Question

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

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

Answer 1

Answer: (A)

$Q=\Delta m\times 931$
$=\left[\left(235 . 0439 + 1 . 0087\right) - \left(98 . 916 + 133 . 9115 + 3 . 0261\right) \times 931\right]$
$Q=184.84MeV$