Question

Binding energy per nucleon of deuteron and Helium nuclei are 1.1 MeV and 7 MeV respectively. If a single Helium nucleus was formed by adding two Deuterons, the energy released is

• A. 23.6 MeV
• B. 32.4 MeV
• C. 28.6 MeV
• D. 13.6 MeV

Hint:

Always convert "per nucleon" values to total binding energy by multiplying by the mass number before subtracting!

Answer 1

The Correct Option is A

Step 1: Energy released idea.
The energy released equals the total binding energy of the product minus the total binding energy of the reactants.

Step 2: How to get total binding energy.
Multiply binding energy per nucleon by the number of nucleons (mass number $A$).

Step 3: Reactant side (two deuterons).
Each deuteron has $A=2$, so its BE $= 2\times 1.1 = 2.2$ MeV. Two of them give $2\times 2.2 = 4.4$ MeV.

Step 4: Product side (helium).
Helium has $A=4$, so its BE $= 4\times 7 = 28$ MeV.

Step 5: Subtract.
\[ \Delta E = 28 - 4.4 = 23.6 \text{ MeV} \]
Step 6: State the answer.
So the energy released is $23.6$ MeV, which is option 1.
\[ \boxed{\Delta E = 23.6 \text{ MeV}} \]