Home Physics Atomic and Nuclear Physics JEE Main 2025 - ( Nuclear Physics ) Energy released when two deuterons fuse to …
Physics Atomic and Nuclear Physics JEE Main 2025 - ( Nuclear Physics ) MCQ (Single Correct)

Energy released when two deuterons fuse to form a helium nucleus is:

(Given: Binding energy per nucleon of and binding energy per nucleon of )

A
26.8 MeV
B
8.1 MeV
C
23.6 MeV
D
5.9 MeV

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Text Solution

Verified by Experts
The correct answer is:
C

To calculate the energy released when two deuterons ( ) fuse to form a helium nucleus , consider the following:

The reaction is represented as:

Given the binding energy per nucleon:
For
For
Calculating Binding Energy
Binding Energy for Reactants:

Each deuteron ( ) has a binding energy of 1.1 MeV per nucleon.

Since there are two nucleons in a deuteron, the total binding energy for one deuteron is .

Therefore, for two deuterons:

Binding Energy for Product:
For : Each of the four nucleons has a binding energy of 7.0 MeV.

Total binding energy for the helium nucleus:

Energy Released (Q)
The energy released, Q, is the difference between the binding energy of the products and the reactants:

Thus, the energy released when two deuterons fuse to form a helium nucleus is 23.6 MeV.

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