Question

The binding energy per nucleon for ${}_{6}{{C}^{12}}$ nucleons is 7.68MeV and that for ${}_{6}{{C}^{13}}$ is 7.47 MeV. Calculate the energy required to remove a neutron from ${}_{6}{{C}^{13}}$ nucleus.

Answer 1

Calculate the binding energies of the nuclei of ${}_{6}{{C}^{13}}$ and ${}_{6}{{C}^{12}}$. Binding energy is equal to the binding energy per nucleon multiplied by the total number of nucleons of the nucleus. Then the difference of both the binding energies is the energy required to remove a neutron of ${}_{6}{{C}^{13}}$.

Formula used:

Binding energy of the Nucleus = Binding energy per nucleon $\times$ Total number of nucleons

Complete step by step answer:

Let us first understand what binding energy of a nucleus is. We know that a nucleus consists of neutrons and protons. The neutrons and protons are called the nucleons of the nucleus. Therefore, the mass of the nucleus should be equal to the sum of masses of the individual nucleons. However, it is found that the mass of the nucleus is not equal to the sum of the masses of the individual nucleons. It is found that the mass of the nucleus is less than the sum of the masses of protons and nucleons.

This happens because some neutrons and protons come together and make up a nucleus. They are held together by a strong force called the nuclear force. In the process of formation of a nucleus, some amount of mass is converted into an energy. This converted mass is called mass defect and the energy released is called binding energy of the nucleus.

The following equation shows the formation of a nucleus.

$an+bp\to {}_{b}{{X}^{a+b}}+E$.

Here, ‘$a$’ neutrons and ‘$b$’ number of protons form a nucleus $X$ with a release of a binding energy E.

This means that if it provides an energy equal to $E$, then the nucleus will break apart into ‘$a$’ neutrons and ‘$b$’ protons.

i.e. ${}_{b}{{X}^{a+b}}+E\to an+bp$

Binding energy per a nucleon is the energy required to remove a nucleon from a nucleus. This can be written as the binding energy of the nucleus divided by the total number of nucleons.

$\text{Binding energy per a Nucleon }=\dfrac{\text{Binding energy of the Nucleus}}{\text{Total number of Nucleons}}$

From the given question, we need to calculate the energy required to remove a neutron from ${}_{6}{{C}^{13}}$ nucleus.

However, we cannot simply remove only one nucleon from a nucleus. Therefore, we must first break apart the whole nucleus and then isolate the one neutron. Then the remaining 12 nucleons will form the nucleus of ${}_{6}{{C}^{12}}$.

Let us find the binding energy to form ${}_{6}{{C}^{13}}$.

The binding energy per nucleon of ${}_{6}{{C}^{13}}$ is $7.47MeV$ and the number of nucleons is 13. Therefore, the binding energy of the nucleus will be $7.47MeV \times 13=97.11MeV$.

The binding energy per nucleon of ${}_{6}{{C}^{12}}$ is $7.68MeV$ and the number of nucleons is $12$. Therefore, the binding energy of the nucleus will be $7.68MeV\times 12=92.16MeV$.

Now, we will provide an energy of $97.11MeV$ to the nucleus ${}_{6}{{C}^{13}}$. Then it will break apart. The $12$ nucleons will form the nucleus of ${}_{6}{{C}^{12}}$ by releasing an energy of $92.16MeV.$

Hence, the net energy given is $97.11-92.16 = 4.95MeV.$

Therefore, the energy required to remove a neutron of ${}_{6}{{C}^{13}}$ is 4.95MeV.

Note:

The stability of a nucleus depends on the binding energy per nucleon of the nucleus. More the binding energy per nucleon, the more stable the nucleus. Note that it is not dependent on the total binding energy.

As you can see in the solution, the binding energy of ${}_{6}{{C}^{13}}$ is more than that of ${}_{6}{{C}^{12}}$. However, the binding energy per nucleon of ${}_{6}{{C}^{12}}$ is more than that of ${}_{6}{{C}^{13}}$. Therefore, ${}_{6}{{C}^{12}}$ is more stable.