Nuclear Chemistry

Composition of nucleus

Nucleus is a positively charged and is positioned centrally in an atom. It is surrounded by a cloud of negatively charged electrons. The nucleus is composed of proton which has +1 charge and mass of 1.00758 a.m.u. and the neutron which has zero charge and mass of 1.00893 a.m.u. These are collectively called nucleons. There are a number of other fundamental particles such as electron, anti-proton, positron and neutrino, photons and the gravitation which are stable. These are listed in Table, below.

Characteristics of certain stable fundamental particles.

Characteristics of certain unstable fundamental particles

Stability of nucleus

The stability of a nucleus depends on the number of protons and neutrons present in it. Elements of atomic number upto 20 have the same number of protons and neutrons and the nuclei of these elemetns are most stable. The ratio N/P = 1. In case of heavier elements with higher atomic numbers, the repulsive force increases at greater rate than the attractive force. In addition of extra neutron the average distance between proton within the nucleus decrease the repulsive force. Elements with atomic number greater than 92 having \(\frac{n}{p}\) ratio greater than 1.6 undergoes spontaneous fission.Stability of nucleus

The graph plotted between number of neutrons with number of protons shows the stable non-radioactive nuclei. The stable nuclei is represented by the shaded area and is called the zone of stabilty. Nuclei that lies above the zone of stability will contain more neutrons while those lying below have more protons which results in instability. Such nuclei would attain stability by undergoing change that would produce a nucleus with n to p ratio within the stability zone. To achieve stability two causes arise depending on \(\frac{n}{p}\) ratio of the nuclei.

1. A nucleus with lower \(\frac{n}{p}\) ratio a would achieve stability by an increase of neutrons, a decrese of protons or both simultaneously.
2. A nucleus with higher \(\frac{n}{p}\) ratio would involve a decrease of neutrons, an increase of proton or a ismultaneous alternation of both particles to attain stability.

As the radioactive elements are unstable its \(\frac{n}{p}\) ratio lie above and below the zone of stability. Therefore spontaneous disintegration of radioactive elements occur with the emission of \(\alpha\) and \(\beta\)- particles. Emission of each \(\alpha\)-particle is equivalent to loss of two protons and two neutrons and loss of each \(\beta\)-particles is equivalent to increase of one proton and decrease of one neutrons.

Modes of decay

Nuclei with \(\frac{N}{P}\) ratio appreciably higher or lower than the stable ratio are radioactive. When they decay nuclei of maximum stability. If the\(\frac{N}{P}\) ratio is high, a nucleus will decay in such a way that it reduces the\(\frac{N}{P}\) ratio and forms a stable arrangement. The ratio can be reduced in two ways.

Beta emission:

Electrons or \(\beta\) radiation may be emitted from the nucleus when a neutron is converted into a proton, an electron and a neutrino. This reduces the \(\frac{N}{P}\) ratio. The neutrinoν is a strange particle. It has zero mass and zero charge and is postulated to balance the spins. Neutrino is emitted in almost all nuclear transformations. The change of a neutron into a proton may be written as:

$$_{1}^{0}\textrm{n}$$ $$\longrightarrow$$$$_{1}^{1}\textrm{p}$$ + -$$_{0}^{-1}\textrm{e}$$ +v

The mass numbers are shown at the top and must be balanced on both sides of this equation. The nuclear charges are shown at the bottom. These too must be balanced on both sides of equation. The loss of an electron from the nucleus in this way decreases the\(\frac{N}{P}\) ratio.

$$_{14}^{6}\textrm{C}$$ $$\longrightarrow$$$$_{14}^{7}\textrm{N}$$ + -$$_{0}^{-1}\textrm{e}$$ +v

$$_{29}^{13}\textrm{Al}$$$$\longrightarrow$$$$_{29}^{14}\textrm{Si}$$ +-$$_{0}^{-1}\textrm{e}$$ +v

In \(\beta\) decay the mass number remains same , but the nuclear charge increases by one unit. Thus when \(\beta\) decay occurs the element moves one place to the right in the periodic table.

Neutron emission:

The emission of neutron from the nucleus decrease\(\frac{N}{P}\) ratio. This form of decay is not common and takes place with highly energetic nuclei. This is binding energy of the neutron in the nucleus is high (about 8 MeV). One of the few examples involves Kr, which can decay either by neutron emission or \(\beta\) decay.

$$_{87}^{36}\textrm{Kr}$$$$\longrightarrow$$$$_{86}^{36}\textrm{Kr}$$ +$$_{1}^{0}\textrm{n}$$

$$_{87}^{36}\textrm{Kr}$$ $$\xrightarrow{\beta}$$ $$_{87}^{37}\textrm{Rb}$$$$\xrightarrow{\beta}$$$$_{87}^{38}\textrm{Sr}$$

If the\(\frac{N}{P}\) ratio is too low. There are two possible modes of decay.

1. Positron emission: Positrons or \(\beta\) radiation (positive electrons) result from the transformation of a proton to a neutron. The positron is ejected from the nucleus together with an anti-neutrino \(\bar{ν}\)

$$_{1}^{1}\textrm{P}$$ $$\longrightarrow$$$$_{1}^{0}\textrm{n}$$ +$$_{0}^{1}\textrm{e}$$ + \(\bar{ν}\)

Some examples of positron emission are:

$$_{19}^{10}\textrm{Ne}$$$$\longrightarrow$$$$_{19}^{9}\textrm{F}$$ +$$_{0}^{1}\textrm{e}$$+ \(\bar{ν}\)

$$_{11}^{6}\textrm{Ne}$$$$\longrightarrow$$$$_{11}^{5}\textrm{B}$$ +$$_{0}^{1}\textrm{e}$$+ \(\bar{ν}\)

2. Orbital or K-electron capture: The nucleus may capture an orbital electron and thus convert a proton into a neutron with the emission of a neutrino:

$$_{1}^{1}\textrm{P}$$ + $$_{0}^{-1}\textrm{e}$$ $$\longrightarrow$$$$_{1}^{0}\textrm{n}$$ + \(\bar{ν}\)

This process increases the\(\frac{N}{P}\) ratio. Usually an electron from the shell closest to nucleus is captured. This is called the K shell, so the process is called K-shell capture. Some examples are:

$$_{7}^{4}\textrm{Be}$$ + $$_{0}^{-1}\textrm{e}$$ $$\longrightarrow$$$$_{3}^{7}\textrm{Li}$$ + ν

$$_{19}^{40}\textrm{K}$$ + $$_{0}^{-1}\textrm{e}$$ $$\longrightarrow$$$$_{18}^{40}\textrm{Ar}$$ + ν

Mass defect and binding energy

In case of hydrogen atom¹₁H, the mass is equal to the mass of one proton and the mass of one electron. But in case of other atoms the mass is less than the sum of the constituent neutrons, protons and electrons. This difference in the mass is called mass defect. The binding energy holding the neutrons and protons together is related to mass defect. The energy of the stable nucleus should be less than that of its constituent particles for its existance.

Here, Mass defect (ΔM) = (Mass of proton + mass of neutron) - Actual mass of nucleus

Now, binding energy can be calculated by using Einstein's mass energy relationship

ΔE =Δm \(\times\) C²where is the velocity of the light

Binding energy per nucleon is called average binding energy

B.E. per nucleon = \(\frac{B.E.}{mass number}\)

Radioactivity

The nuclei of heavy elements like U, Th, Ra and Po are unstable and keep on emitting spontaneously invisible rays or radiations like \(\alpha\), \(\beta\) or \(\gamma\) rays and give more stable elements. These heavy elements which emit \(\alpha\), \(\beta\) or \(\gamma\) rays are called radioactive elements and the property of emitting these rays is called radioactivity of the element. The word “radioactivity” means “Ray emitting – activity”.

The rays produced in the radioactivity are called radioactive rays. The spontaneous emission of radioactive rays from the nucleus of a radioactive element is called radioactive disintegration or radioactive decay of the element.

There are two types of radioactivity natural and artificial.

1. Natural radioactivity: It occurs in nature and is spontaneous. The nuclei of heavy atoms disintegrate naturally forming more stable and lighter nuclei emitting \(\alpha\), \(\beta\) or \(\gamma\) radiations. It cannot be controlled i.e. cannot be slowed down or accelerated by any means.
2. Artificial radioactivity: It does not occur naturally so it is not spontaneous. For artificial radioactivity to occur the nuclei of the atoms should be bombarded by fast moving particles like \(\alpha\)-particles, neutrons, protons, deuterons etc. It can be controlled with the increase and decrease of speed of the bombarding particles. This can be done in light elements also.

Nuclear fission reaction:

In this reaction, the nuclei of heavy atom split into two nuclei with lighter atoms. This process result in the release of large amount of energy which is about 8 \(\times\) 10⁹kJmol^-1. This process occurs due to excitation of neutrons. In the case of , several different primary products are formed, depending on exactly how the nucleus splits up. Here, the common three reactions are:

Nuclear fusion reaction:

In this reaction, two or more light nuclei are made to combine to give heavier nuclei. Duringm this process also some mass is lost and is converted in to energy. This reaction is the source of energy of the sun i.e. the fusion of hydrogen nuclei to from helium nuclei releasing energy. For example, the fusion or combination of deuterium and tritium gives a stable nucleus gives a stable nucleus of helium and energy is liberated.

Half life determination

Half life period is the time taken for half the radioactive sample to decay. This is a characteristics of a particular isotope. It is denoted by t_{\(\frac{1}{2}\).}

Let initially, t = 0, there are 'a' amount of radioactive substance 'A'. After 't' second certain number of atoms X will have decay leaving (a-x).

The rate of disintegration \(\frac{dx}{dt}\) can be given as,

\(\frac{dx}{dt}\)∝ (a-x)

or,\(\frac{dx}{dt}\) = k(a-x) where k = disintegration constant (decay constant)

or, \(\frac{dx}{(a-x)}\) = k.dt

Taking the integration we get,

- \(\ln\) (a-x) = kt + c...................(i)

when, t = 0, x= 0

then -\(\ln\)a= c

putting c =-\(\ln\)a in equation (i) we get,

- \(\ln\) (a-x) = kt - \(\ln\)a

or, kt= -\(\ln\) (a-x) +\(\ln\)a

or, kt = \(\ln\)\(\frac{a}{(a-x)}\)

or, k = \(\frac{1}{t}\) \(\ln\)\(\frac{a}{(a-x)}\)

or, k = \(\frac{2.303}{t}\) \(\log\)\(\frac{a}{(a-x)}\)...........(ii)

Suppose after time t_{\(\frac{1}{2}\)}, when half of the atoms have decayed i.e. x = \(\frac{1}{2}\)

∴ k = \(\frac{2.303}{t_\frac{1}{2}}\) \(\log\)\(\frac{a}{a-\(\frac{a}{2}\)

or, k =\(\frac{2.303}{t_\frac{1}{2}}\) \(\log\)\(\frac{2a}{a}\) = \(\frac{0.693}{t_\frac{1}{2}}\)

∴t_{\(\frac{1}{2}\)}= \(\frac{0.693}{k}\) which is required expression for the half life period.

Radioactivedisplacement law

1. Emission of an \(\alpha\) particle produces an element which is four mass units lighter and the atomic number decreases by two units. The daughter element is therefore two places to the left from the parent in the periodic table.

$$_{a}^{b}\textrm{M}$$ $$\xrightarrow{-\alpha}$$$$_{a-2}^{b-4}\textrm{M}$$

Where the metal M has atomic number 'a' and mass number 'b'. The element is shifted two position to the left in the periodic table as a result ofα-emission,

2. The atomic number of an element is increased by one unit while the mass number remains unchanged when it emits a \(\beta\)-paticle.

$$_{a}^{b}\textrm{M}$$ $$\xrightarrow{-\beta}$$$$_{a+1}^{b}\textrm{M}$$

The position of the element is shifted one position to the right in the periodic table as a result of \(\beta\) emission.

These changes are shown in the following series

$$_{88}^{233}\textrm{Ra}$$$$\xrightarrow{\alpha}$$$$_{86}^{219}\textrm{Rn}$$$$\xrightarrow{\alpha}$$$$_{84}^{215}\textrm{Po}$$

$$\xrightarrow{\alpha}$$$$_{82}^{211}\textrm{Pb}$$$$\xrightarrow{\beta}$$$$_{83}^{211}\textrm{Bi}$$$$\xrightarrow{\alpha}$$ $$_{81}^{207}\textrm{Tl}$$$$\xrightarrow{\alpha}$$$$_{82}^{207}\textrm{Pb}$$

Radioactive Decay Series

The heavy radioactive elements may be grouped into four decay series. The common radioactive elements thorium, uranium and actinium occur naturally and belong to three different series named after them . They are the parent members of their respective sries and have the longest half life periods. They decay by a series of \(\alpha\) and \(\beta\) emissions, and produce radioactive elements which are successively more stable until finally a stable isotope is reached. All three series terminate with lead ($$_{82}^{206}\textrm{Pb}$$,$$_{82}^{207}\textrm{Pb}$$,$$_{82}^{208}\textrm{Pb}$$)

Following the discovery of the artificial post-uranium elements, the neptunium series has been added, which ends with bismuth,$$_{92}^{209}\textrm{Pb}$$

Thorium (4n) series

Neptunium (4n + 1) series

Uranium (4n+ 2) series

Actinium (4n + 3) series

The numbers in brackets indicate that the parent and all the members of a particular series have mass mumbers exactly divisibly by four, or divisible by four with a remainder of one, two or three. There is no natural cross-linking between the four series, although this can be performed artificially.

References

Lee, J.D. Conscise Inorganic Chemistry. 5th. John Wiley and Sons Inc., 2007.

wikipedia. n.d. <https://en.wikipedia.org/wiki/Stable_nuclide>.

Bodner research web. n.d. <http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch23/modes.php>.