Physics > Advanced Modern Physics > 4.0 Nuclear binding energy
Advanced Modern Physics
1.0 X-Rays
2.0 Moseley's Law.
3.0 Nuclear Structure
4.0 Nuclear binding energy
4.1 Binding energy per nucleon
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
5.0 Radioactivity
6.0 Radioactive decay law
4.2 Variation of Binding energy per nucleon with mass number
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
The binding energy per nucleon is a characteristic property of elements.
The graph in the figure shows the variation of binding energy per nucleon with the mass number of all the elements of periodic table.
In the graph, we can see that the binding energy per nucleon increase rapidly for nuclei with small masses and reaches a maximum of approximately 8.8 $MeV/nucleon$ for iron(${}_Z^AFe$).
The nuclei with mass numbers greater or less than 60 are not as strongly bound as those near the middle of the periodic table.
This fact allows energy to be released in fission and fusion reactions. The curve is slowly varying for $A$>40, which suggests that the nuclear force saturates.
In other words, a particular nucleon can interact with only a limited number of other nucleons, which can be viewed as the "nearest neighbors" in the close packed structure illustrated in Fig.7
For greater nucleon numbers, the binding energy per nucleon decrease gradually.
Later, the binding energy per nucleon decreases enough so there is insufficient binding energy to hold the nucleon together in the nucleus.
It is observed that nuclei with $A$>209 are unstable and hence radioactive.
In Fig. 8 we can see that binding energy per nucleon for ${}^4He,{\text{ }}{}^{12}C,{\text{ and }}{}^{16}O$ are relatively higher as compared to their neighboring elements or these elements have nuclei which are relatively more stable than their neighbors. This is because of the existence of nuclear energy levels in the nucleus.
Each nuclear energy level can contain two neutrons of opposite spins or two protons of opposite spins.
Energy levels in nuclei are filled in sequence, just as energy levels in atoms, to achieve configurations of minimum energy and therefore maximum stability.
Similar to the case of atomic orbitals, here also the configuration of opposite spins in nucleons with pairs in nuclear energy levels are more stable.
This concept can be used to explain the reason of more stability of ${}^4He,{\text{ }}{}^{12}C,{\text{ and }}{}^{16}O$ compared to their neighboring elements.
Important observations from the graph:
- The maximum binding energy per nucleon is 8.8 $MeV/nucleon$ and it is for ${}_{26}^{56}Fe$. Thus, the iron nucleus is the most stable and the most tightly bound.
- The fact that ($BE$/$A$) varies by less than 10% above $A$=10 suggests that each nucleon in the nucleus interacts only with the neighbors, independently of the total number of nucleons present in the nucleus.
- A small decrease after $A$=56 is due to the destabilizing effect of the long range repulsive Coulomb force.
- Not only the Coulomb force reduces the $BE$, it also shifts the neutron-proton ratio in heavy nuclei toward the neutron excess, by an amount that will increase with $A$.
Question 9. Calculate the binding energy for nucleon of ${}_6^{12}C$ nucleus, if mass of proton $m_p=1.0078\ u$, mass of neutron $m_n=1.0087\ u$, mass of ${C_{12}}$, $m_c=12.0000\ u$.
Solution:
In the nucleus of carbon $C$, there are 6 protons and 6 neutrons.
Now, $$\begin{equation} \begin{aligned} 6\left( {{m_p}} \right) = 6 \times 1.0078 = 6.0468\;u \\ 6\left( {{m_n}} \right) = 6 \times 1.0087 = 6.05224\;u \\\end{aligned} \end{equation} $$
Total mass, $$\begin{equation} \begin{aligned} m = 6\left( {{m_p}} \right) + 6\left( {{m_n}} \right) \\ m = \left( {6.0468\; + 6.0522} \right)\;u \\ m = 12.0990\;u \\\end{aligned} \end{equation} $$
Mass defect, $$\begin{equation} \begin{aligned} \Delta m = m - {m_c} \\ \Delta m = 0.0990\;u \\\end{aligned} \end{equation} $$
So, the binding energy $(BE)$ is given by, $$\begin{equation} \begin{aligned} BE = \Delta m \times 931.5 \\ BE = 92.2\;MeV \\\end{aligned} \end{equation} $$
While writing nuclear reactions we should always check two things.
1. Conservation of nucleon number (Mass number and nucleon number must be balanced)
2. Conservation of energy
Example: Let us consider the following reaction
$${}_1^2H + {}_7^{14}N \to {}_6^{12}C + {}_2^4He$$
Here, $$\begin{equation} \begin{aligned} \Rightarrow \Delta m = \left[ {{M_{_1^2H}} + {M_{_7^{14}N}}} \right] - \left[ {{M_{_6^{12}C}} + {M_{_2^4He}}} \right] \\ \Rightarrow \left[ {2.014102 + 14.003074} \right] - \left[ {12.000000 + 4.002602} \right] \\ \Rightarrow 16.017176 - 16.002602 \\ \Rightarrow 0.0145774 \\\end{aligned} \end{equation} $$
This energy is converted into kinetic energy of the nuclei present after the reaction. $$\begin{equation} \begin{aligned} \Rightarrow \Delta E = 0.0145774 \times 931.5{\text{ }}MeV\; \\ \Rightarrow \Delta E = 13.576{\text{ }}MeV \\\end{aligned} \end{equation} $$
The energy that is required to balance the equation(energetically) is called the $Q$ value of the reaction.
For the above nuclear reaction $Q$ value is 13.576 $Mev$.
However, if the reacting species have an initial kinetic energy (say about 5 $Mev$), we must have to add it to the energy released due to the mass effect that has occurred during the reaction to completely balance the equation in terms of energy.
Now, consider a nuclear reaction,
$${}_2^4He + {}_7^{17}N \to {}_8^{17}O + {}_1^1H$$
This reaction $Q$ value of -1.194 $Mev$. Such a reaction will not take place unless the incoming particle has at least enough kinetic energy to overcome the energy deficit.
So nuclear reactions can be exothermic(+ve $Q$) or endothermic(-ve $Q$).
In the first place, it might look like that if an incoming alpha particle having kinetic energy equal to 1.194 $Mev$ then the reaction will take place. But then the products created will have zero kinetic energy(as per law of conservation of energy) and hence zero velocity. As the initial alpha particle had a velocity, it can clearly be seen that law of conservation of linear momentum is defined.
So, the reaction will not take place.
So, in order to conserve both linear momentum and energy, the incoming alpha particle must have a minimum kinetic energy given by,
$$K{E_{\min }} = \left( {1 + \frac{m}{M}} \right)\left| Q \right|$$
where
$m$= mass of the incident particle
$M$= mass of the target particle
Calculating for the above reaction we get, $$\begin{equation} \begin{aligned} \Rightarrow K{E_{\min }} = \left( {1 + \frac{m}{M}} \right)\left| Q \right| \\ \Rightarrow \left( {1 + \frac{{4.00262}}{{14.003074}}} \right)\left| { - 1.194{\text{ }}MeV} \right| \\ \Rightarrow 1.535{\text{ }}MeV \\\end{aligned} \end{equation} $$
The reaction will occur if and only if the incoming alpha particle have a $KE$ of 1.535 $Mev$ or higher.
The minimum kinetic energy of the incoming particle is called the threshold energy.
