Physics Q38 – JEE-MAIN 2025

Energy released when two deuterons (

_{1} H^{2}

) fuse to form a helium nucleus (

_{2} H e^{4}

) is : (Given : Binding energy per nucleon of

_{1} H^{2} = 1.1

MeV and binding energy per nucleon of

_{2} H e^{4} = 7.0

MeV)

Clue 1 of 3: Nuclear Fusion Energy

Energy released in a nuclear fusion reaction is due to the difference in the total binding energies of the products and reactants.

🧠 Full Solution Path

Step 1: Calculate total binding energy of reactants✦ Active

The reactants are two deuterons ( $_{1} H^{2}$ ). Each deuteron has 2 nucleons. The binding energy per nucleon for $_{1} H^{2}$ is $1.1 MeV$ .

{Binding energy of one}_{1} H^{2} = 2 nucleons \times 1.1 MeV/nucleon = 2.2 MeV

Total binding energy of two $_{1} H^{2}$ nuclei:

Total BE (reactants) = 2 \times 2.2 MeV = 4.4 MeV

Step 2: Calculate total binding energy of products○ Expand

The product is one helium nucleus ( $_{2} H e^{4}$ ). It has 4 nucleons. The binding energy per nucleon for $_{2} H e^{4}$ is $7.0 MeV$ .

Total BE (products) = 4 nucleons \times 7.0 MeV/nucleon = 28.0 MeV

Step 3: Calculate the energy released○ Expand

The energy released in the fusion reaction is the difference between the total binding energy of the products and the total binding energy of the reactants.

Energy released = Total BE (products) - Total BE (reactants)

Energy released = 28.0 MeV - 4.4 MeV = 23.6 MeV

💡 Teacher's Secret Hint

Ensure to multiply the binding energy per nucleon by the correct number of nucleons for each nucleus.

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Nuclear Fusion Energy

🧠 Full Solution Path