What is the total energy emitted for given fission reaction
$_{0}^{1} n +_{92}^{235} U ⟶_{92}^{236} U ⟶_{40}^{98} Z r +_{52}^{136} T e + 2_{0}^{1} n$
The daughter nuclei are unstable therefore they decay into stable end products ${}_{42}^{98}M o and {}_{54}^{136}X e$ by successive emission of β-particles. Let mass of $_{0}^{1} n = 1.0087$ amu, mass of $_{92}^{235} U = 236.0526$ amu, mass of $_{54}^{136} Xe = 135.9170$ amu, mass of ${}_{42}^{98}M o$ = 97.9054

Question

What is the total energy emitted for given fission reaction

$_{0}^{1} n +_{92}^{235} U ⟶_{92}^{236} U ⟶_{40}^{98} Z r +_{52}^{136} T e + 2_{0}^{1} n$

The daughter nuclei are unstable therefore they decay into stable end products ${}_{42}^{98}M o and {}_{54}^{136}X e$ by successive emission of β-particles. Let mass of $_{0}^{1} n = 1.0087$ amu, mass of $_{92}^{235} U = 236.0526$ amu, mass of $_{54}^{136} Xe = 135.9170$ amu, mass of ${}_{42}^{98}M o$ = 97.9054

Infinity Learn · Accepted Answer

$$01n+92235U$$⟶$$4298Mo+54136Xe+201n(1.0087+235.0439)=(97.9054+135.917+2.0174)$$Δ$$m=0.2128$$∴  Total energy released during a fission $$reaction=0.2128$$×$$931MeV=198MeV$$

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