission event of ${\displaystyle {}_{92}^{238}\text{U}}$ by fast-moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are ${\displaystyle {}_{58}^{140}\text{Ce}}$ and ${\displaystyle {}_{44}^{99}\text{Ru}}$ Calculate Q for this process. Neglect the masses of electrons/positrons emitted during the intermediate steps.

Given:

${\displaystyle \text{m}\left({}_{92}^{238}\text{U}\right)}$ = 238.05079 u

${\displaystyle \text{m}\left({}_{58}^{140}\text{Ce}\right)}$ = 139.90543 u

${\displaystyle \text{m}\left({}_{44}^{99}\text{Ru}\right)}$ = 98.90594 u

${\displaystyle \text{m}\left({}_0^1\text{n}\right)}$ = 1.008665 u

Answer 1

In the fission of ${\displaystyle {}_{92}^{238}\text{U}}$, 10 β⁻ particles decay from the parent nucleus. The nuclear reaction can be written as:

$${\phantom{A}}_{\phantom{92}}^{\phantom{238}}{\phantom{A}}_{\phantom{2}92}^{\phantom{2}238}\mathrm{U}+{\phantom{A}}_{\phantom{0}}^{\phantom{1}}{\phantom{A}}_{\phantom{2}0}^{\phantom{2}1}\mathrm{n}⟶{\phantom{A}}_{\phantom{58}}^{\phantom{140}}{\phantom{A}}_{\phantom{2}58}^{\phantom{2}140}\mathrm{Ce}+{\phantom{A}}_{\phantom{44}}^{\phantom{99}}{\phantom{A}}_{\phantom{2}44}^{\phantom{2}99}\mathrm{Ru}+10{\phantom{A}}_{-1}^0\mathrm{e}$$

It is given that:

Mass of a nucleus ${\displaystyle \text{m}\left({}_{92}^{238}\text{U}\right)}$ m₁ = 238.05079 u

Mass of a nucleus ${\displaystyle \text{m}\left({}_{58}^{140}\text{Ce}\right)}$m₂ = 139.90543 u

Mass of a nucleus${\displaystyle \text{m}\left({}_{44}^{99}\text{Ru}\right)}$, m₃ = 98.90594 u

Mass of a neutron ${\displaystyle \text{m}\left({}_0^1\text{n}\right)}$, m₄ = 1.008665 u

Q-value of the above equation,

${\displaystyle \text{Q}=[\text{m'}\left({}_{92}^{238}\text{U}\right)+\text{m}\left({}_0^1\text{n}\right)-\text{m'}\left({}_{28}^{140}\text{Ce}\right)-\text{m'}\left({}_{44}^{99}\text{Ru}\right)-10{\text{m}}_{\text{e}}]{\text{c}}^2}$

Where,

m’ = Represents the corresponding atomic masses of the nuclei

${\displaystyle \text{m'}\left({}_{92}^{238}\text{U}\right)}$ = m₁ − 92m_e

${\displaystyle \text{m'}\left({}_{58}^{140}\text{Ce}\right)}$= m₂ − 58m_e

${\displaystyle \text{m'}\left({}_{44}^{99}\text{Ru}\right)}$ = m₃ − 44m_e

${\displaystyle \text{m}\left({}_0^1\text{n}\right)}$= m₄

${\displaystyle \text{Q}=[{\text{m}}_1-92{\text{m}}_{\text{e}}+{\text{m}}_4-{\text{m}}_2+58{\text{m}}_{\text{e}}-{\text{m}}_3+44{\text{m}}_{\text{e}}-10{\text{m}}_{\text{e}}]{\text{c}}^2}$

${\displaystyle =[{\text{m}}_1+{\text{m}}_4-{\text{m}}_2-{\text{m}}_3]{\text{c}}^2}$

${\displaystyle =[238.05079+1.008665-139.90543-98.90594]{\text{c}}^2}$

${\displaystyle =[0.248535 {\text{c}}^2]\text{u}}$

But 1 u = ${\displaystyle 931.5\text{MeV}/{\text{c}}^2}$

${\displaystyle \text{Q}=0.248535\times 931.5=231.510 \text{MeV}}$

Hence, the Q-value of the fission process is 231.510 MeV.