The nuclear reaction H $$11$$ $$+$$ H $$11$$ $$=$$ He $$24$$ (mass$$ of deuteron $$=$$ $$2.0141$$ a.m.u. and mass of He $$=$$ $$4.0024$$ a.m.u.$$) is

Correct option is 1Fusion reaction releasing 24 MeV energy

Questions

The nuclear reaction ${}_{1}^{1}H + {}_{1}^{1}H = {}_{2}^{4}{He}$ (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is

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Fusion reaction releasing 24 MeV energy

Fusion reaction absorbing 24 MeV energy

Fission reaction releasing 0.0258 MeV energy

Fission reaction absorbing 0.0258 MeV energy

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detailed solution

Correct option is A

Total mass of reactants= (2.0141) x 2 = 4.0282 amuTotal mass of products = 4.0024 amuMass defect = 4.0282 amu — 4.0024 amu = 0.0258 amu∴Energy released E = 931 x 0.0258= 24 MeV

Similar Questions

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