Soon after the discovery of the sub-atomic particles of an atom, scientists were eager to figure out the distribution of these particles within the atom. Several atomic models were proposed to explain the structure of the atom. However, a lot of them could not explain the stability of the atom. Let’s learn about two of these atomic models that have led to our current concept of the atom.
Thomson’s Atomic Model
In 1898, J. J. Thomson proposed the first of many atomic models to come. He proposed that an atom is shaped like a sphere with a radius of approximately 10-10m, where the positive charge is uniformly distributed. The electrons are embedded in this sphere so as to give the most stable electrostatic arrangement.Doesn’t the figure above remind you of a cut watermelon with seeds inside? Or, you can also think of it as a pudding with the electrons being the plum or the raisins in the pudding. Therefore, this model is also referred to as the watermelon model, the plum pudding model or the raisin pudding model.
An important aspect of this model is that it assumes that the mass of the atom is uniformly distributed over the atom. Thomson’s atomic model was successful in explaining the overall neutrality of the atom. However, its propositions were not consistent with the results of later experiments. In 1906, J. J. Thomson was awarded the Nobel Prize in physics for his theories and experiments on electricity conduction by gases.
Rutherford’s Atomic Model
The second of the atomic models was the contribution of Ernest Rutherford. To come up with their model, Rutherford and his students – Hans Geiger and Ernest Marsden performed an experiment where they bombarded very thin gold foil with α-particles. Let’s understand this experiment.
α-Particle Scattering Experiment
In this experiment, high energy α-particles from a radioactive source were directed at a thin foil (about 100nm thickness) of gold. A circular, fluorescent zinc sulfide screen was present around the thin gold foil. A tiny flash of light was produced at a point on the screen whenever α-particles struck it.
Based on Thomson’s model, the mass of every atom in the gold foil should be evenly spread over the entire atom. Therefore, when α-particles hit the foil, it is expected that they would slow down and change directions only by small angles as they pass through the foil. However, the results from Rutherford’s experiment were unexpected –
- Most of the α-particles passed undeflected through the foil.
- A small number of α-particles were deflected by small angles.
- Very few α-particles (about 1 in 20,000) bounced back.
Conclusions of the α-scattering experiment
Based on the above results, Rutherford made the following conclusions about the structure of the atom:
- Since most of the α-particles passed through the foil undeflected, most of the space in the atom is empty.
- The deflection of a few positively charged α-particles must be due to the enormous repulsive force. This suggests that the positive charge is not uniformly spread throughout the atom as Thomson had proposed. The positive charge has to be concentrated in a very small volume to deflect the positively charged α-particles.
- Rutherford’s calculations show that the volume of the nucleus is very small compared to the total volume of the atom and the radius of an atom is about 10-10m, while that of the nucleus is 10-15m.
Nuclear Model Of The Atom
Based on his observations and conclusions, Rutherford proposed his model of the structure of the atom. According to this model –
- Most of the mass of the atom and the positive charge is densely concentrated in a very small region in the atom. Rutherford called this region the nucleus.
- Electrons surround the nucleus and move around it at very high speeds in circular paths called orbits. This arrangement also resembles the solar system, where the nucleus forms the sun and the electrons are the revolving planets. Therefore, it is also referred to as the Planetary Model.
- Electrostatic forces of attraction hold the nucleus and electrons together.
Drawbacks Of Rutherford’s Atomic Model
- According to Rutherford’s atomic model, the electrons (planets) move around the nucleus (sun) in well-defined orbits. Since a body that moves in an orbit must undergo acceleration, the electrons, in this case, must be under acceleration. According to Maxwell’s electromagnetic theory, charged particles when accelerated must emit electromagnetic radiation. Therefore, an electron in an orbit will emit radiation and eventually the orbit will shrink. If this is true, then the electron will spiral into the nucleus. But this does not happen. Thus, Rutherford’s model does not explain the stability of the atom.
- Contrarily, let’s consider that the electrons do not move and are stationary. Then the electrostatic attraction between the electrons and the dense nucleus will pull the electrons into the nucleus to form a miniature version of Thomson’s model.
- Rutherford’s model also does not state anything about the distribution of the electrons around the nucleus and the energies of these electrons.
Thus, Thomson and Rutherford’s atomic models revealed key aspects of the structure of the atom but failed to address some critical points. Now that we know the two atomic models, let’s try to understand a few concepts.
Atomic Number And Mass Number
As we know now, a positive charge on the nucleus is due to the protons. Also, the charge on the proton is equal but opposite to that of the electron. Atomic Number (Z) is the number of protons present in the nucleus. For example, the number of protons in sodium is 11 whereas it is 1 in hydrogen, Therefore, the atomic numbers of sodium and hydrogen are 11 and 1, respectively.
Also, to maintain electrical neutrality, the number of electrons in an atom is equal to the number of protons (atomic number, Z). Therefore, the number of electrons in sodium and hydrogen is 11 and 1, respectively.
Atomic number = the number of protons in the nucleus of an atom
= the number of electrons in a neutral atom
The positive charge on the nucleus is due to protons, but the mass of the atom is due to protons and neutrons. They are collectively known as nucleons. Mass number (A) of the atom is the total number of nucleons.
Mass number (A) = the number of protons (Z) + the number of neutrons (n)
Therefore, the composition of an atom is represented using the element symbol (X) with the mass number (A) as super-script on the left and atomic number (Z) as sub-script on the left – AZX.
Learn about Atomic number here in more detail here.
Isobars and Isotopes
Isobars are atoms with the same mass number but a different atomic number. For example, 146C and 147N.
Learn about Isobars here in more detail here.
Isotopes, on the other hand, are atoms with the same atomic number but a different mass number. This means that the difference in the isotopes is due to the presence of a different number of neutrons in the nucleus. Let’s understand this using hydrogen as an example –
- 99.985% of hydrogen atoms contain only one proton. This isotope is protium (11H).
- The isotope containing one proton and one neutron is deuterium (21D).
- The isotope with one proton and two neutrons is tritium (31T). This isotope exists in trace amounts on earth.
Other common isotopes are – carbon atoms with 6 protons and 6, 7, or 8 neutrons (126C, 136C, 146C) and chlorine atoms with 17 protons and 18 or 20 neutrons (3517Cl, 3717Cl).
Note: Chemical properties of atoms are under the influence of the number of electrons, which are dependent on the number of protons in the nucleus. The number of neutrons has a very little effect on the chemical properties of an element. Therefore, all isotopes of an element show same chemical behaviour.
Learn about Isotopes here in more detail here.
Solved Examples For You
Question 1: Match the columns:
|1. Mass number||a. Nuclear Model of Atom|
|2. J.J. Thomson||b. Number of protons|
|3. Rutherford||c. Number of nucleons|
|4. Atomic number||d. Plum pudding model|
Solution: 1 → c, 2 → d, 3 → a, 4 → b.
Question 2: Calculate the number of protons, neutrons, and electrons in 5626Fe.
Solution: In 5626Fe, atomic number (Z) = 26, mass number (A) = 56.
Number of protons = number of electrons = Z = 26.
Number of neutrons = A – Z = 56 – 26 = 30.