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The Binding energy per nucleon of ${}_{3}^{7}{Li}$ and ${}_{2}^{4}{He}$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ${}_{3}^{7}{Li} + {}_{1}^{1}H \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + Q$ the value of energy Q released is

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8.4 MeV

17.3 MeV

19.6 MeV

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detailed solution

Correct option is C

B.E of Li 7 3 =7 x 5.60 Mev=39.2 MevB.E of He 4 2 =4 x 7.06 Mev=28.24 MevQ=2B.E of He 4 2 -B.E of Li 7 3 =(2 x 28.24) – 39.2=17.28 Mev

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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