Subfield: Modern Physics

3. Simple model of an atomic nucleus 

Introduction

Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below.

Task 1 - Atomic nucleus as closely packed system of nucleons

In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter.

<img_4362>
Context question:
a) Calculate what would be the packing factor $f$ if nucleons were arranged in a "simple cubic" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].
Context answer:
\boxed{$\frac{\pi}{6}$}


Extra Supplementary Reading Materials:

Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$.
Context question:
b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$.
Context answer:
\boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$}


Extra Supplementary Reading Materials:

Task 2 - Binding energy of atomic nuclei - volume and surface terms
Context question:
Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction.
Context answer:
\boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$}


Extra Supplementary Reading Materials:

Task 3 - Electrostatic (Coulomb) effects on the binding energy

The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$.
Context question:
a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks.
Context answer:
\boxed{$U_{c}=\frac{3 Z(Z-1) e^{2}}{20 \pi \varepsilon_{0} R}$}


Context question:
b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction.
Context answer:
\boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left(\frac{A^{5 / 3}}{4}-\frac{A^{2 / 3}}{2}\right)$}


Extra Supplementary Reading Materials:

Task 4 - Fission of heavy nuclei

Fission is a nuclear process in which a nucleus splits into smaller parts (lighter nuclei). Suppose that a nucleus with $A$ nucleons splits into only two equal parts as depicted in Fig. 2.
Context question:
a) Calculate the total kinetic energy of the fission products $E_{\text {kin }}$ when the centers of two lighter nuclei are separated by the distance $d \geq 2 R(A / 2)$, where $R(A / 2)$ is their radius. The large nucleus was initially at rest.
Context answer:
\boxed{$E_{k i n}(d)=-3 f^{1 / 3} A^{2 / 3} a_{V}\left(2^{1 / 3}-1\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\left(2^{2 / 3}-1\right)-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left[\frac{A^{5 / 3}}{4}\left(2^{-2 / 3}-1\right)-\frac{A^{2 / 3}}{2}\left(2^{1 / 3}-1\right)\right]-\frac{1}{4 \pi \varepsilon_{0}} \frac{A^{2} e^{2}}{16 d}$}


Context question:
b) Assume that $d=2 R(A / 2)$ and evaluate the expression for $E_{k i n}$ obtained in part a) for $A=$ 100, 150, 200 and 250 (express the results in units of MeV). Estimate the values of $A$ for which fission is possible in the model described above?

<img_4423>
Context answer:
Numerically one gets:
$A=100 \ldots E_{k i n}=-33.95 \mathrm{MeV}$,
$A=150 \ldots E_{k i n}=-30.93 \mathrm{MeV}$,
$A=200 \ldots E_{k i n}=-14.10 \mathrm{MeV}$,
$A=250 \ldots E_{\text {kin }}=+15.06 \mathrm{MeV}$.

and $A \geq 227$


Extra Supplementary Reading Materials:

Task 5 - Transfer reactions

Question: a) In modern physics, the energetics of nuclei and their reactions is described in terms of masses. For example, if a nucleus (with zero velocity) is in an excited state with energy $E_{\text {exc }}$ above the ground state, its mass is $m=m_{0}+E_{e x c} / c^{2}$, where $m_{0}$ is its mass in the ground state at rest. The nuclear reaction ${ }^{16} \mathrm{O}+{ }^{54} \mathrm{Fe} \rightarrow{ }^{12} \mathrm{C}+{ }^{58} \mathrm{Ni}$ is an example of the so-called "transfer reactions", in which a part of one nucleus ("cluster") is transferred to the other (see Fig. 3). In our example the transferred part is a ${ }^{4} \mathrm{He}-\mathrm{cluster}(\alpha$-particle). The transfer reactions occur with maximum probability if the velocity of the projectile-like reaction product (in our case: ${ }^{12} \mathrm{C}$ ) is equal both in magnitude and direction to the velocity of projectile (in our case: ${ }^{16} \mathrm{O}$ ). The target ${ }^{54} \mathrm{Fe}$ is initially at rest. In the reaction, ${ }^{58} \mathrm{Ni}$ is excited into one of its higher-lying states. Find the excitation energy of that state (and express it units of $\mathrm{MeV}$ ) if the kinetic energy of the projectile ${ }^{16} \mathrm{O}$ is $50 \mathrm{MeV}$. The speed of light is $\mathrm{c}=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$.

| 1. | $\mathrm{M}\left({ }^{16} \mathrm{O}\right)$ | 15.99491 a.m.u. |
| ---: | ---: | :--- |
| 2. | $\mathrm{M}\left({ }^{54} \mathrm{Fe}\right)$ | 53.93962 a.m.u. |
| 3. | $\mathrm{M}\left({ }^{12} \mathrm{C}\right)$ | 12.00000 a.m.u. |
| 4. | $\mathrm{M}\left({ }^{58} \mathrm{Ni}\right)$ | 57.93535 a.m.u. |
| { Table 1. The rest masses of the reactants in their ground states. 1 a.m.u. $=1.6605 \cdot 10^{-27} \mathrm{~kg}$}. |  |  |

Solution:
Non-relativistic solution

First one has to find the amount of mass transferred to energy in the reaction (or the energy equivalent, so-called Q-value):
$$
\begin{aligned}
\Delta m & =(\text { total mass })_{\text {after reaction }}-(\text { total mass })_{\text {before reaction }}= \\
& =(57.93535+12.00000) \text { a.m.u. }-(53.93962+15.99491) \text { a.m.u. }= \\
& =0.00082 \text { a.m.u. }= \\
& =1.3616 \cdot 10^{-30} \mathrm{~kg} .
\end{aligned}
\tag{19}
$$
Using the Einstein formula for equivalence of mass and energy, we get:
$$
\begin{aligned}
Q & =(\text { total kinetic energy })_{\text {after reaction }}-(\text { total kinetic energy })_{\text {before reaction }}= \\
& =-\Delta m \cdot c^{2}= \\
& =-1.3616 \cdot 10^{-30} \cdot 299792458^{2}=-1.2237 \cdot 10^{-13} \mathrm{~J}
\end{aligned}
\tag{20}
$$
Taking into account that $1 \mathrm{MeV}$ is equal to $1.602 \cdot 10^{-13} \mathrm{~J}$, we get:
$$
Q=-1.2237 \cdot 10^{-13} / 1.602 \cdot 10^{-13}=-0.761 \mathrm{MeV}
\tag{21}
$$
This exercise is now solved using the laws of conservation of energy and momentum. The latter gives (we are interested only for the case when ${ }^{12} \mathrm{C}$ and ${ }^{16} \mathrm{O}$ are having the same direction so we don't need to use vectors):
$$
m\left({ }^{16} \mathrm{O}\right) v\left({ }^{16} \mathrm{O}\right)=m\left({ }^{12} \mathrm{C}\right) v\left({ }^{12} \mathrm{C}\right)+m\left({ }^{58} \mathrm{Ni}\right) v\left({ }^{58} \mathrm{Ni}\right)
\tag{22}
$$
while the conservation of energy gives:
$$
E_{k}\left({ }^{16} \mathrm{O}\right)+Q=E_{k}\left({ }^{12} \mathrm{C}\right)+E_{k}\left({ }^{58} \mathrm{Ni}\right)+E_{x}\left({ }^{58} \mathrm{Ni}\right)
\tag{23}
$$

where $E_{x}\left({ }^{58} \mathrm{Ni}\right)$ is the excitation energy of ${ }^{58} \mathrm{Ni}$, and $Q$ is calculated in the first part of this task. But since ${ }^{12} \mathrm{C}$ and ${ }^{16} \mathrm{O}$ have the same velocity, conservation of momentum reduced to:
$$
\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right] v\left({ }^{16} \mathrm{O}\right)=m\left({ }^{58} \mathrm{Ni}\right) v\left({ }^{58} \mathrm{Ni}\right)
\tag{24}
$$
Now we can easily find the kinetic energy of ${ }^{58} \mathrm{Ni}$ :
$$
\begin{aligned}
E_{k}\left({ }^{58} \mathrm{Ni}\right) & =\frac{m\left({ }^{58} \mathrm{Ni}\right) v^{2}\left({ }^{58} \mathrm{Ni}\right)}{2}=\frac{\left[m\left({ }^{58} \mathrm{Ni}\right) v\left({ }^{58} \mathrm{Ni}\right)\right]^{2}}{2 m\left({ }^{58} \mathrm{Ni}\right)}= \\
& =\frac{\left.\left[ m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right] v\left({ }^{(6} \mathrm{O}\right)\right]^{2}}{2 m\left({ }^{58} \mathrm{Ni}\right)}= \\
& =E_{k}\left({ }^{16} \mathrm{O}\right) \frac{\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right]^{2}}{m\left({ }^{58} \mathrm{Ni}\right) m\left({ }^{16} \mathrm{O}\right)}
\end{aligned}
\tag{25}
$$
and finally the excitation energy of ${ }^{58} \mathrm{Ni}$ :
$$
\begin{aligned}
E_{x}\left({ }^{58} \mathrm{Ni}\right) & =E_{k}\left({ }^{16} \mathrm{O}\right)+Q-E_{k}\left({ }^{12} \mathrm{C}\right)-E_{k}\left({ }^{58} \mathrm{Ni}\right)= \\
& =E_{k}\left({ }^{16} \mathrm{O}\right)+Q-\frac{m\left({ }^{12} \mathrm{C}\right) v^{2}\left({ }^{16} \mathrm{O}\right)}{2}-E_{k}\left({ }^{16} \mathrm{O}\right) \frac{\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right]^{2}}{m\left({ }^{58} \mathrm{Ni}\right) m\left({ }^{16} \mathrm{O}\right)}= \\
& =Q+E_{k}\left({ }^{16} \mathrm{O}\right)-E_{k}\left({ }^{16} \mathrm{O}\right) \cdot \frac{m\left({ }^{12} \mathrm{C}\right)}{m\left({ }^{16} \mathrm{O}\right)}-E_{k}\left({ }^{16} \mathrm{O}\right) \frac{\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right]^{2}}{m\left({ }^{58} \mathrm{Ni}\right) m\left({ }^{16} \mathrm{O}\right)}= \\
& =Q+E_{k}\left({ }^{16} \mathrm{O}\right)\left[1-\frac{m\left({ }^{12} \mathrm{C}\right)}{m\left({ }^{16} \mathrm{O}\right)}-\frac{\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right]^{2}}{m\left({ }^{58} \mathrm{Ni}\right) m\left({ }^{16} \mathrm{O}\right)}\right]= \\
& =Q+E_{k}\left({ }^{16} \mathrm{O}\right) \frac{\left[m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right] \cdot\left[m\left({ }^{58} \mathrm{Ni}\right)-m\left({ }^{16} \mathrm{O}\right)+m\left({ }^{12} \mathrm{C}\right)\right]}{m\left({ }^{58} \mathrm{Ni}\right) m\left({ }^{16} \mathrm{O}\right)}
\end{aligned}
\tag{26}
$$
Note that the first bracket in numerator is approximately equal to the mass of transferred particle (the ${ }^{4} \mathrm{He}$ nucleus), while the second one is approximately equal to the mass of target nucleus ${ }^{54} \mathrm{Fe}$. Inserting the numbers we get:
$$
\begin{aligned}
E_{x}\left({ }^{58} \mathrm{Ni}\right) & =-0.761+50 \cdot \frac{(15.99491-12 .)(57.93535-15.99491+12 .)}{57.93535 \cdot 15.99491}= \\
& =10.866 \mathrm{MeV}
\end{aligned}
\tag{27}
$$
Relativistic solution 

In the relativistic version, solution is found starting from the following pair of equations (the first one is the law of conservation of energy and the second one the law of conservation of momentum):
$$
m\left({ }^{54} \mathrm{Fe}\right) \cdot c^{2}+\frac{m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}=\frac{m\left({ }^{12} \mathrm{C}\right) \cdot c^{2}}{\sqrt{1-v^{2}\left({ }^{12} \mathrm{C}\right) / c^{2}}}+\frac{m^{*}\left({ }^{58} \mathrm{Ni}\right) \cdot c^{2}}{\sqrt{1-v^{2}\left({ }^{58} \mathrm{Ni}\right) / c^{2}}}

\frac{m\left({ }^{16} \mathrm{O}\right) \cdot v\left({ }^{(16} \mathrm{O}\right)}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}=\frac{\left.m\left({ }^{12} \mathrm{C}\right) \cdot v^{(12} \mathrm{C}\right)}{\sqrt{1-v^{2}\left({ }^{12} \mathrm{C}\right) / c^{2}}}+\frac{m^{*}\left({ }^{58} \mathrm{Ni}\right) \cdot v\left({ }^{58} \mathrm{Ni}\right)}{\sqrt{1-v^{2}\left({ }^{58} \mathrm{Ni}\right) / c^{2}}} \quad
\tag{28}
$$
All the masses in the equations are the rest masses; the ${ }^{58} \mathrm{Ni}$ is NOT in its ground-state, but in one of its excited states (having the mass denoted with $m^{*}$ ). Since ${ }^{12} \mathrm{C}$ and ${ }^{16} \mathrm{O}$ have the same velocity, this set of equations reduces to:
$$
\begin{aligned}
& m\left({ }^{54} \mathrm{Fe}\right)+\frac{m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}=\frac{m^{*}\left({ }^{58} \mathrm{Ni}\right)}{\sqrt{1-v^{2}\left({ }^{58} \mathrm{Ni}\right) / c^{2}}} \\
& \frac{\left(m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right) \cdot v\left({ }^{16} \mathrm{O}\right)}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}=\frac{m^{*}\left({ }^{58} \mathrm{Ni}\right) \cdot v\left({ }^{58} \mathrm{Ni}\right)}{\sqrt{1-v^{2}\left({ }^{58} \mathrm{Ni}\right) / c^{2}}}
\end{aligned}
\tag{29}
$$
Dividing the second equation with the first one gives:
$$
v\left({ }^{58} \mathrm{Ni}\right)=\frac{\left(m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right) \cdot v\left({ }^{16} \mathrm{O}\right)}{\left(m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right)+m\left({ }^{54} \mathrm{Fe}\right) \sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}
\tag{30}
$$
The velocity of projectile can be calculated from its energy:
$$
\begin{aligned}
& E_{k i n}\left({ }^{16} \mathrm{O}\right)=\frac{m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}}-m\left({ }^{16} \mathrm{O}\right) \cdot c^{2} \\
& \sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}=\frac{m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}{E_{k i n}\left({ }^{16} \mathrm{O}\right)+m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}} \\
& v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}=1-\left(\frac{m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}{E_{k i n}\left({ }^{16} \mathrm{O}\right)+m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}\right)^{2} \\
& v\left({ }^{16} \mathrm{O}\right)=\sqrt{1-\left(\frac{m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}{E_{k i n}\left({ }^{16} \mathrm{O}\right)+m\left({ }^{16} \mathrm{O}\right) \cdot c^{2}}\right)^{2}} \cdot c
\end{aligned}
\tag{31}
$$
For the given numbers we get:
$$
\begin{aligned}v\left({ }^{16} \mathrm{O}\right) & =\sqrt{1-\left(\frac{15.99491 \cdot 1.6605 \cdot 10^{-27} \cdot\left(2.9979 \cdot 10^{8}\right)^{2}}{50 \cdot 1.602 \cdot 10^{-13}+15.99491 \cdot\left(2.9979 \cdot 10^{8}\right)^{2}}\right)^{2}} \cdot c= \\ & =\sqrt{1-0.99666^{2} \cdot c=0.08172 \cdot c=2.4498 \cdot 10^{7} \mathrm{~km} / \mathrm{s}}\end{aligned}
\tag{32}
$$
Now we can calculate:
$$
v\left({ }^{58} \mathrm{Ni}\right)=\frac{(15.99491-12.0) \cdot 2.4498 \cdot 10^{7} \mathrm{~km} / \mathrm{s}}{(15.99491-12.0)+53.93962 \sqrt{1-0.08172^{2}}}=1.6946 \cdot 10^{6} \mathrm{~km} / \mathrm{s}
\tag{33}
$$

The mass of ${ }^{58} \mathrm{Ni}$ in its excited state is then:
$$
\begin{aligned}
m^{*}\left({ }^{58} \mathrm{Ni}\right) & =\left(m\left({ }^{16} \mathrm{O}\right)-m\left({ }^{12} \mathrm{C}\right)\right) \frac{\sqrt{1-v^{2}\left({ }^{58} \mathrm{Ni}\right) / c^{2}}}{\sqrt{1-v^{2}\left({ }^{16} \mathrm{O}\right) / c^{2}}} \cdot \frac{v\left({ }^{16} \mathrm{O}\right)}{v\left({ }^{58} \mathrm{Ni}\right)}= \\
& =(15.99491-12.0) \frac{\sqrt{1-\left(1.6945 \cdot 10^{6} / 2.9979 \cdot 10^{8}\right)^{2}}}{\sqrt{1-0.08172^{2}}} \cdot \frac{2.4498 \cdot 10^{7}}{1.6945 \cdot 10^{6}} \text { a.m.u. }= \\
& =57.9470 \text { a.m.u. }
\end{aligned}
\tag{34}
$$
The excitation energy of ${ }^{58} \mathrm{Ni}$ is then:
$$
\begin{aligned}
E_{x} & =\left[m^{*}\left({ }^{58} \mathrm{Ni}\right)-m\left({ }^{58} \mathrm{Ni}\right)\right] \cdot c^{2}=(57.9470-57.93535) \cdot 1.6605 \cdot 10^{-27}\left(2.9979 \cdot 10^{8}\right)^{2} \\
& =2.00722 \cdot 10^{-12} / 1.602 \cdot 10^{-13} \mathrm{MeV} / \mathrm{J}=10.8636 \mathrm{MeV}
\end{aligned}
\tag{35}
$$
The relativistic and non-relativistic results are equal within $2 \mathrm{keV}$ so both can be considered as correct -we can conclude that at the given beam energy, relativistic effects are not important.

Final answer: 10.8636 MeV