Bohr's model and physics of the atom

Flashcards for topic Bohr's model and physics of the atom

Intermediate47 cardsphysics

Preview Cards

Card 1

Front

In Rutherford's alpha particle scattering experiment, approximately what fraction of alpha particles were deflected by more than 90°, and what did this reveal about atomic structure?

Back

  • About 1 in 8000 alpha particles was deflected by more than 90° (turned back)
  • This revealed:
    • The existence of a very heavy, positively charged nucleus
    • The nucleus contains virtually all mass of the atom
    • The nucleus occupies only about 10^-15 of the atom's volume
    • The atom must contain mostly empty space
  • This observation led to the nuclear model of the atom, replacing Thomson's model
Card 2

Front

Explain how Bohr's model accounts for the discrete spectral lines observed in hydrogen's emission spectrum, including the mathematical relationship for wavelengths.

Back

  • Spectral lines are produced when electrons transition between energy states

  • When an electron jumps from higher state m to lower state n, it emits a photon with:

    • Energy: EmEn=hc/λE_m - E_n = hc/λ
    • Wavelength: 1λ=RZ2(1n21m2)\frac{1}{λ} = RZ^2(\frac{1}{n^2} - \frac{1}{m^2})
  • Rydberg constant: R=me48ε02h3c1.097×107m1R = \frac{me^4}{8\varepsilon_0^2h^3c} ≈ 1.097 × 10^7 m^{-1}

  • Spectral series:

    • Lyman series (n=1): ultraviolet region
    • Balmer series (n=2): visible region
    • Paschen series (n=3): infrared region
  • The discrete nature comes from quantized energy levels, explaining why hydrogen emits only specific wavelengths

Card 3

Front

What are Bohr's four postulates for his atomic model, and how do they specifically address the shortcomings of classical physics?

Back

Bohr's postulates:

  1. Electrons revolve around the nucleus in circular orbits

    • Addresses: Provides stability mechanism classical physics couldn't explain
  2. Only certain special orbits with discrete radii are allowed ("stationary orbits")

    • Addresses: Explains discrete spectra and prevents radiation collapse
  3. Electrons don't radiate energy while in stationary orbits

    • Addresses: Directly contradicts Maxwell's theory to solve the stability problem
  4. Angular momentum is quantized: mvr=nh2πmvr = n\frac{h}{2\pi} (where n is a positive integer)

    • Addresses: Provides mathematical basis for allowed orbits and energy levels

These postulates represent a revolutionary break from classical physics by introducing quantization.

Card 4

Front

Calculate the excitation potential, ionization potential, and binding energy for the hydrogen atom, and explain the relationship between these quantities.

Back

  • Ionization potential: 13.6 V

    • Energy needed to remove electron from ground state to infinity
    • Corresponds to transition from n=1 to n=∞
    • Calculated from: Ei=E1=13.6E_i = |E_1| = 13.6 eV
  • Excitation potential (to first excited state): 10.2 V

    • Energy needed to excite electron from ground state to first excited state
    • Corresponds to transition from n=1 to n=2
    • Calculated from: Eex=E1E2=13.63.4=10.2E_{ex} = E_1 - E_2 = 13.6 - 3.4 = 10.2 eV
  • Binding energy: 13.6 eV

    • Energy released when electron and proton form hydrogen atom
    • Equal to ionization energy for ground state
    • Represents the stability of the atom

Relationship: Ionization potential = Binding energy > Excitation potential

Card 5

Front

Why did Rutherford's experiment with alpha particles necessitate the replacement of Thomson's model with the nuclear model, and why was this inconsistent with classical electromagnetic theory?

Card image

Back

Rutherford's experiment necessitated replacing Thomson's model because:

  1. Thomson's model predicted all alpha particles would pass through with minimal deflection since positive charge was diffusely spread throughout the atom
  2. Large-angle deflections observed required concentrated positive charge and mass in a tiny nucleus
  3. Most alpha particles passing straight through indicated atoms were mostly empty space

Inconsistency with classical electromagnetic theory:

  • According to classical electromagnetism, orbiting electrons should continuously emit radiation
  • This radiation would cause electrons to lose energy and spiral into the nucleus
  • Atoms would collapse in approximately 10⁻¹⁰ seconds
  • This contradiction created a fundamental problem that was later resolved by Bohr's quantum model with its stationary orbits and quantized angular momentum

This fundamental contradiction ultimately led to the development of quantum mechanics to explain atomic stability.

Card image
Card 6

Front

How can you identify which series a particular hydrogen spectral line belongs to, and calculate its wavelength?

Card image

Back

To identify and calculate hydrogen spectral lines:

  1. Series identification: Determine by wavelength region and corresponding transition:

    • Ultraviolet region (90-121.6 nm): Lyman series (transitions to n=1)
    • Visible region (365-656.3 nm): Balmer series (transitions to n=2)
    • Infrared region (820-1875 nm): Paschen series (transitions to n=3)
    • Longer infrared: Brackett (n=4), Pfund (n=5), etc.
  2. Wavelength calculation using Rydberg formula: 1λ=R(1n21m2)\frac{1}{\lambda} = R\left(\frac{1}{n^2} - \frac{1}{m^2}\right)

    Where:

    • R = 1.097 × 10⁷ m⁻¹ (Rydberg constant)
    • n = lower energy level (fixed for a series)
    • m = higher energy level (m > n)
  3. Line identification within a series:

    • First line (m = n+1): Longest wavelength in series
    • Series limit (m = ∞): Shortest wavelength in series
    • Spacing between lines decreases as wavelength decreases

Example: For Balmer series first line (n=2, m=3): 1λ=1.097×107(1419)=1.097×107×536=1.52×106\frac{1}{\lambda} = 1.097 × 10⁷ \left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 × 10⁷ × \frac{5}{36} = 1.52 × 10⁶ λ=6.56×107 m=656 nm\lambda = 6.56 × 10^{-7} \text{ m} = 656 \text{ nm}

Card image
Card 7

Front

What factors determine the degree of collimation (directional precision) in a laser beam?

Back

Factors determining laser beam collimation:

  1. Cavity geometry:

    • Length-to-width ratio of resonator
    • Mirror curvature and alignment
    • Larger ratio results in narrower beam divergence
  2. Diffraction effects:

    • Governed by λ/D (wavelength divided by aperture diameter)
    • Fundamental limit to collimation
    • Minimum divergence angle θ ≈ λ/D
  3. Mode structure:

    • TEM₀₀ (fundamental mode) has best collimation
    • Higher-order modes increase divergence
  4. Gain medium properties:

    • Homogeneity of refractive index
    • Thermal gradients affecting beam path
  5. Output coupler design:

    • Transmission percentage
    • Surface quality (λ/10 or better)

A perfectly collimated beam is physically impossible due to diffraction, but laser beams approach this ideal, allowing them to travel to the moon and back with minimal spreading.

Card 8

Front

Explain how stimulated emission leads to light amplification in a laser using rate equations.

Back

Light amplification via stimulated emission:

Rate equation approach:

  1. Define N₁ and N₂ as populations of lower and upper energy states
  2. Rate of stimulated emission = B₂₁·N₂·ρ(ν)
    • B₂₁ is Einstein B coefficient
    • ρ(ν) is energy density at frequency ν
  3. Rate of absorption = B₁₂·N₁·ρ(ν)
  4. Net rate of photon generation = ρ(ν)·(B₂₁·N₂ - B₁₂·N₁)

For amplification:

  • Must have B₂₁·N₂ > B₁₂·N₁
  • Since B₂₁ = B₁₂ (Einstein relation), need N₂ > N₁
  • Population inversion is necessary condition

As light passes through medium:

  • Intensity grows exponentially: I(z) = I₀·e^(γz)
  • Gain coefficient γ = (N₂-N₁)·B₂₁·hν/c
  • Amplification occurs when γ > losses in system

Example: In He-Ne laser, when neon atoms in E₂ state outnumber those in E₁, each stimulated emission event adds a coherent photon, leading to exponential growth of the light wave.

Card 9

Front

What is the operating principle of a He-Ne laser and what specific energy level interactions make it work?

Card image

Back

A He-Ne laser operates through:

  1. Pumping mechanism: Electrons collide with helium atoms, exciting them from ground state (E=0) to metastable state (E₃=20.61 eV)

  2. Energy transfer: Excited helium atoms collide with neon atoms, transferring energy because:

    • He metastable state (20.61 eV) closely matches Ne excited state (E₂=20.66 eV)
    • He returns to ground state while Ne is excited to E₂
  3. Population inversion: This process creates more Ne atoms in E₂ than E₁ (18.70 eV)

  4. Stimulated emission: Ne atoms transition from E₂ to E₁, emitting 632.8 nm red laser light

  5. Depletion of lower level: Ne atoms rapidly decay from E₁ to ground state via spontaneous emission, maintaining population inversion

The energy difference E₂-E₁ = 1.96 eV corresponds precisely to the 632.8 nm wavelength of the characteristic red He-Ne laser light.

Card image
Card 10

Front

How does the He-Ne laser achieve and maintain population inversion, and why is the specific energy level structure of both gases critical?

Card image

Back

Population inversion in He-Ne laser:

  1. Critical energy match: Helium's metastable state (20.61 eV) nearly resonates with neon's upper laser level (20.66 eV), enabling efficient energy transfer through collisions

  2. Selective pumping mechanism:

    • Electric discharge excites helium to metastable state
    • He-Ne collisions preferentially populate Ne's E₂ level (20.66 eV)
    • This process continuously pumps atoms to E₂, bypassing E₁
  3. Fast lower level depletion:

    • Ne's E₁ level (18.70 eV) has short lifetime (~10 ns)
    • Atoms quickly decay from E₁ to ground state
    • This prevents E₁ from accumulating population
  4. Metastability advantage:

    • He's metastable state has long lifetime (~1 ms), acting as energy reservoir
    • This allows continuous, efficient pumping of Ne atoms
  5. Gas mixture optimization:

    • 90% He / 10% Ne ratio maximizes collision probability
    • Low pressure ensures excited atoms can travel without premature collisions

The gas mixture essentially creates a four-level laser system that inherently maintains more atoms in E₂ than E₁, sustaining population inversion.

Card image

Showing 10 of 47 cards. Add this deck to your collection to see all cards.