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This deck focuses on The Bohr Model Of Atomic Structure, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.
Study The Bohr Model Of Atomic Structure in AP Physics 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is a0 in the Bohr Model formula for orbit radius?
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The Bohr radius, approximately 5.29×10−11 m. The radius of the first electron orbit in hydrogen.
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This deck focuses on The Bohr Model Of Atomic Structure, giving you a quick way to review the definitions, rules, and examples that matter most for AP Physics 2.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: The Bohr radius, approximately 5.29×10−11 m. The radius of the first electron orbit in hydrogen.
Answer: −13.6 eV. This is the ionization energy of hydrogen with opposite sign.
Answer: Approximately 2.18×106 m/s. Calculated from v1=ħke2 for n=1.
Answer: Bohr Model includes quantized orbits. Classical physics predicted continuous energy and orbital decay.
Answer: 2.55 eV. Energy difference: E4−E2=−0.85−(−3.4)=2.55 eV.
Answer: Quantization. Energy and angular momentum can only have discrete values.
Answer: Lyman series. UV region transitions from higher levels to ground state.
Answer: Electrons orbit the nucleus in fixed energy levels. This quantization prevents classical electromagnetic radiation losses.
Answer: Hydrogen. Its simple one-electron structure made calculations feasible.
Answer: Balmer series. Visible region transitions producing the familiar hydrogen spectrum.
Answer: Angular momentum is quantized. Only integer multiples of 2πh are allowed.
Answer: vn=ħnke2. Where e is electron charge and ħ is reduced Planck's constant.
Answer: En=−n213.6 eV. Energy decreases as n2 increases, with ground state at n=1.
Answer: n=3 to n=2 transition. Smallest energy difference produces longest wavelength photon.
Answer: Paschen series. Infrared region transitions from higher levels to n=3.
Answer: 5.29×10−11 m. Using r1=12×a0 where a0=5.29×10−11 m.
Answer: Fine structure and Zeeman effect. These require relativistic effects and spin considerations.
Answer: Paschen series. Infrared region transitions from higher levels to n=3.
Answer: f=hEi−Ef. Energy difference between levels divided by Planck's constant.
Answer: It only accurately describes hydrogen-like atoms. Multi-electron atoms involve complex electron-electron interactions.
Answer: Planck's constant. Fundamental constant linking energy and frequency: 6.626×10−34 J·s.
Answer: An electron transition between energy levels. Electrons jump instantly between allowed energy states.
Answer: vn=ℏnke2. Where e is electron charge and ℏ is reduced Planck's constant.
Answer: Electrons orbit the nucleus in fixed energy levels. This quantization prevents classical electromagnetic radiation losses.
Answer: Hydrogen emission spectrum. Discrete spectral lines match calculated energy differences.
Answer: Radius increases as n2. Higher energy levels correspond to larger orbital radii.
Answer: Quantized energy levels prevent spiraling into nucleus. Only specific orbits are allowed, preventing energy loss.
Answer: Bohr Model includes quantized orbits. Classical physics predicted continuous energy and orbital decay.
Answer: λ1=RH×(n121−n221). Rydberg formula relating wavelength to quantum number transitions.
Answer: It only accurately describes hydrogen-like atoms. Multi-electron atoms involve complex electron-electron interactions.
Answer: Used to calculate wavelengths of spectral lines. Essential for predicting hydrogen's emission and absorption spectra.
Answer: rn=n2×a0. Where n is the principal quantum number and a0 is the Bohr radius.
Answer: 5.29×10−11 m. Using r1=12×a0 where a0=5.29×10−11 m.
Answer: Rydberg constant. Empirical constant: 1.097×107 m⁻¹ for hydrogen.
Answer: Approximately 4.57×1014 Hz. Using f=hE3−E2=h1.89 eV.
Answer: Transitions between energy levels. Each line corresponds to a specific energy level change.
Answer: f=hEi−Ef. Energy difference between levels divided by Planck's constant.
Answer: Coulomb's constant. Electrostatic force constant: 8.99×109 N·m²/C².
Answer: Planck's constant. Fundamental constant linking energy and frequency: 6.626×10−34 J·s.
Answer: Electrons have fixed energies corresponding to orbits. No intermediate energies are allowed between levels.
Answer: The Bohr radius, approximately 5.29×10−11 m. The radius of the first electron orbit in hydrogen.
Answer: Planck's constant, h. It appears in energy quantization and angular momentum conditions.
Answer: L=n×2πh. Where ħ=2πh is reduced Planck's constant.
Answer: Electrons move in circular orbits. Classical planetary model with quantization constraints.
Answer: Quantization. Energy and angular momentum can only have discrete values.
Answer: Balmer series. Visible region transitions producing the familiar hydrogen spectrum.
Answer: Accurate prediction of hydrogen spectral lines. Quantized energy levels perfectly matched observed spectral lines.
Answer: Angular momentum is quantized. Only integer multiples of 2πh are allowed.
Answer: Rydberg constant. Empirical constant: 1.097×107 m⁻¹ for hydrogen.
Answer: Radius increases as n2. Higher energy levels correspond to larger orbital radii.
Answer: Hydrogen. Its simple one-electron structure made calculations feasible.
Answer: An electron transition between energy levels. Electrons jump instantly between allowed energy states.
Answer: Planck's constant, h. It appears in energy quantization and angular momentum conditions.
Answer: λ1=RH×(n121−n221). Rydberg formula relating wavelength to quantum number transitions.
Answer: Used to calculate wavelengths of spectral lines. Essential for predicting hydrogen's emission and absorption spectra.
Answer: Hydrogen emission spectrum. Discrete spectral lines match calculated energy differences.
Answer: The principal quantum number. It determines the energy level and orbit radius of the electron.
Answer: −13.6 eV. This is the ionization energy of hydrogen with opposite sign.
Answer: 2.55 eV. Energy difference: E4−E2=−0.85−(−3.4)=2.55 eV.
Answer: Coulomb's constant. Electrostatic force constant: 8.99×109 N·m²/C².
Answer: Electrons move in circular orbits. Classical planetary model with quantization constraints.
Answer: Accurate prediction of hydrogen spectral lines. Quantized energy levels perfectly matched observed spectral lines.
Answer: rn=n2×a0. Where n is the principal quantum number and a0 is the Bohr radius.
Answer: Approximately 2.18×106 m/s. Calculated from v1=ħke2 for n=1.
Answer: Lyman series. UV region transitions from higher levels to ground state.
Answer: Electrons have fixed energies corresponding to orbits. No intermediate energies are allowed between levels.
Answer: Approximately 486 nm. Using Rydberg formula with n1=2, n2=4.
Answer: n=3 to n=2 transition. Smallest energy difference produces longest wavelength photon.
Answer: −3.4 eV. Using E2=−2213.6=−3.4 eV.
Answer: −3.4 eV. Using E2=−2213.6=−3.4 eV.
Answer: Fine structure and Zeeman effect. These require relativistic effects and spin considerations.
Answer: Approximately 486 nm. Using Rydberg formula with n1=2, n2=4.
Answer: Quantized energy levels prevent spiraling into nucleus. Only specific orbits are allowed, preventing energy loss.
Answer: L=n×2πh. Where ħ=2πh is reduced Planck's constant.
Answer: The principal quantum number. It determines the energy level and orbit radius of the electron.
Answer: En=−n213.6 eV. Energy decreases as n2 increases, with ground state at n=1.
Answer: Approximately 4.57×1014 Hz. Using f=hE3−E2=h1.89 eV.
Answer: Transitions between energy levels. Each line corresponds to a specific energy level change.