Atomic mass, size, composition of nucleus, practice problems, FAQs

The structure of atoms was always a subject of intrigue for scientists. Many, including JJ Thomson and Neils Bohr had suggested various models to explain the motion of electrons inside an atom; Rutherford, along with his disciples Geiger and Marsden, conducted the famous gold foil experiment, in which they observed that most of the alpha particles(helium nuclei) went undeflected; meaning that majority of the space in an atom is empty; the entire positive charge appears to be concentrated at the core of an atom called its nucleus. Understanding the structure, size and volume of nucleus is vital; it finds various applications in nuclear phenomena like nuclear fission and nuclear fusion. In this article, we will explore more on the size, composition and density of a nucleus.

Table of contents

What is nucleus?

Composition of the nucleus
Size of the nucleus
Nuclear density
Isotopes, isobars and isotones
Practice problems
FAQs

What is nucleus?

The entire positive charge of an atom is concentrated at the center called its nucleus. It consists of protons(+vely charged) and neutrons(no charge). The nuclear force which exists between them is charge independent; i.e. $F_{p p} = F_{p n} = F_{n n}$ .

Composition of the nucleus

Nucleus is made up of positively charged protons and neutrons; the electrons revolve around the nucleus.The mass of the proton (m_p) and neutron (m_n) are as follows;

$m_{p} = 1.67262 \times 10^{- 27} k g$

$m_{n} = 1.67493 \times 10^{- 27} k g$

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Fig showing an atom with the nucleus at the center

Nucleus in an atom

Atomic Number

The atomic number is defined as the number of protons; which in turn is equal to the number of electrons. It is represented by the letter Z.

Z=No. of protons=No. of electrons

Mass number

The total number of neutrons and protons which are present in the nucleus is called the nucleons or mass number.(A)

A=Number of nucleons=Number of neutrons+protons

Any element X, can be written in the form

$_{Z}^{} X^{A}$

The total charge in the nucleus is equal to Ze, where e= $1.6 \times 10^{- 19} C$ is the electronic charge; in other words it is the charge carried by one electron.

Size of the nucleus

After much experimentation, it was found that the volume of the nucleus is directly proportional to the number of nucleons i.e. the mass number. If R represents the radius of the nucleus, having mass number A , then

$V = \frac{4}{3}$ $π R^{3} \propto A$

$R = R_{0} A^{\frac{1}{3}}$ ; $R_{0} = 1.2 \times 10^{- 15} m$ is an empirical constant called nuclear unit radius.

Size of the nucleus $\approx 10^{- 15} m$

Size of the atom $\approx 10^{- 10} m$

Note: 1 fermi $(f m) = 10^{- 15} m$

Nuclear density

So now you have an idea about the nucleus, its size, and what actually constitutes a nucleus. Now let us talk about nuclear density, which gives us an idea about how closely protons and neutrons are packed.

Density of nucleus = $\frac{M a s s o f n u c l e u s}{V o l u m e o f n u c l e u s}$

Let m indicate the mass of one nucleon $\approx 1.67 \times 10^{- 27} k g)$ and A is the mass number, then mass of the nucleus=mA, where m indicates the mass of a nucleon, A is the mass number. If R indicates the radius of the nucleus, then

Volume of nucleus $= \frac{4}{3} π$ = $\frac{4}{3} π$ ${(R_{0} A^{\frac{1}{3}})}^{3}$ = $= \frac{4}{3} π$ ${(R_{0}}^{3} A)$

∴Density of the nucleus, $ρ = \frac{m A}{\frac{4}{3} π {(R_{0}}^{3} A)}$ $\Rightarrow ρ = \frac{m}{\frac{4}{3} π {R_{0}}^{3}}$

I.e. the nuclear density is independent of the mass number A.

Isotopes, isobars and isotones

Isotopes

Atoms of the same element which have same atomic number but different atomic mass are called isotopes

eg. $_{1} H^{1},_{1} H^{2}$

Isobars

Atoms of different elements having same mass number( A ) but different atomic number ( Z ) are called isobars

eg. $_{1} H^{3}$ and $_{2} {H e}^{3}$

Isotones

The atoms of different elements having equal number of neutrons A-Z are isotones

eg. $_{3} {L i}^{7}$ and $_{4} {B e}^{8}$ both have 4 neutrons.

Practice problems

Q1. Calculate the approx. nuclear radius of Fe¹²⁵, if that of Al²⁷ is 6.4 fermi?
Solution) We know that

$R = R_{0} A^{\frac{1}{3}} \Rightarrow \frac{R_{F e}}{R_{A l}}$ = ${(\frac{A_{F e}}{A_{A l}})}^{\frac{1}{3}}$ $= {(\frac{125}{27})}^{\frac{1}{3}}$

$R_{F e} = \frac{5}{3} R_{A l} = \frac{5}{3} \times 6.4 \approx 10 f m$

Q2. The nucleus of an atom has a mass number of 24. Then it has;

(a)11 protons and 13 electrons.
(b)11 protons and 13 neutrons.
(c)11 electrons, 13 protons, 11 neutrons.
(d)13 electrons, 11 protons, 13 neutrons.

Solution) b

Mass number=Number of protons+Number of neutrons=11+13=24.

Q3. The radius of $_{29} {C u}^{64}$ in fermi is ( $R_{0} = 1.2 \times 10^{- 15} m$ )

(a)1.2
(b)4.8
(c)7.7
(d)9.6

Solution) b

$R = R_{0} {(A)}^{\frac{1}{3}} \Rightarrow R = 1.2 \times 10^{- 15} \times {(64)}^{\frac{1}{3}} = 4.8$

Q4. In $_{88} {R a}^{226}$ nucleus, there are

(a)138 neutrons and 88 protons
(b)138 protons and 88 neutrons
(c)226 neutrons and 88 electrons
(d)226 neutrons and 138 electrons

Solution)a

No. of neutrons=226-88=138

No. of protons =88

FAQs

Q1. Define mass number.
Ans) Mass number is defined as the number of nucleons(protons+neutrons) that are present in the nucleus.

Q2. Define atomic number
Ans) Atomic number (Z) is the number of protons in an atomic nucleus which is equal to the number of electrons.

Q3. Write the expression for nuclear density
Ans) Nuclear density, $ρ = \frac{m}{\frac{4}{3} π {R_{0}}^{3}}$ where $R_{0} = 1.2 \times 10^{- 15} m$ is a constant and m is the mass of one nucleon.