Atomic Mass and Composition of Nucleus

To measure various objects, we use a gram or a kilogram as the standards. However, the mass of an atom is really small as compared to a kilogram. To give you a perspective, the mass of a carbon atom 12C, is 1.992647 × 10–26 kg. Imagine writing the mass of atoms in kilograms – not very convenient is it? Hence, Atomic Mass is expressed using different units. Let’s find more about the composition of a nucleus in the section below.

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Atomic Mass and Isotopes

One Atomic Mass unit (u) is defined as 1/12th of the mass of the carbon atom (12C). Therefore,

1u = (mass of one 12C atom)/12 = (1.992647 × 10–26)/12 = 1.660539 10–27 kg … (1)

Now, the atomic mass of different elements expressed in atomic units (u) is nearly equal to being integral multiples of the mass of a hydrogen atom. However, it is important to note that there are many exceptions to this rule.

The next point is how do we measure atomic mass accurately? The answer is – by using a mass spectrometer. An interesting observation made by measuring atomic masses is that there are many atoms of the same element which have different masses but exhibit the same chemical properties. Such atoms are called ‘Isotopes’.

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Further observation revealed that nearly all elements have a mixture of many isotopes. However, the number of isotopes can vary, depending on the element. Let’s look at some examples:

Example 1

Chlorine has two isotopes having masses 34.98u and 36.98u. These are nearly the integral multiples of the atomic mass of a hydrogen atom. Also, the relative abundance of these two isotopes is 75.4 % and 24.6% respectively. So, the average mass of chlorine is calculated by finding the weighted average of the masses of these two isotopes; which is,

= {(75.4 x 34.98) + (24.6 x 36.98)}/100 = 35.47u = the atomic mass of chlorine.

Example 2

Atomic Mass

[Source: Wikipedia]

Hydrogen, on the other hand, has three isotopes having masses 1.0078 u, 2.0141 u, and 3.0160 u. Of these, the lightest isotope has a relative abundance of 99.985% and is called the Proton. Now, the mass of a proton is,

mp = 1.00727 u = 1.67262 10– 27 kg … (2)

If we take the mass of a hydrogen atom and subtract the mass of an electron from it, we get

1.00783u – 0.00055u = 1.00728u = mp

The isotope having mass 2.0141 u is deuterium and the one having mass 3.0160 u is tritium. Important to note here that tritium nuclei are unstable and do not occur naturally. They are produced artificially in laboratories.

Important Note

A proton is stable and carries one unit of fundamental charge. It is the positive charge in the nucleus. Post the Quantum theory, it was agreed that the electrons lie outside the nucleus (earlier there was a lot of debate about them being inside the nucleus). The number of electrons is equal to the atomic number of the element – Z.

Hence, the total charge of the atomic electrons is –Ze. Since an atom is electrically neutral, the nucleus carries a charge of +Ze. By this, we can conclude that the number of protons is equal to the number of electrons in an atom = the atomic number Z.

Discovery of Neutron

Deuterium and tritium are the isotopes of hydrogen. Hence, the must contain one proton each. However, there is a clear difference in their atomic masses. The ratio of the atomic mass of hydrogen, deuterium, and tritium is 1:2:3. Hence, there has to be some more matter in these isotopes adding to their atomic masses. Also, this additional matter needs to be electrically neutral since the protons and electrons are in balance.

In this case, the additional matter is present in multiples of the mass of the proton (deuterium has additional matter equal to the mass of one proton and tritium has matter equal to the mass of two protons). Hence, it is easy to conclude that the nuclei of atoms contain neutral matter, in addition to protons, in multiples of a basic unit.

In 1932, James Chadwick observed the emission of a neutral radiation when a Beryllium nucleus was bombarded with alpha particles. Now, at that time, the only known neutral radiation was photons or electromagnetic radiation. If the neutral radiation consisted of protons, then

The energy of photons >> The energy available from the bombardment of the Beryllium nuclei with alpha particles

Chadwick hypothesised that the neutral radiation consisted of a new type of neutral particles – Neutrons. By further calculations, he proved that the mass of a neutron is nearly equal to the mass of a proton. Now, the mass of a neutron known to a high degree of accuracy is,

mn = 1.00866 u = 1.6749×10–27 kg ~ mp

Composition of Nucleus

A free neutron is unstable. It decays into a proton, an electron and an antineutrino (an elementary particle) with a mean life of around 1000s. However, it is stable inside the nucleus. Therefore, the composition of a nucleus is described as follows:

A = Z + N


  • Z – atomic number = number of protons
  • N – neutron number = number of neutrons
  • A – mass number = = total number of protons and neutrons

Protons or Neutrons are also called Nucleons. Hence, the mass number (A) of an atom is the total number of nucleons in it. A typical nuclide of an atom is AZX, where X is the chemical symbol of the atom.


The nuclide of gold is denoted by 19779Au. Hence, we can conclude that there are 197 nucleons in a nucleus of gold. Out of these 79 are protons and 118 (197-79) are neutrons.

  • Deuterium is denoted as 21H … it has one proton and one neutron
  • Tritium is denoted as 31H … it has one proton and two neutrons

Isobars and Isotones

  • Isobars – Nuclides with same mass number A. For Eg. 31H and 32H
  • Isotones – Nuclides with the same number of neutrons (N) but different atomic numbers (Z). For e.g. 19880Hg and 19779Au.

Solved Examples for You

Question: Two stable isotopes of lithium 63Li and 73Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512u and 7.01600u, respectively. Find the atomic mass of lithium.

Answer: Mass of the first lithium isotope 63Li = m1 = 6.01512u

Mass of the second lithium isotope 73Li = m2 = 7.01600u

Abundance of 63Li = n1 = 7.5%

Abundance of 73Li = n2 = 92.5%

The atomic mass (m) of lithium is

m = (m1n1 + m2n2) / (n1 + n2)

= {(6.01512u x 7.5) + (7.01600u x 92.5)} / (7.5 + 92.5)

= 6.940934u

Hence, the atomic mass of lithium is 6.940934u.


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