NUCLEI

Introduction

In the previous section, we have discussed about various preliminary atom models, Rutherford’s alpha particle scattering experiment and Bohr atom model. These played a vital role to understand the structure of the atom and the nucleus. In this section, the structure of the nuclei and their properties, classifications are discussed.

Atoms have a nucleus surrounded by electrons. The nucleus contains protons and neutrons. The neutrons are electrically neutral (q = 0) and the protons have positive charge (q = + e) equal in magnitude to the charge of the electron (q = –e). The number of protons in the nucleus is called the atomic number and it is denoted by Z. The number of neutrons in the nucleus is called neutron number (N). The total number of neutrons and protons in the nucleus is called the mass number and it is denoted by A. Hence, A = Z+N.

The two constituents of nucleus namely neutrons and protons, are collectively called as nucleons. The mass of a proton is 1.6726x10^-27 kg which is roughly 1836 times the mass of the electron. The mass of a neutron is slightly greater than the mass of the proton and it is equal to 1.6749×10 ^-27 kg.

To specify the nucleus of any element, we use the following general notation \[ _{Z}^{A}\textrm{X} \]

where X is the chemical symbol of the element, A is the mass number and Z is the atomic number. For example, the nitrogen nucleus is represented by 15_N_7 . It implies that nitrogen nucleus contains 15 nucleons of which 7 are protons (Z = 7) and 8 are neutrons (N = A – Z = 8). Note that once the element is specified, the value of Z is known and subscript Z is sometimes omitted. For example, nitrogen nucleus is simply denoted as 15_N_ and we call it as ‘nitrogen fifteen’.

Since the nucleus is made up of positively charged protons and electrically neutral neutrons, the overall charge of the nucleus is positive and it has the value +Ze. But the atom is electrically neutral which implies that the number of electrons in the atom is equal to the number of protons in the nucleus.

Isotopes:

In nature, there are atoms of a particular element whose nuclei have same number of protons but different number of neutrons. These kinds of atoms are called isotopes. In other words, isotopes are atoms of the same element having same atomic number Z, but different mass number A. For example, hydrogen has three isotopes and they are represented as \( _{1}^{1}\textrm{H} (hydrogen), _{2}^{1}\textrm{H} (deuterium),and _{3}^{1}\textrm{H} (tritium)\) . Note that all the three nuclei have one proton and, hydrogen has no neutron, deuterium has 1 neutron and tritium has 2 neutrons.

The number of isotopes for the particular element and their relative abundances (percentage) vary with each element. For example, carbon has four main isotopes: \( _{6}^{11}\textrm{C},_{6}^{12}\textrm{C},_{6}^{13}\textrm{C}, and _{6}^{14}\textrm{C},\) . But in nature, the percentage of \(_{6}^{12}\textrm{C}\) is approximately 98.9%, that of \(_{6}^{13}\textrm{C}\) is 1.1% and that of \(_{6}^{14}\textrm{C}\) is 0.0001%. The other carbon isotope \(_{6}^{11}\textrm{C}\) , does not occur naturally and it can be produced only in nuclear reactions in the laboratory or by cosmic rays.

Since the chemical properties of any atom are determined only by electrons, the isotopes of any element have same electronic structure and same chemical properties. So the isotopes of the same element are placed in the same location in the periodic table.

Isobars:

Isobars are the atoms of different elements having the same mass number A, but different atomic number Z. In other words, isobars are the atoms of different chemical elements which have same number of nucleons. For example \(_{16}^{40}\textrm{S},_{17}^{40}\textrm{Cl},_{18}^{40}\textrm{Ar},_{19}^{40}\textrm{K}, and _{20}^{40}\textrm{Ca}\) are isobars having same mass number 40 but different atomic numbers. Unlike isotopes, isobars are chemically different elements. They have different physical and chemical properties.

Isotones:

Isotones are the atoms of different elements having same number of neutrons. \(_{5}^{12}\textrm{B}\) and \(_{6}^{13}\textrm{C}\) are examples of isotones with 7 neutrons each.

The mass of nuclei is very small (about 10^-25 kg or less). Therefore, it is more convenient to express it in terms of another unit namely, the atomic mass unit (u). One atomic mass unit (u) is defined as the (1/12)th of the mass of the isotope of carbon \(_{6}^{12}\textrm{C}\) which is more abundant in naturally occurring isotope of carbon.

In other words \[ 1u = \frac{{\text{{mass of }} ^{12}_6C \text{{ atom}}}}{{12}} = \frac{{1.9926 \times 10^{-26}}}{{12}} = 1.660 \times 10^{-27} \, \text{{kg}}\]

In terms of this atomic mass unit, the mass of the neutron = 1.008665 u, the mass of the proton =1.007276 u, the mass of the hydrogen atom = 1.007825 u and the mass of \(_{6}^{12}\textrm{C}\) =12 u. Note that usually mass specified is the mass of the atom, not mass of the nucleus. To get the nuclear mass of particular nucleus, the mass of electrons has to be subtracted from the corresponding atomic mass. Experimentally the atomic mass is determined by the instrument called Bainbridge mass spectrometer. If we determine the atomic mass of the element without considering the effect of its isotopes, we get the mass averaged over different isotopes weighted by their abundances.

EXAMPLE 9.6

Calculate the average atomic mass of chlorine if no distinction is made between its different isotopes?

Solution

The element chlorine is a mixture of 75.77% of \(_{17}^{35}\textrm{Cl}\) and 24.23% of \(_{17}^{37}\textrm{Cl}\) . So the average atomic mass will be

In fact, the chemist uses the average atomic mass or simply called chemical atomic weight (35.453 u for chlorine) of an element. So it must be remembered that the atomic mass which is mentioned in the periodic table is basically averaged atomic mass.

The alpha particle scattering experiment and many other measurements using different methods have been carried out on the nuclei of various atoms. The nuclei of atoms are found to be approximately spherical in shape. It is experimentally found that radius of nuclei for Z > 10, satisfies the following empirical formula \[ R = R_0 A^{1/3} \] (9.19)

Here A is the mass number of the nucleus and the constant \( R_0\) = 1.2 F, where 1 F = 1 × 10^–15 m.The unit fermi (F) is named after Enrico Fermi.

EXAMPLE 9.7

Calculate the radius of \(_{79}^{197}\textrm{Au}\) nucleus.

Solution

According to the equation (9.19),

R = 1.2 × 10^-15 × (197)^(1/3)= 6.97 × 10^-15 m

Or R = 6.97F

EXAMPLE 9.8

Calculate the density of the nucleus with mass number A.

Solution

From equation (9.19), the radius of the nuclecus, \( R = R_0 A^{1/3} \) . Then the volume of the nucleus \[ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_0^3 A \] By ignoring the mass difference between the proton and neutron, the total mass of the nucleus having mass number A is equal to A.m where m is mass of the proton and is equal to 1.6726 x 10^-27 kg.

Nuclear density,

\[ \rho = \frac{\text{mass of the nuclei}}{\text{Volume of the nuclei}} \] \[ \rho = \frac{A \cdot m}{\frac{4}{3}\pi R_0^3 A} \] \[ \rho = \frac{m}{\frac{4}{3}\pi R_0^3} \] The above expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z > 10) have the same density and it is an important characteristic property of all nuclei.

We can calculate the numerical value of this density by substituting the corresponding values.

\[ \rho = \frac{1.67 \times 10^{-27}}{\frac{4}{3} \pi x (1.2 \times 10^{-15})^3} \] \[ \rho \approx 2.3 \times 10^{17} \, \text{kg m}^{-3} \] It implies that nucleons are extremely tightly packed or compressed state in the nucleus and compare this density with the density of water which is 103 kg m-3.

Do You Know

A single teaspoon of nuclear matter would weigh about trillion tons.

It is experimentally found out that the mass of any nucleus is always less than the sum of the masses of its individual constituent particles. For example, consider the carbon-12 nucleus which is made up of 6 protons and 6 neutrons. Mass of 6 neutrons = 6×1.00866 u =6.05196_u_ Mass of 6 protons =6 ×1.00727_u_ = 6.04362_u_ Mass of 6 electrons =6 × 0.00055_u_ = 0.0033_u_

The expected mass of carbon-12 nucleus =6.05196u + 6.04362_u_ = 12.09558_u_

But using mass spectroscopy, the atomic mass of carbon-12 atom is found to be 12 u. So if we subtract the mass of 6 electrons (0.0033 u) from 12 u, we get the nuclear mass of carbon-12 atom which is equal to 11.9967 u. Hence the experimental mass of carbon-12 nucleus is less than the total mass of its individual constituents by ∆_m u_= 0 09888. . This difference in mass ∆ m is called mass defect. In general, if M, m_p, and m_n are mass of the nucleus \( _{Z}^{A}\textrm{X} \) , the mass of a proton and the mass of a neutron respectively, then the mass defect is given by

\[\Delta m = (Zm_p + Nm_n)-M \] (9.20)

Where has this mass disappeared? The answer was provided by Albert Einstein with the help of famous mass-energy relation (E=mc²). According to this relation, the mass can be converted into energy and energy can be converted into mass. In the case of the carbon-12 nucleus, when 6 protons and 6 neutrons combine to form carbon-12 nucleus, mass equal to mass defect disappears and an energy equivalent to missing mass. This energy is called the binding energy of the nucleus (BE) and is equal to (∆m)c² . In fact, to separate the carbon-12 nucleus into individual constituents, we must supply the energy equal to binding energy of the nucleus.

We can write the equation (9.20) in terms of binding energy \[BE = (Zm_p + Nm_n - M)c^2 \] (9.21)

It is always convenient to work with the mass of the atom rather than with the mass of the nucleus. Hence by adding and subtracting the mass of the Z electrons, we get

where m_p + m_e = m_H (mass of hydrogen atom)

Here M + Zm_e = M_A, where M_A is the mass of the atom of an element \( _{Z}^{A}\textrm{X} \) . Finally, the binding energy in terms of the atomic masses is given by \[ BE =[Zm_H + Nm_n - M_A]c^2 \] (9.24)

Note:

Using Einstein’s mass-energy equivalence,the energy equivalent of one atomic mass unit

1u =1.66×10^-27×(3×10⁸)²= 14.94×10^-11≈931 MeV

EXAMPLE 9.9

Compute the binding energy of \( ^{4}_2He\) nucleus using the following data: Atomic mass of Helium atom, M_A(He) = 4.00260 u and that of hydrogen atom,m_H=1.00785 u.

Solution:

Binding energy \( BE =[Zm_H + Nm_n - M_A]c^2 \)

For helium nucleus, Z = 2, N = A–Z = 4–2 = 2

Mass defect

∆m =[(2×1.00785u)+(2×1.008665u) - 4.00260u]

∆m = 0.03043u

B.E = 0.03043u × c²

B.E = 0.03043× 931MeV = 28.33MeV

[1uc²= 931 MeV]

The binding energy of the \( ^{4}_2{He}\) nucleus is 28.33 MeV.

In the previous section, the origin of the binding energy is discussed. Now we can find the average binding energy per nucleon BE . It is given by

\[\overline{BE} = \frac{[Zm_H + Nm_n - M_A]c^2}{A}\] (9.25)

The average binding energy per nucleon is the average energy required to separate single nucleon from the particular nucleus. When BE is plotted against A of all known nuclei. We get BE aveage curve as shown in Figure 9.24.

Important inferences from of the average binding energy curve:

(1) The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.

(2) The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number lying between A = 40 and 120. These elements are comparatively more stable and not radioactive.

(3) For higher mass numbers, the curve drops slowly and BE for uranium is about 7.6 MeV. Such nuclei are unstable and exhibit radioactive.

From Figure 9.24, if two light nuclei with A<28 combine with a nucleus with A<56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.

(4) If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled. Fission is explained in the section 9.7

EXAMPLE 9.10

Compute the binding energy per nucleon of \( ^{4}_2He\) nucleus.

Solution

From Example 9.9, we found that the BE of \( ^{4}_2He\) =28.33 Mev

Binding energy per nucleon = B.E = 28.33 MeV/4 = 7 MeV.