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Structure of Atom - Atomic Models (Thomson and Rutherford)

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Thomson's Atomic Model (1898): Proposed that an atom consists of a uniform sphere of positive charge (radius approximately 10−10m10^{-10} m) in which electrons are embedded. It is also known as the 'plum pudding' or 'watermelon' model.

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Rutherford's α\alpha-particle Scattering Experiment: High energy α\alpha-particles (24He2+^4_2He^{2+}) were directed at a thin gold foil (100nm100 nm thickness). Observations showed most particles passed undeflected, some were deflected by small angles, and very few (11 in 20,00020,000) were deflected by nearly 180∘180^\circ.

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Rutherford's Nuclear Model: Concluded that the positive charge and most of the mass of the atom are densely concentrated in an extremely small region called the nucleus.

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Atomic Dimensions: The radius of the atom is about 10−10m10^{-10} m, while the radius of the nucleus is about 10−15m10^{-15} m. This implies the volume of the atom is about 101510^{15} times the volume of the nucleus.

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Stability Drawback: According to Maxwell’s electromagnetic theory, an accelerating electron (moving in a circular orbit) should continuously emit radiation. This loss of energy would cause the electron to spiral into the nucleus, making the atom unstable.

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Failure to explain Line Spectra: Rutherford's model could not explain the discrete frequencies of light emitted by atoms (atomic spectra).

📐Formulae

R=R0A1/3 (where R0≈1.3×10−15m and A is the mass number)R = R_0 A^{1/3} \text{ (where } R_0 \approx 1.3 \times 10^{-15} m \text{ and } A \text{ is the mass number)}

r0=14πϵ02Ze2K (Distance of closest approach)r_0 = \frac{1}{4 \pi \epsilon_0} \frac{2 Z e^2}{K} \text{ (Distance of closest approach)}

F=14πϵ0q1q2r2 (Coulombic force between nucleus and electron)F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \text{ (Coulombic force between nucleus and electron)}

V=43πr3 (Volume of the atom or nucleus)V = \frac{4}{3} \pi r^3 \text{ (Volume of the atom or nucleus)}

💡Examples

Problem 1:

Calculate the ratio of the volume of a gold atom to the volume of its nucleus, assuming the radius of the atom is 10−10m10^{-10} m and the radius of the nucleus is 10−15m10^{-15} m.

Solution:

The volume of a sphere is given by V=43πr3V = \frac{4}{3} \pi r^3. The ratio is calculated as: VatomVnucleus=43πratom343πrnucleus3=(10−10m10−15m)3=(105)3=1015\frac{V_{atom}}{V_{nucleus}} = \frac{\frac{4}{3} \pi r_{atom}^3}{\frac{4}{3} \pi r_{nucleus}^3} = \left( \frac{10^{-10} m}{10^{-15} m} \right)^3 = (10^5)^3 = 10^{15}.

Explanation:

This result shows that the volume of the atom is 101510^{15} times larger than the nucleus, supporting Rutherford's conclusion that most of the space within an atom is empty.

Problem 2:

An α\alpha-particle with kinetic energy KK approaches a gold nucleus (Z=79Z=79) head-on. Write the expression for the distance of closest approach (r0r_0).

Solution:

At the distance of closest approach, the initial kinetic energy (KK) of the α\alpha-particle is entirely converted into electrostatic potential energy. Thus, K=14πϵ0q1q2r0K = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_0}. For an α\alpha-particle, q1=2eq_1 = 2e, and for a nucleus, q2=Zeq_2 = Ze. Therefore, r0=14πϵ02Ze2Kr_0 = \frac{1}{4 \pi \epsilon_0} \frac{2 Z e^2}{K}.

Explanation:

The distance of closest approach provides an upper limit for the size of the nucleus, as the particle stops and reverses direction due to electrostatic repulsion.

Atomic Models (Thomson and Rutherford) Revision - Class 11 Chemistry CBSE