Tutorial on Nuclear Chemistry

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michele_laino · Answer

here is the PDF file:

michele_laino · Answer

$\mathbf{FOUNDATIONS \; \;OF \; \;NUCLEAR\; \; CHEMISTRY}$

                                   by  Michele Laino

                                           ${\mathbf{Abstract}}$
In this brief tutorial, some of basics of nuclear chemistry are shown. Such tutorial it is mainly focused on binding energy of nuclei and on the balancing of the nuclear reactions, so traditional topics, like carbon-14 dating and the disintegration kinetics, have been omitted. Finally a note on the units of measure: all nuclear masses, and all  particles masses as well, are measured in $amu$.

$\mathbf{Basic \;\; Definitions\;\; and\;\; the \;\;Einstein's\;\; Equation}$
As it is well known, atomic nucleus is composed by $protons$ and $neutrons$, the so called $nucleons$ as the scientist $W. Heisenberg$ has suggested for first. Furthermore, the mass of an atom is due, in last analysis, to the mass of such protons and neutrons, since the mass of an $electron$ is about $1/1840$ times the mass of a nucleon. The subsequent table show the main properties of the particles of an atom, together with the commonly used symbols, which can be used to refer to them.

$$\begin{gathered}
  \boxed{\begin{array}{*{20}{c}}
  {electron}&{{e^ - },{\beta ^ - },\beta }&{0.0005486}&{ - 1} \ \
  {positron}&{{e^ + },{\beta ^ + }}&{0.0005486}&{ + 1} \ \
  {proton}&{_1^1{\text{H}}{\text{, }}_{\text{1}}^{\text{1}}{\text{p}}{\text{, p}}}&{1.00783}&{ + 1} \ \
  {neutron}&{_{\text{0}}^{\text{1}}{\text{n}}{\text{, n}}}&{1.00867}&0 \ \
  {deuteron}&{_{\text{1}}^{\text{2}}{\text{D}}{\text{, }}_{\text{1}}^{\text{2}}{\text{H}}{\text{, d}}}&{2.01410}&{ + 1} \ \
  {\alpha  - particle}&{_2^4{\text{He}}{\text{, }}_2^4\alpha {\text{, }}\alpha }&{4.00260}&{ + 2} 
\end{array}} \hfill \
  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Table\;1 \hfill \ 
\end{gathered} $$

As we can see, at the third column are reported the mass of the listed particles, whereas the corresponding electrical charges are written at the fourth column.

The $atomic \;number$ $Z$, is defined as the number of protons inside a nucleus of an atom.
The $mass\; number$ $A$, is defined as the sum of the number of protons with the number of neutrons, inside of the nucleus of an atom. Furthermore, the closest whole number to the mass of the nucleus of an element, can be considered another definition of the mass number, as well.
 In general, if we compute the sum of all masses of protons and neutrons, of a nucleus, we get, as result, a mass which is greater than the mass of such nucleus, as showed in the subsequent
$\mathbf{Example 1}$
Let's consider the nucleus of Helium; such nucleus is composed by 2 protons and 2 neutrons, so, using the data of table 1, the sum of such masses is:

$$\sum {{m_N} = } 2 \cdot \left( {1.00783 + 1.00867} \right) = 4.033$$
wherein $m_N$ represents the mass of the involved nucleon.
On the other hand we see, from the Periodic Table, that the mass of an atom of Helium is:
$${m_{He}} = 4.0026$$
the corresponding difference $\Delta m$:
$$\Delta m = \sum {{m_N} - } {m_{He}} = 4.033 - 4.0026 = 0.0304$$
is called the $defect \;of \;mass$, and according to the $Einstein's \;equation$ $E=\Delta m \cdot c^2$, it represents the the fraction of the original mass, which has been transformed in energy needed to to make such atom of Helium.
In other words, the nucleons of the atom of Helium are bounded each to other by means of such energy, therefore the quantity:

$$E = \Delta m \cdot {c^2} = 0.0304 \cdot 9 \cdot {10^{20}} \simeq 2.74 \cdot {10^{19}}\;erg$$
can be considered the \emph{binding energy} for the nucleus of Helium.

Another quantity, which is interesting for nuclear physicist, is the so called \emph{average binding energy for nucleon} $\overline E $, which is simply given dividing the binding energy of a nucleus, by the number of its nucleons; so, in the case of the nucleus of Helium, we have:
$$\overline E  = E/N = 2.74 \cdot {10^{19}}/4 = 6.85 \cdot {10^{18}}\;erg$$

An useful formula, which can be used in order to measure the binding energy in $kcal$ is:

$$E = 2.15 \cdot {10^{10}} \cdot \Delta m\;kcal$$
and recalling that:
$$1{\text{MeV}} = {10^6}{\text{eV}} = 1.6 \cdot {10^{ - 13}}joules = 3.827 \cdot {10^{ - 14}}cal$$
we have:
$$1\;{\text{MeV}}/mol = 3.827 \cdot {10^{ - 14}} \cdot 6.023 \cdot {10^{23}} = 2.3 \cdot {10^{10}}cal = 2.3 \cdot {10^7}kcal$$
where $mol$ refers to a mole of nuclei of a specified chemical element.

Thus, returning to the example above, we get the subsequent value for the average binding energy $\overline E$:

$$\overline E  = 7.1\;{\text{MeV}}$$
which is in agreement with the accepted average value of $8\;\text{MeV}/\text{nucleon}$.

$\mathbf{Example 2}$

In this example, we will compute the average binding energy, or the binding energy for nucleon, of the uranium isotope, whose nucleus possesses 92 protons and 238 neutrons. In order to represent such nuclear specie or \textbf{nuclide}, the physicists and the chemists use this symbol:

$$_{92}^{238}{\text{U}}$$

 Namely, the mass number is located at left upper corner, whereas the atomic number is reported at the lower left corner of the corresponding chemical symbol. More generally:
|dw:1452282360379:dw|

 Now the mass of such isotope of Uranium, is:
$${m_{_{92}^{238}U}} = 238.05$$
so the corresponding defect of mass, is:
$$\Delta m = \left( {92 \cdot 1.00783} \right) + \left\{ {\left( {238 - 92} \right) \cdot 1.00867} \right\} - 238.05 = 1.9362$$
therefore the binding energy is:
$$\begin{gathered}
  E = 1.9362 \cdot 2.15 \cdot {10^{10}} = 4.1628 \cdot {10^{10}}\;kcal \hfill \
   \hfill \
  E = 1809.9\;{\text{MeV/atom}} \hfill \ 
\end{gathered} $$
and the binding energy for nucleon is:
$$\overline E  = \frac{E}{A} = 7.6\;{\text{MeV/nucleon}}$$
As we can see, the obtained value agrees with the accepted mean value of $8\;\text{MeV}/\text{nucleon}$.

michele_laino · Answer

$\mathbf{Nuclear\; Reactions}$

Nuclear reactions can be distinguished in $\mathbf{induced}$ and $\mathbf{spontaneous}$ reactions. The spontaneous nuclear processes comprise all phenomena labeled as radioactivity, for example the nucleus of an atom (radioisotope) transforms into another nucleus by means of an emission of charged particles of $\alpha$ or $\beta$ type, or electromagnetic radiation, usually denoted by the greek letter $\gamma$, furthermore, a more heavy nucleus disintegrates into more light nuclei with an emission of some neutrons. The induced nuclear processes can be obtained hitting a nucleus by means of elementary particles, or light nuclei. Usually, in such reactions, the products can be other radioactive nuclei, which in turn they transform spontaneously, by means of a subsequent radioactive decay.

In order to balance the corresponding equation of reaction of a nuclear process, we can apply these two rules:

1) $\mathbf{the\; conservation\; of \;mass\; number}:$ so, the sum of the $mass\; numbers$ of the nuclei and particles at the left side, has to be equal to the sum of the $mass\; numbers$ of the nuclei and particles at the right side of an equation;

2) $\mathbf{the \;conservation\; of\; electrical\; charge}:$ the sum of the $atomic \;numbers$ of the nuclei and particles at the left side, has to be equal to the sum of the $atomic\; numbers$ of the nuclei and particles at the right side of an equation.

Usually, nuclear reactions are represented in a short way, for example between the main nuclei, are written, between parentheses, the hitting particle, and the scattered particle, like below:
$$\begin{gathered}
  1)\quad A + a \to B + b + Q \hfill \
  {\text{or:}} \hfill \
  A\left( {p,p} \right)A \to {\text{elastic scattering of a proton by a nucleus A}}{\text{,}} \hfill \
   \hfill \
  2)\quad A + n \to B + \gamma  + Q \hfill \
  {\text{or:}} \hfill \
  A\left( {n,\gamma } \right)B \to {\text{capture of a neutron with a subsequent emission of gamma rays}}{\text{,}} \hfill \ 
\end{gathered} $$

 wherein $Q$, is the energy involved in such specific reaction, and in the second example, $A$ is the initial nucleus, and $B$ is the final nucleus.
More specifically, other examples can be these ones:
$$\begin{gathered}
  3)\quad _7^{14}{\text{N}} + _1^1{\text{p}} \to _6^{11}{\text{C}} + _2^4\alpha \quad \quad {\text{or:}}\quad _7^{14}{\text{N}} + \left( {{\text{p}},\alpha } \right)_6^{11}{\text{C}}\quad  \hfill \
   \hfill \
  4)\quad _{25}^{55}{\text{Mn}} + _1^2{\text{H}} \to _{26}^{55}{\text{Fe}} + 2_0^1{\text{n}}\quad {\text{or:}}\quad _{25}^{55}{\text{Mn}}\left( {{\text{d}},2{\text{n}}} \right)_{26}^{56}{\text{Fe}} \hfill \ 
\end{gathered} $$
As we can see in the third example, the sum of mass numbers at the left side is $14+1=\mathbf{15}$, and the sum of atomic numbers is $7+1=\mathbf{8}$; whereas, at the right side, the corresponding quantities are: $11+4=\mathbf{15}$ for total mass number, and $6+2=\mathbf{8}$ for total atomic number. Such reaction is, therefore, a balanced reaction.
Similarly, looking at the fourth example, we see that the sum of mass numbers and the sum of atomic numbers at the left side, are: $55+2=\mathbf{57}$, and $25+1=\mathbf{26}$ respectively, whereas the sum of mass numbers, and the sum of atomic numbers at the right side are: $55+(2 \cdot 1)=\mathbf{57}$, and $26+0=\mathbf{26}$ respectively. Again we have a balanced nuclear reaction.

$\mathbf{Balancing \;of\; Nuclear\; Reactions}$
Let's suppose to have to balance the subsequent nuclear reaction:

$$_{95}^{241}{\text{Am}} + ... \to _{97}^{243}{\text{Bk}} + 2_0^1{\text{n}}$$
as we can see, the sum of mass numbers at the right side is: $243+(2\cdot 1)=245$, whereas the sum of atomic numbers at the same side is: $97+0=97$. Then, we have to require that the sum of mass numbers and the sum of atomic numbers, have to be the same quantity, respectively. Now, if I call with $x$ the missing mass number, and with $y$ the missing atomic number, then we can write the subsequent algebraic system:
$$\left\{ \begin{gathered}
  241 + x = 245 \hfill \
  95 + y = 97 \hfill \ 
\end{gathered}  \right.$$
whose solution is: $x=4,\;y=2$
so, looking at table 1, the only way to balance such equation of reaction, is to add an $\alpha-$particle:

$$\boxed{_{95}^{241}{\text{Am}} + _2^4\alpha  \to _{97}^{243}{\text{Bk}} + 2_0^1{\text{n}}}$$
As other example, let's consider the subsequent nuclear decay:
$$_{92}^{239}{\text{U}} \to _{93}^{239}{\text{Np}} + ...$$
again, if I call with $x$ the missing mass number, and with $y$ the missing atomic number, I can write this algebraic system:
$$\left\{ \begin{gathered}
  239 = 239 + x \hfill \
  92 = 93 + y \hfill \ 
\end{gathered}  \right.$$
whose solution is: $x=0,\;y=-1$. Looking at table 1, we can balance such equation, if we add one electron:

$$\boxed{_{92}^{239}{\text{U}} \to _{93}^{239}{\text{Np}} + \;_{ - 1}^0{{\text{e}}^ - }}$$
Next, let's consider this nuclear reaction:
$$4_1^1{\text{H}} \to _2^4{\text{He}} + ...$$
again, with the same meaning for the variables $x,\;y$, we can write the subsequent algebraic system:
$$\left\{ \begin{gathered}
  4 = 4 + x \hfill \
  4 = 2 + y \hfill \ 
\end{gathered}  \right.$$
The corresponding solution, is: $x=0,\;y=2$, so, from table 1, we see that the balancing of such equation can be obtained, if we add 2 positrons at the right side:
$$\boxed{4_1^1{\text{H}} \to _2^4{\text{He}} + 2{\beta ^ + }}$$
or more precisely:
$$4_1^1{\text{H}} \to _2^4{\text{He}} + 2{\beta ^ + } + 26.7\;{\text{MeV}}$$
Such last process is well known, it is the nuclear fusion, which happens in the Sun and in all other stars.

michele_laino · Answer

$\mathbf{Final\; Remarks}$

In this brief and elementary tutorial, we have seen the meaning of the Einstein's equation, when it is applied to nuclear processes, so we have been led to the fundamental quantity, namely the binding energy for nucleon, which is interesting for nuclear physicists and chemists. We have understood the procedure which is applied in order to write balanced equations for many nuclear processes. Of course there are several nuclear processes, it is suffice to think to the ones from which we get the chemical artificial elements, which are located inside the $actinide\; series$ for example.
Therefore, before concluding this tutorial, I leave to the reader some incomplete nuclear reactions which they have to be balanced:
$$\begin{gathered}
  {\text{1)}}\;_4^9{\text{Be}} + _2^4{\text{He}} \to _6^{12}{\text{C}} + ... \hfill \
   \hfill \
  2)\;_{26}^{54}{\text{Fe}} + _1^2{\text{H}} \to _{27}^{55}{\text{Co}} + ... \hfill \
   \hfill \
  3)\;_{15}^{30}{\text{P}} \to ... + {\beta ^ + } \hfill \
   \hfill \
  4)\;_1^3{\text{H}} \to _2^3{\text{He}} + ... \hfill \
   \hfill \
  5)\;_{92}^{238}{\text{U}} + _2^4\alpha  \to ... + 20_1^1{\text{H}} + 35_0^1{\text{n}}\;. \hfill \ 
\end{gathered} $$

photon336 · Answer

Great work, I like that you made a PDF of this too

michele_laino · Answer

thanks! for your appreciation to my work :) @Photon336

nincompoop · Answer

Epic awesomeness.  Just as it would appear in a college textbook, this tutorial is friendly in its approach.

rvc · Answer

Woohooo!
Michele at CHEM!!!!!!!!!!!!!!
YAY!
Thank you so much :)

michele_laino · Answer

thanks!! :D @nincompoop  and @rvc

abhisar · Answer

That's neat Michele!
Thanks for sharing it with us.

michele_laino · Answer

thanks!! for your appreciation to my work :) @Abhisar