Atomic Models

Introduction to Atomic Models and Fundamental Particles

The discovery of sub-atomic particles (electrons, protons, and neutrons) shattered Dalton's idea of an indivisible atom. This breakthrough left scientists with several major challenges:

1. To account for the stability of the atom.
2. To compare the physical and chemical properties of different elements.
3. To explain the formation of molecules from different atoms.
4. To understand the origin and nature of electromagnetic radiation absorbed or emitted by atoms.

Millikan's Oil Drop Method

To understand the atom, scientists first needed to measure the properties of its constituents. R.A. Millikan devised the Oil Drop Method to measure the charge of an electron.

In this experiment, oil droplets (mist) produced by an atomiser enter a chamber through a tiny hole. A telescope with a micrometer eyepiece is used to view their downward motion, allowing Millikan to measure the mass of the droplets.

X-rays are passed through the chamber to ionize the air. The oil droplets collide with these gaseous ions and acquire an electrical charge. By applying a voltage across electrical condenser plates, the fall of these charged droplets can be retarded, accelerated, or made stationary depending on the charge and the applied voltage.

[JEE Main Tip] Millikan concluded that the magnitude of electrical charge ($q$) on the droplets is always an integral multiple of the elementary electrical charge ($e$).

$$q = n e$$

where $n = 1, 2, 3...$

Properties of Fundamental Particles

Before diving into atomic models, it is crucial to know the exact properties of the fundamental particles that make up the atom.

Name	Symbol	Absolute charge / C	Relative charge	Mass / kg	Mass / u	Approx. mass / u
Electron	e	$-1.602176 \times 10^{-19}$	$-1$	$9.109382 \times 10^{-31}$	$0.00054$	$0$
Proton	p	$+1.602176 \times 10^{-19}$	$+1$	$1.6726216 \times 10^{-27}$	$1.00727$	$1$
Neutron	n	$0$	$0$	$1.674927 \times 10^{-27}$	$1.00867$	$1$

Key Point: Notice that the mass of a neutron is slightly greater than the mass of a proton, while the mass of an electron is negligible in comparison.

Thomson Model of Atom and Discovery of Radiations

Thomson Model of Atom

Proposed by J.J. Thomson in 1898, this was one of the earliest attempts to describe atomic structure.

Thomson suggested that an atom is a sphere of positive charge (with a radius of approximately $10^{-10}$ m) in which the positive charge is uniformly distributed. The electrons are embedded into it to give the most stable electrostatic arrangement.

This model is famously known by several names:

• Plum pudding model
• Raisin pudding model
• Watermelon model

[School Exam Focus]

• Success: It successfully explained the overall electrical neutrality of the atom.
• Failure: It assumed that the mass of the atom is uniformly distributed, which was later disproved by Rutherford's scattering experiment.

Discovery of X-rays and Radioactivity

In the late 19th century, new types of rays were discovered that paved the way for better atomic models.

1. X-rays (Wilhelm Röentgen, 1895): When electrons strike a dense metal anode (target) in a cathode ray tube, they produce rays that cause fluorescence.

• They are not deflected by electric or magnetic fields.
• They have very high penetrating power.
• They have very short wavelengths ($\sim 0.1$ nm) and possess electromagnetic character.

2. Radioactivity (Henri Becquerel): Certain elements emit radiation on their own. This phenomenon is called radioactivity. Rutherford and others studied these emissions and categorized them into three types:

• $\alpha$-rays: High-energy particles carrying two units of positive charge and four units of atomic mass. Rutherford concluded they are helium nuclei (since combining an $\alpha$-particle with two electrons yields helium gas). They have the least penetrating power.
• $\beta$-rays: Negatively charged particles similar to electrons. Penetrating power is 100 times that of $\alpha$-particles.
• $\gamma$-rays: High-energy radiations like X-rays. They are neutral, do not consist of particles, and have the highest penetrating power (1000 times that of $\alpha$-particles).

Rutherford's Nuclear Model of Atom

Rutherford's $\alpha$-Particle Scattering Experiment

Ernest Rutherford, along with Hans Geiger and Ernest Marsden, bombarded a very thin gold foil (thickness $\sim 100$ nm) with high-energy $\alpha$-particles from a radioactive source. A circular fluorescent zinc sulphide (ZnS) screen surrounded the foil to detect the particles via tiny flashes of light.

According to Thomson's model, the mass and positive charge should be evenly spread, so the heavy $\alpha$-particles should pass through with only minor deflections. However, the results were shocking:

Observations:

1. Most of the $\alpha$-particles passed through the gold foil undeflected.
2. A small fraction of the $\alpha$-particles was deflected by small angles.
3. A very few $\alpha$-particles ($\sim 1$ in 20,000) bounced back, i.e., were deflected by nearly $180^\circ$.

Conclusions:

1. Empty Space: Most of the space in the atom is empty (since most particles passed undeflected).
2. Nucleus: The positive charge is not spread throughout the atom. It is concentrated in a very small volume that repelled the positively charged $\alpha$-particles. Rutherford called this dense center the nucleus.
3. Size Comparison: The volume of the nucleus is negligibly small compared to the atom.

• Radius of atom $\approx 10^{-10}$ m
• Radius of nucleus $\approx 10^{-15}$ m
• Analogy: If the nucleus is a cricket ball, the radius of the atom would be about 5 km!

Rutherford's Nuclear Model

Based on these conclusions, Rutherford proposed a new model:

1. The positive charge and most of the mass are densely concentrated in the extremely small nucleus.
2. Electrons move around the nucleus with very high speed in circular paths called orbits (resembling the solar system).
3. Electrons and the nucleus are held together by electrostatic forces of attraction.

[Exception Alert / Drawback] While Rutherford's model successfully explained the nucleus, it faced issues when classical mechanics was applied. In a solar system, gravitational force is $G \frac{m_1 m_2}{r^2}$. But applying classical mechanics to charged particles in orbits raised questions about the stability of the atom, which led to the need for Bohr's model (discussed in later sections).

Atomic Number, Mass Number, Isobars, and Isotopes

Atomic Number and Mass Number

The positive charge of the nucleus is due entirely to protons. To maintain electrical neutrality, a neutral atom must have an equal number of electrons and protons.

Atomic Number ($Z$): $$Z = \text{number of protons in the nucleus}$$ $$Z = \text{number of electrons in a neutral atom}$$

Mass Number ($A$): Protons and neutrons are collectively known as nucleons. The mass of the atom is primarily due to these nucleons. $$A = \text{number of protons (Z)} + \text{number of neutrons (n)}$$

Isobars and Isotopes

The composition of an atom is represented by the notation: $_{Z}^{A}\text{X}$ where X is the element symbol, A is the mass number (superscript), and Z is the atomic number (subscript).

Isobars: Atoms with the same mass number ($A$) but different atomic number ($Z$). Example: $_{6}^{14}\text{C}$ and $_{7}^{14}\text{N}$

Isotopes: Atoms with identical atomic number ($Z$) but different mass number ($A$). The difference in mass is due to a different number of neutrons.

[NEET Important] Isotopes of Hydrogen:

1. Protium ($_{1}^{1}\text{H}$): 1 proton, 0 neutrons (99.985% abundance).
2. Deuterium ($_{1}^{2}\text{D}$): 1 proton, 1 neutron (0.015% abundance).
3. Tritium ($_{1}^{3}\text{T}$): 1 proton, 2 neutrons (trace amounts).

Other common isotopes:

• Carbon: $_{6}^{12}\text{C}$, $_{6}^{13}\text{C}$, $_{6}^{14}\text{C}$
• Chlorine: $_{17}^{35}\text{Cl}$, $_{17}^{37}\text{Cl}$

Key Point: Chemical properties of atoms are controlled by the number of electrons (which depends on protons). Since isotopes have the same number of protons and electrons, all isotopes of a given element show the same chemical behaviour.

Memory Capsule

Quick Revision Formula Sheet:

• Quantization of Charge: $q = n e$ (Millikan)
• Atomic Number: $Z = p = e$ (for neutral atoms)
• Mass Number: $A = Z + n$ (Nucleons)
• Number of Neutrons: $n = A - Z$
• Gravitational Force: $F = G \frac{m_1 m_2}{r^2}$

Key Dimensions to Memorize:

• Radius of an atom: $\sim 10^{-10}$ m
• Radius of a nucleus: $\sim 10^{-15}$ m
• Wavelength of X-rays: $\sim 0.1$ nm
• Thickness of Gold Foil: $\sim 100$ nm

Radiation Penetrating Power: $\alpha$-rays $<$ $\beta$-rays $<$ $\gamma$-rays (Ratio roughly $1 : 100 : 1000$)

Isotopes vs Isobars Mnemonic:

• Isotopes: Top number ($A$) is different, bottom number ($Z$) is same. (Same element).
• Isobars: Bar (weight/mass $A$) is same, bottom number ($Z$) is different. (Different elements).

Example 1: Calculating Subatomic Particles (Book Problem)

Calculate the number of protons, neutrons and electrons in $_{35}^{80}\text{Br}$.

Solution:

1. Identify the notation: The species is represented as $_{Z}^{A}\text{X}$.

• Atomic number ($Z$) = 35
• Mass number ($A$) = 80

1. Check for charge: The species is neutral (no charge indicated).

2. Calculate protons and electrons:

• Number of protons = $Z = 35$
• Number of electrons = $Z = 35$ (since it is neutral)

1. Calculate neutrons:

• Number of neutrons ($n$) = $A - Z$
• $n = 80 - 35 = 45$

Takeaway: Always verify if the species is neutral or an ion before equating electrons to protons.

Example 2: Identifying an Ion (Book Problem)

The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.

Solution:

1. Identify the element:

• The atomic number ($Z$) is equal to the number of protons = 16.
• The element with $Z = 16$ is Sulphur (S).

1. Calculate the mass number ($A$):

• $A = \text{number of protons} + \text{number of neutrons}$
• $A = 16 + 16 = 32$

1. Determine the charge:

• The species is not neutral because protons (16) $\neq$ electrons (18).
• It has an excess of electrons, making it an anion.
• Charge = $\text{Protons} - \text{Electrons} = 16 - 18 = -2$.

1. Write the symbol:

• Combining the element symbol, $Z$, $A$, and charge: $_{16}^{32}\text{S}^{2-}$.

Takeaway: The charge of an ion is always calculated as (Protons - Electrons). A negative result means it's an anion.

Example 3: Millikan's Charge Quantization

In Millikan's oil drop experiment, an oil drop is found to carry a charge of $-4.806528 \times 10^{-19}$ C. Calculate the number of excess electrons on the oil drop.

Solution:

1. Identify given values:

• Total charge on drop, $q = -4.806528 \times 10^{-19}$ C
• Charge of one electron, $e = -1.602176 \times 10^{-19}$ C (from Table 2.1)

1. Apply quantization formula:

• $q = n e$

1. Solve for $n$:

• $n = \frac{q}{e}$
• $n = \frac{-4.806528 \times 10^{-19}}{-1.602176 \times 10^{-19}}$
• $n = 3$

Takeaway: The number of electrons ($n$) must always be an integer. If your calculation yields a fraction, recheck your math or the given values.

Example 4: Comparing Atom and Nucleus Size

Based on Rutherford's estimates, the radius of an atom is $10^{-10}$ m and the radius of its nucleus is $10^{-15}$ m. Calculate the ratio of the volume of the atom to the volume of the nucleus, assuming both are spherical.

Solution:

1. Identify formulas:

• Volume of a sphere, $V = \frac{4}{3} \pi r^3$

1. Set up the ratio:

• $\frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \frac{\frac{4}{3} \pi (r_{\text{atom}})^3}{\frac{4}{3} \pi (r_{\text{nucleus}})^3}$
• $\frac{V_{\text{atom}}}{V_{\text{nucleus}}} = \left( \frac{r_{\text{atom}}}{r_{\text{nucleus}}} \right)^3$

1. Substitute values:

• Ratio = $\left( \frac{10^{-10}}{10^{-15}} \right)^3$
• Ratio = $(10^5)^3 = 10^{15}$

Takeaway: The atom is $10^{15}$ times larger in volume than its nucleus, mathematically proving Rutherford's conclusion that most of the atom is empty space.

Example 5: Identifying Isotopes and Isobars

Given the following species: $_{18}^{40}\text{Ar}$, $_{19}^{40}\text{K}$, $_{20}^{40}\text{Ca}$, and $_{6}^{12}\text{C}$, $_{6}^{14}\text{C}$. Classify them into isobars and isotopes.

Solution:

1. Define the terms:

• Isobars: Same mass number ($A$), different atomic number ($Z$).
• Isotopes: Same atomic number ($Z$), different mass number ($A$).

1. Analyze the first group ($_{18}^{40}\text{Ar}$, $_{19}^{40}\text{K}$, $_{20}^{40}\text{Ca}$):

• Mass numbers ($A$) are all 40.
• Atomic numbers ($Z$) are 18, 19, and 20 (different).
• Therefore, they are Isobars.

1. Analyze the second group ($_{6}^{12}\text{C}$, $_{6}^{14}\text{C}$):

• Atomic numbers ($Z$) are both 6.
• Mass numbers ($A$) are 12 and 14 (different).
• Therefore, they are Isotopes.

Takeaway: Isobars belong to different chemical elements, while isotopes belong to the same chemical element.

Example 6: Composition of an Alpha Particle

Rutherford concluded that $\alpha$-particles are helium nuclei. Calculate the absolute charge and approximate mass (in u) of an $\alpha$-particle using Table 2.1.

Solution:

1. Identify the composition:

• A helium atom has 2 protons and 2 electrons.
• A helium nucleus ($\alpha$-particle) has 2 protons and 2 neutrons (no electrons).

1. Calculate absolute charge:

• Charge is due to 2 protons.
• Charge = $2 \times (+1.602176 \times 10^{-19} \text{ C})$
• Charge = $+3.204352 \times 10^{-19} \text{ C}$

1. Calculate approximate mass:

• Mass = (2 $\times$ approx mass of proton) + (2 $\times$ approx mass of neutron)
• Mass = $(2 \times 1 \text{ u}) + (2 \times 1 \text{ u}) = 4 \text{ u}$

Takeaway: $\alpha$-particles carry two units of positive charge and four units of atomic mass, making them relatively heavy and highly charged compared to $\beta$-particles.

Example 7: Neutrons in Hydrogen Isotopes

Calculate the total number of neutrons present in one molecule of heavy water (Deuterium oxide, $\text{D}_2\text{O}$). Assume the oxygen isotope is $_{8}^{16}\text{O}$.

Solution:

1. Analyze Deuterium (D):

• Deuterium is $_{1}^{2}\text{D}$.
• Number of neutrons in one D atom = $A - Z = 2 - 1 = 1$.
• Since there are two D atoms in $\text{D}_2\text{O}$, neutrons from D = $2 \times 1 = 2$.

1. Analyze Oxygen (O):

• Oxygen is $_{8}^{16}\text{O}$.
• Number of neutrons in one O atom = $A - Z = 16 - 8 = 8$.

1. Calculate total neutrons:

• Total neutrons = Neutrons from D + Neutrons from O
• Total neutrons = $2 + 8 = 10$.

Takeaway: Unlike normal water ($\text{H}_2\text{O}$) which has 8 neutrons (since Protium has 0 neutrons), heavy water has 10 neutrons per molecule.

Example 8: Rutherford's Scattering Fraction

In a hypothetical replication of Rutherford's experiment, $10^6$ $\alpha$-particles are fired at a gold foil. Based on Rutherford's historical observation ratio, approximately how many particles would you expect to bounce back (deflected by nearly $180^\circ$)?

Solution:

1. Identify the historical ratio:

• According to NCERT, a very few $\alpha$-particles, approximately 1 in 20,000, bounced back.
• Fraction = $\frac{1}{20000}$

1. Calculate expected number:

• Expected number = Total particles $\times$ Fraction
• Expected number = $10^6 \times \frac{1}{20000}$
• Expected number = $\frac{1000000}{20000} = \frac{100}{2} = 50$

Takeaway: The extreme rarity of $180^\circ$ deflections (only 50 out of a million) highlights just how incredibly small the nucleus is compared to the total volume of the atom.

Example 9: Nucleons in an Isotope

An atom of an element contains 17 protons and 20 neutrons. Identify the element, write its isotopic symbol, and state its mass number.

Solution:

1. Identify the element:

• Number of protons ($Z$) = 17.
• The element with atomic number 17 is Chlorine (Cl).

1. Calculate mass number ($A$):

• Mass number ($A$) = Total number of nucleons
• $A = \text{protons} + \text{neutrons} = 17 + 20 = 37$.

1. Write the isotopic symbol:

• Format: $_{Z}^{A}\text{X}$
• Symbol: $_{17}^{37}\text{Cl}$

Takeaway: Chlorine naturally exists as two major isotopes: $_{17}^{35}\text{Cl}$ and $_{17}^{37}\text{Cl}$. Both have identical chemical properties because they have the same number of electrons (17).

Example 10: Classical Mechanics Formula Application

According to the drawbacks of Rutherford's model, the solar system analogy relies on classical mechanics. Write the expression for the gravitational force between a planet of mass $m_1$ and the sun of mass $m_2$ separated by distance $r$.

Solution:

1. Identify the formula from the text:

• The gravitational force expression given in the text is based on Newton's laws of motion.

1. Write the expression:

• Force = $G \frac{m_1 m_2}{r^2}$
• Where $G$ is the gravitational constant.

Takeaway: While this formula perfectly describes macroscopic planetary orbits, applying similar classical mechanics to microscopic charged particles (electrons orbiting a nucleus) fails to explain atomic stability.