Question

Calculate the ground state Q value of the induced fission reaction in the equation
$n{+}_{92}^{235}U{\to }_{92}^{236}{U}^{\ast }{\to }_{40}^{99}Ze{+}_{52}^{134}Te+3n$
if the neutron is thermal. A thermal neutron is in the thermal equilibrium with its environment; it has an average kinetic energy given by (3/2) kT. Given: $m(n)=1.0087 amu, M(^{235}U)=235.0439 amu, M(^{99}Zr)=98.916 amu M(^{134}Te)=133.9115 amu$

Answer 1

Q value is equal to mass-energy equivalence

$Q=△m{c}^2△m=m_{\left(n\right)}+m_{{(}^{235}U)}-m_{{(}^{99}Ze)}-m_{{(}^{134}Te)}-3m_{\left(n\right)}=m_{{(}^{235}U)}-m_{{(}^{99}Ze)}-m_{{(}^{134}Te)}-2m_{\left(n\right)}=235.0439-98.916-133.9115-2\times 1.0087=0.199amuQ=0.199\times 931MeVQ=185.269MeV$