Question

The binding energy of deuteron $(H_1^2)$ is $1.15MeV$ per nucleon and an alpha particle $(He_2^4)$ has a binding energy of $7.1{\text{ }}MeV$ per nucleon. Then in the reaction, ${\text{H}}_1^2 + {\text{H}}_1^2 \ to {\text{He}}_2^4 + {\text{Q}}$ the energy Q is

(A) $33.0{\text{ }}MeV$

(B) $28.4{\text{ }}MeV$

(C) $23.8{\text{ }}MeV$

(D) $4.6{\text{ }}MeV$

Answer 1

Calculate the binding energy of the alpha particle first and then find the total binding energy for the deuterons given.Then finally we can find the value of the energy released using the above two values calculated.

Complete Step by step solution

We know that the binding energy is equal to the amount of energy released in forming the nucleus.

Given that the binding energy of a deuteron $(H_1^2)$ is $1.15MeV$ per nucleon and an alpha particle $(He_2^4)$ has a binding energy of $7.1{\text{ }}MeV$ per nucleon.

The given reaction is ${\text{H}}_1^2 + {\text{H}}_1^2 \to {\text{He}}_2^4 + {\text{Q}}$

We have the total binding energy of the deuterons $= {\text{ }}4\left( {1.15} \right){\text{ }} = {\text{ }}4.60{\text{ }}MeV$

We have the total binding energy of the alpha particles $= {\text{ }}4\left( {7.1} \right){\text{ }} = {\text{ }}28.4{\text{ }}MeV$

Therefore the energy released in the process is equal to the total binding energy of the alpha particle minus the total binding energy of the deuterons.

Then the energy released in the process $= {\text{ }}28.4{\text{ }}-{\text{ }}4.60{\text{ }} = {\text{ }}23.8{\text{ }}MeV$

So the correct option is C.

Note We need to consider the total binding energy of the deuterons and not only the value of the binding energy of the single deuteron and similarly, but we also need to consider the total binding energy of the alpha particles to get the correct value of the energy released.