The amount of energy released in the fusion of $_{1} H^{2}$ to form a $_{2} H e^{4}$ nucleus will be :
[Binding energy per nucleon of $_{1} H^{2} = 1.1 M e V$ and $_{2} H e^{4} = 7 M e V$ ]

Question

The amount of energy released in the fusion of 1H2 to form a 2He4 nucleus will be -Binding energy per nucleon of 1H2-1-1 MeV and 2He4-7 MeV-

Toppr Admin · Accepted Answer

Nuclear reaction- -xA0-2-xA0-1H2-x2192-xA0-2He4-xA0-Energy released - Binding energy of products -x2212- Binding energy of reactants-xA0-Total Binding energy - nucleons x binding energy per nucleon4-xD7-7-x2212-2-xD7-2-xD7-1-1-23-6-xA0-MeV

The amount of energy released in the fusion of 1H2 to form a 2He4 nucleus will be :[Binding energy per nucleon of 1H2=1.1 MeV and 2He4=7 MeV]

Nuclear reaction: 2 1H2→ 2He4 Energy released = Binding energy of products − Binding energy of reactants. Total Binding energy = nucleons x binding energy per nucleon4×7−2×2×1.1=23.6 MeV

The amount of energy released in the fusion of $_{1} H^{2}$ to form a $_{2} H e^{4}$ nucleus will be :
[Binding energy per nucleon of $_{1} H^{2} = 1.1 M e V$ and $_{2} H e^{4} = 7 M e V$ ]

Nuclear reaction: $2_{1} H^{2} \to_{2} H e^{4}$

Energy released $=$ Binding energy of products $-$ Binding energy of reactants.

Total Binding energy $=$ nucleons x binding energy per nucleon

$4 \times 7 - 2 \times 2 \times 1.1 = 23.6 M e V$