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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.