Calculate the energy released in MeV in the following nuclear reaction - 12H-12H- 23He-01nAssume that the masses of 12H- 23He and neutron n respectively are 2-020- 3-0160 and 1-0087 in amu-

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. In the fusion reaction

_{1}^{2} H e +_{1}^{2} H \to_{2}^{3} H e +_{0}^{1} n

, the masses of deuteron, helium, and neutron expressed in amu are

2.015, 3.017

and

1.009

, respectively. If

1 k g

of deuterium undergoes complete fusion, find the amount of total energy released.

(1 a m u = 931.5 m e V / c^{2})

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Calculate the energy released in MeV in the following nuclear reaction :21H+21H→32He+10nAssume that the masses of 21H, 32He and neutron (n) respectively are 2.020, 3.0160 and 1.0087 in amu.

The correct option is A 14.25 MeVThe mass defect is the difference between the mass of reactants and the mass of products.Δm=2×m 12H−m 23He−mn=(2×2.020)−(3.0160+1.0087)=4.040−4.0247=0.0153The energy released during the nuclear reaction is ΔE=Δm×931.48MeV=14.25MeV

Calculate the energy released in MeV in the following nuclear reaction :
$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H,_{2}^{3} He$ and neutron (n) respectively are 2.020, 3.0160 and 1.0087 in amu.