Final Exam Review | Source: Knight, Ch. 29.1–29.4
Difficulty: Intermediate | Prerequisites: Ch. 28.4–28.6 study notes (de Broglie wavelength, quantised energy levels, photon emission/absorption).
Chapter 28 showed that confining a particle quantises its energy. This chapter applies that idea to a real physical system: the hydrogen atom. Bohr's model treats the electron as orbiting the nucleus at specific allowed radii, each with a definite energy. Transitions between these levels produce the sharp spectral lines that spectroscopy measures, including the visible Balmer series that you can observe with a simple diffraction grating. This is where the abstract quantum rules become experimentally measurable predictions.
Hydrogen emits light at specific wavelengths described by the Balmer formula. Bohr explained this by proposing that electrons orbit at quantised radii r_n = n² a_B with quantised energies E_n = −13.60/n² eV, and that photons are emitted or absorbed when the electron jumps between orbits.
Spectroscopy
The study of light emitted or absorbed by atoms and molecules, used to identify elements and measure energy levels. Think of it as reading an element's fingerprint by looking at the colours of light it produces.
Balmer formula
An empirical equation that predicts the visible wavelengths of hydrogen: λ = 91.1 nm / (1/m² – 1/n²), where m and n are positive integers with n > m. For the visible series, m = 2.
Balmer series
The set of spectral lines produced when hydrogen electrons fall to the n = 2 level. The visible lines are at 656 nm (red), 486 nm (blue-green), 434 nm (violet), and 410 nm (violet).
Stationary state
A particular arrangement of electrons in allowed orbits within an atom. Each stationary state has a definite, well-defined energy labelled by quantum number n. In simple terms, it is a stable configuration the atom can sit in without radiating.
Bohr radius (a_B)
The radius of the smallest allowed orbit in hydrogen: a_B = 5.29 × 10⁻¹¹ m = 0.0529 nm. All other allowed radii are integer-squared multiples of this value.
Collisional excitation
The process by which a collision with another particle (electron, atom) transfers kinetic energy to an atom, bumping it to a higher energy level. Think of it as a billiard ball knock that pushes the electron to a higher orbit.
Alpha particle
A helium-4 nucleus (2 protons, 2 neutrons) ejected during radioactive decay. Mass = 6.64 × 10⁻²⁷ kg.
Hydrogen emission spectrum
Hydrogen gas in a discharge tube emits light at only specific wavelengths, not a continuous rainbow.
The visible lines (the Balmer series) are at 656 nm, 486 nm, 434 nm, and 410 nm.
The Balmer formula
Predicts hydrogen wavelengths: λ = 91.1 nm / (1/m² – 1/n²).
m = 1, 2, 3, ... and n can be any integer greater than m.
For the visible series, set m = 2 and let n = 3, 4, 5, 6, ... to reproduce 656, 486, 434, 410 nm.
Setting m = 1 gives the ultraviolet Lyman series; m = 3 gives the infrared Paschen series.
Alpha-particle speed from kinetic energy
A nucleus ejects an alpha particle with a known kinetic energy (often given in MeV).
To find the speed, convert MeV to joules: 1 MeV = 10⁶ eV = 1.6 × 10⁻¹³ J.
Then solve ½ mv² = KE for v, using m_alpha = 6.64 × 10⁻²⁷ kg.
Worked example: alpha particle with KE = 8.3 MeV
8.3 × 10⁶ eV × (1.6 × 10⁻¹⁹ J/eV) = 1.3 × 10⁻¹² J.
½ × 6.64 × 10⁻²⁷ × v² = 1.3 × 10⁻¹².
v = 2.0 × 10⁷ m/s (about 7% of the speed of light).
Key postulates
Electrons can exist only in certain allowed orbits (stationary states).
Each stationary state has a discrete energy E_n. These are labelled by quantum number n in order of increasing energy: E_1 < E_2 < E_3 < ...
The lowest energy state (n = 1) is the ground state. It is stable and the atom can remain there indefinitely.
Higher states (n = 2, 3, ...) are excited states.
Photon emission
An electron in an excited state jumps down to a lower state and emits a photon.
The photon's frequency: f = ΔE / h. Its wavelength: λ = c / f.
Photon absorption
An incoming photon is absorbed only if its energy exactly matches the gap between two allowed levels.
In a cold gas, absorption transitions start from the ground state (n = 1).
Collisional excitation
An approaching particle (electron or atom) collides with the atom, transfers kinetic energy, and bumps the electron to a higher orbit.
The approaching particle loses kinetic energy equal to the gap it excited.
The excited atom soon drops back down, emitting photons.
Worked example: atom with E_1 = 0.0 eV, E_2 = 2.0 eV, E_3 = 5.0 eV
Absorption from n = 1 to n = 2: ΔE = 2.0 eV. f = 2.0 / 4.141 × 10⁻¹⁵ = 4.8 × 10¹⁴ Hz. λ = 620 nm.
Absorption from n = 1 to n = 3: ΔE = 5.0 eV. f = 1.2 × 10¹⁵ Hz. λ = 250 nm.
Emission can also occur from n = 3 to n = 2: ΔE = 3.0 eV. λ ≈ 415 nm.
The absorption spectrum has two lines (from n = 1 only). The emission spectrum has three.
Allowed orbital radii
The electron's orbit must satisfy r_n = n² a_B, where a_B = 0.0529 nm is the Bohr radius.
n = 1: r = 0.053 nm. n = 2: r = 0.212 nm. n = 3: r = 0.476 nm. n = 4: r = 0.847 nm.
The hydrogen atom can exist only if the orbital radius takes one of these values.
Orbital speed
The condition for a stable circular orbit: v² = Ke² / (mr), where K is the Coulomb constant, e is the electron charge, and m is the electron mass.
Higher orbits have lower speeds: n = 1 gives v = 2.18 × 10⁶ m/s; n = 4 gives v = 0.55 × 10⁶ m/s.
Allowed energies
Total energy (KE + PE): E_n = –E_1 / n² = −13.60 eV / n².
Energies are negative because the electron is bound to the nucleus. Zero energy means the electron is free.
n = 1: E = −13.60 eV. n = 2: E = −3.40 eV. n = 3: E = −1.51 eV. n = 4: E = −0.85 eV.
Summary table
n | r_n (nm) | v_n (m/s) | E_n (eV) |
|---|---|---|---|
1 | 0.053 | 2.18 × 10⁶ | −13.60 |
2 | 0.212 | 1.09 × 10⁶ | −3.40 |
3 | 0.476 | 0.73 × 10⁶ | −1.51 |
4 | 0.847 | 0.55 × 10⁶ | −0.85 |
Formula | Meaning |
|---|---|
λ = 91.1 nm / (1/m² – 1/n²) | Balmer formula for hydrogen wavelengths (n > m) |
v² = Ke² / (mr) | Orbital speed condition for the Bohr atom |
r_n = n² a_B | Allowed orbital radii (a_B = 0.0529 nm) |
E_n = −13.60 eV / n² | Allowed energies of the Bohr hydrogen atom |
f = ΔE / h | Photon frequency for a transition |
λ = c / f | Wavelength from frequency |
1 MeV = 10⁶ eV = 1.6 × 10⁻¹³ J | Energy unit conversion for nuclear problems |
m_alpha = 6.64 × 10⁻²⁷ kg | Mass of an alpha particle |
The Balmer series is how astronomers determine the chemical composition, temperature, and velocity of stars. Each element has its own unique set of spectral lines, so matching observed wavelengths to known patterns identifies the element even across billions of light-years.
Alpha-particle energies and speeds matter in nuclear engineering and medical physics. Radiation therapy dose calculations depend on knowing precisely how fast and how far these particles travel in tissue.
Students often think Bohr's model applies to all atoms. It works well only for hydrogen and hydrogen-like ions (one electron). Multi-electron atoms require more advanced quantum mechanics.
A common error is treating the Bohr energies as positive. They are negative (bound states). E = 0 means the electron has escaped the atom entirely.
Students sometimes confuse the Balmer series with all hydrogen spectral lines. The Balmer series is only the visible set (transitions ending at n = 2). Other series (Lyman, Paschen) cover UV and infrared.
Students occasionally apply the Balmer formula with m = n or n < m. Remember: n must be strictly greater than m.
⚠️ Using the Balmer formula is a standard exam question. Be comfortable plugging in m = 2 and n = 3, 4, 5, 6 for the visible lines.
⚠️ Calculating the energy of a Bohr level and the photon emitted in a transition is almost guaranteed on a PHYS116 final. Know E_n = −13.60/n² cold.
⚠️ Distinguishing absorption from emission spectra: absorption starts from n = 1 (ground state), giving fewer lines than emission.
⚠️ Unit conversions between eV, MeV, and joules appear frequently. The factor 1 eV = 1.6 × 10⁻¹⁹ J needs to be automatic.
Fill in the blank: The four visible wavelengths of hydrogen (Balmer series) are 656, 486, 434, and ______ nm.
Answer: 410.
True or false: The Bohr radius a_B is the radius of the n = 2 orbit.
Answer: False. a_B is the radius of the n = 1 (ground state) orbit.
Fill in the blank: The energy of the hydrogen atom at n = 2 is ______ eV.
Answer: −3.40.
True or false: In Bohr's model, an electron can orbit at any radius it likes.
Answer: False. Only radii r_n = n² a_B are allowed.
True or false: Collisional excitation requires a photon.
Answer: False. It involves a collision with another particle (electron, atom), not a photon.
Q: What is the wavelength of light emitted when a hydrogen atom transitions from n = 3 to n = 2?
A: Use the Balmer formula with m = 2, n = 3: λ = 91.1 / (1/4 – 1/9) = 91.1 / (5/36) = 91.1 / 0.1389 ≈ 656 nm. This is the red H-alpha line.
Q: What is the ionisation energy of hydrogen (the energy needed to remove the electron from the ground state)?
A: The ground state is at E_1 = −13.60 eV. Removing the electron means reaching E = 0. So the ionisation energy is 13.60 eV.
Q: An alpha particle has a kinetic energy of 5.0 MeV. What is its speed?
A: KE = 5.0 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J. v = √(2 KE / m) = √(2 × 8.0 × 10⁻¹³ / 6.64 × 10⁻²⁷) = 1.6 × 10⁷ m/s.
Q: How many spectral lines does the emission spectrum of an atom with four energy levels have?
A: N(N–1)/2 = 4(3)/2 = 6 lines.
This material builds directly on Ch. 28.4–28.6 (matter waves and quantised energy). The particle-in-a-box model there is the simplified version of what Bohr does for hydrogen: replace rigid walls with the Coulomb force, and the energy formula changes from E_n = n² E_1 to E_n = −13.60/n².
Looking ahead, Bohr's model is a stepping stone to the full quantum-mechanical treatment of the atom (Schrödinger equation, orbitals, quantum numbers l and m_l). The spectral-line framework also connects to molecular spectroscopy and astrophysics.
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