Animated Solution for Physics - Atoms and Nuclei: If the binding energy per nucleon in ${}_3^7\text{Li}$ and ${}_2^4\text{He}$ nuclei are $5.60\text{ MeV}$ and $7.06\text{ MeV}$ respectively, then in the reaction $p+{}_3^7\text{Li}\to 2{}_2^4\text{He}$ energy of proton must be

Imagine a microscopic game of billiards, but with atomic nuclei. In this problem, we are looking at a specific nuclear reaction: a single proton ($p$) is fired at a stationary Lithium-7 (${}_3^7\text{Li}$) nucleus. When they collide, they don't just bounce off each other; they undergo a nuclear transformation, splitting into two identical Helium-4 nuclei (${}_2^4\text{He}$), which are also known as alpha particles.

The question asks us to find the exact kinetic energy this incoming proton must possess to make this reaction possible. To solve this, we need to dive into the energies that hold these nuclei together.

In any nuclear reaction, the total energy must be conserved. This means the total energy of the reactants (the proton and the Lithium nucleus) must perfectly equal the total energy of the products (the two Helium nuclei).

But what kind of energy are we talking about? It's the Binding Energy. The binding energy is the energy required to completely disassemble a nucleus into its constituent protons and neutrons. Conversely, it's the energy released when those nucleons bind together.

The problem gives us the binding energy per nucleon. To find the total binding energy of a nucleus, we simply multiply this value by the total number of nucleons it contains, which is its mass number ($A$).

For Lithium-7, the mass number is $7$, and the binding energy per nucleon is $5.60\text{ MeV}$.

For Helium-4, the mass number is $4$, and the binding energy per nucleon is $7.06\text{ MeV}$.

Now, let's set up our energy conservation equation. The energy of the reactants is the kinetic energy of the proton ($E_p$) plus the binding energy of the Lithium nucleus. The energy of the products is the binding energy of the two Helium nuclei.

Substituting the values we calculated:

Let's do the arithmetic. Seven times $5.60$ is $39.20$. Four times $7.06$ is $28.24$, and since there are two Helium nuclei, we multiply that by $2$ to get $56.48$.

Finally, to find the energy of the proton, we subtract the binding energy of Lithium from the total binding energy of the Helium nuclei:

This tells us that the proton must bring exactly $17.28\text{ MeV}$ of kinetic energy to the table to compensate for the difference in binding energies and allow this nuclear reaction to proceed.