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Atom, Origin of Spectra: Bohr's Theory of Hydrogen Atom

ICSE · Class 12 · Physics

Flashcards for Atom, Origin of Spectra: Bohr's Theory of Hydrogen Atom — ICSE Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

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Card 1Thomson model

State the discovery and main idea of J.J. Thomson's atomic model.

Answer

Electron was discovered by J.J. Thomson in 1897. Thomson presented the plum pudding atomic model. In this model, the atom is neutral and negatively charged electrons are embedded in a uniformly distri

Card 2Rutherford model

Why did the alpha-particle scattering experiment suggest that the atom is mostly hollow?

Answer

Most alpha-particles passed straight through the gold foil without deflection. This means most of the atom is empty space, so the positive charge and most of the mass must occupy a very small central

Card 3Rutherford experiment

A thin gold foil is used in alpha-particle scattering. Why is gold foil chosen?

Answer

Gold foil can be beaten very thin, which helps ensure single collision of alpha-particles. Its heavy nucleus produces large deflection of alpha-particles.

Card 4Scattering laws

State the relation between foil thickness and the number of scattered alpha-particles for a fixed scattering direction.

Answer

Thickness-scattering relation: N/t = constant, i.e., N1/N2 = t1/t2, where N is the number of scattered alpha-particles and t is the foil thickness.

Card 5Scattering laws

If the foil thickness is doubled, what happens to the number of alpha-particles scattered at the same angle?

Answer

Using N/t = constant, N is directly proportional to t. Step 1: N1/N2 = t1/t2 Step 2: If t2 = 2t1, then N2 = 2N1 Answer: The number of scattered alpha-particles doubles.

Card 6Scattering laws

State the angular dependence of alpha-particle scattering count.

Answer

Scattering angle dependence: N ∝ 1/sin^4(theta/2), where N is the number scattered at angle theta.

Card 7Nucleus size

At ordinary conditions, only about 1 out of 20000 alpha-particles is scattered beyond 90°. What does this show?

Answer

This shows the nucleus is extremely small compared to the atom. Only a very small fraction of alpha-particles comes very close to the nucleus and gets strongly deflected.

Card 8Closest approach

State the formula for distance of closest approach of an alpha-particle to the nucleus.

Answer

Distance of closest approach: r0 = (1/4πε0)(2Ze^2/K), where Z is atomic number, e is electron charge, and K is initial kinetic energy of the alpha-particle.

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