Calculate the energy released in joules and MeV in the following nuclear reaction.
$_{1}^{2} H +_{1}^{2} H ⟶_{2}^{3} H e +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H$ , $_{2}^{3} H e$ and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu.

Question

Calculate the energy released in joules and MeV in the following nuclear reaction.
$_{1}^{2} H +_{1}^{2} H ⟶_{2}^{3} H e +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H$ , $_{2}^{3} H e$ and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu.

Vedantu Content Team · Accepted Answer

Hint: Nuclear reactions are the reactions in which one or more nuclides are produced from the collisions of 2 atomic nuclei or one subatomic particle and an atomic nucleus .Complete step by step solution: There are 4 types of nuclear reactions: Fission, fusion, nuclear decay, and transmutation. The type of nuclear reaction present in the reaction given is fusion reaction. In this reaction two or more atomic nuclei combine to form a single nuclei. Now, the mass defect is given as:$Mass\quad defect,\quad \Delta m\quad =\quad Mass\quad of\quad reactant\quad -\quad mass\quad of\quad product$The mass of the reactant is 2*2.0141 and the mass of the product is 3.0160 + 1.0087. Substituting these values in the above equation, we get$Mass\quad defect,\quad \Delta m\quad =\quad (2\quad \times \quad 2.0141)\quad -\quad (3.0160\quad +\quad 1.0087)$$\implies Mass\quad defect,\quad \Delta m\quad =\quad 4.0282\quad -\quad 4.0247$$\implies Mass\quad defect,\quad \Delta m\quad =\quad 0.0035\quad amu\quad =\quad 3.5\quad \times \quad { 10 }^{ -3 }\quad amu$Now, as we can see the mass defect is coming out to be positive which means that the mass of the reactant is greater than the mass of the product. This shows that some energy has been released in the reaction.$_{ 1 }^{ 2 }{ H }\quad +\quad _{ 1 }^{ 2 }{ H }\quad \longrightarrow \quad _{ 2 }^{ 3 }{ He }\quad +\quad _{ 0 }^{ 1 }{ n }\quad +\quad \Delta E$Now, we know that 1 amu = 931.5 MeVSo, now let us calculate the energy which is given by:$Energy\quad =\quad 3.5\quad \times \quad { 10 }^{ -3 }\quad amu\quad =\quad 931.5\quad \times \quad 3.5\quad \times \quad { 10 }^{ -3 }\quad MeV$$\implies Energy\quad =\quad 3.26025\quad MeV$Also,  $1\quad MeV\quad =\quad 1.602\quad \times \quad { 10 }^{ -13 }\quad Joules$Therefore, $ Energy\quad in\quad Joules\quad =\quad 3.26025\quad \times \quad 1.602\quad \times \quad { 10 }^{ -13 }\quad Joules$$\implies Energy\quad in\quad Joules\quad =\quad 5.2229205\quad \times \quad { 10 }^{ -13 }\quad Joules$Thus, the energy released in MeV and Joules is 3.26025 MeV and $5.22292005 \times {10}^{-13}$ Joules respectively.Note: While calculating the quantities like energy in joules and MeV, do remember the conversion factor. Because, if you don't use the conversion factor correctly you might end up getting the wrong answers.

Calculate the energy released in joules and MeV in the following nuclear reaction.12H+12H⟶23He+01nAssume that the masses of 12H, 23He and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu.

Repeaters Course for NEET 2022 - 23

Calculate the energy released in joules and MeV in the following nuclear reaction.
$_{1}^{2} H +_{1}^{2} H ⟶_{2}^{3} H e +_{0}^{1} n$
Assume that the masses of $_{1}^{2} H$ , $_{2}^{3} H e$ and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu.