Question

Calculate the ground state $Q$ value of the induced fission reaction in the equation
$n{+}_{92}{U}^{23}\to \_{40}^{}Z{r}^{99}+\_{52}^{}T{e}^{134}+3n$
$m\left(n\right)=1.0087u,M\left(U^{235}\right)=235.0439u,$
$M\left(Zr\right)=98.916uM\left(T{e}^{134}\right)=133.9115u$

• A. $184.84MeV$
• B. $200MeV$
• C. $130MeV$
• D. $300MeV$

Answer 1

Answer: (A)

$Q=\Delta m\times 931$
$=\left[\left(235.0439+1.0087\right)-\left(98.916+133.9115+3.0261\right)\times 931\right]$
$Q=184.84MeV$