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Atoms - Alpha-particle Scattering Experiment

Grade 12CBSEPhysics

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

🔑Concepts

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The Geiger-Marsden experiment involved bombarding a thin gold foil (2.1×10−7 m2.1 \times 10^{-7} \text{ m} thick) with high-energy α\alpha-particles (24He^4_2\text{He} nuclei) from a radioactive source like 83214Bi^{214}_{83}\text{Bi}.

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Observations: Most α\alpha-particles passed through undeflected; about 0.14%0.14\% were deflected by more than 1∘1^{\circ}; and roughly 11 in 80008000 were deflected by more than 90∘90^{\circ}.

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Inferences: The atom has a tiny, dense, positively charged core called the nucleus. Most of the atom's volume is empty space.

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Distance of Closest Approach (r0r_0): When an α\alpha-particle is directed head-on toward a nucleus, it comes to rest momentarily at r0r_0 where its initial Kinetic Energy (KK) equals the Electrostatic Potential Energy.

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Impact Parameter (bb): The perpendicular distance of the initial velocity vector of the α\alpha-particle from the center of the nucleus. For large bb, the scattering angle θ\theta is small. For b=0b = 0, the particle is scattered back at θ=180∘\theta = 180^{\circ}.

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Trajetory: The trajectory of the α\alpha-particle depends on the impact parameter bb. Under the influence of the Coulombic force, the path is hyperbolic.

📐Formulae

K=12mv2K = \frac{1}{2} m v^2

r0=14πϵ02Ze2Kr_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}

b=14πϵ0Ze2cot⁡(θ/2)Kb = \frac{1}{4\pi\epsilon_0} \frac{Ze^2 \cot(\theta/2)}{K}

N(θ)∝1sin⁡4(θ/2)N(\theta) \propto \frac{1}{\sin^4(\theta/2)}

💡Examples

Problem 1:

Calculate the distance of closest approach for an α\alpha-particle of energy 7.7 MeV7.7 \text{ MeV} approaching a gold nucleus (Z=79Z = 79).

Solution:

Given K=7.7 MeV=7.7×106×1.6×10−19 JK = 7.7 \text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19} \text{ J}, Z=79Z = 79, and 14πϵ0=9×109 Nm2C−2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}. Using the formula r0=14πϵ02Ze2Kr_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}: r0=9×109×2×79×(1.6×10−19)27.7×1.6×10−13r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 1.6 \times 10^{-13}} r0≈3.0×10−14 mr_0 \approx 3.0 \times 10^{-14} \text{ m}

Explanation:

The distance of closest approach is found by equating the initial kinetic energy of the α\alpha-particle to the electrical potential energy at distance r0r_0 from the nucleus.

Problem 2:

An α\alpha-particle is scattered through an angle of θ=60∘\theta = 60^{\circ} when it passes through a gold foil. What is the impact parameter if the kinetic energy is 5 MeV5 \text{ MeV}?

Solution:

Using the formula b=14πϵ0Ze2cot⁡(θ/2)Kb = \frac{1}{4\pi\epsilon_0} \frac{Ze^2 \cot(\theta/2)}{K}. Here θ/2=30∘\theta/2 = 30^{\circ} and cot⁡(30∘)=3\cot(30^{\circ}) = \sqrt{3}. b=9×109×79×(1.6×10−19)2×35×1.6×10−13b = \frac{9 \times 10^9 \times 79 \times (1.6 \times 10^{-19})^2 \times \sqrt{3}}{5 \times 1.6 \times 10^{-13}} b≈3.94×10−14 mb \approx 3.94 \times 10^{-14} \text{ m}

Explanation:

The impact parameter determines the scattering angle; a larger impact parameter results in a smaller deflection.

Alpha-particle Scattering Experiment Revision - Class 12 Physics CBSE