Quantum Mechanics and Hydrogen Atom - Vocabulary

Hydrogen energy levels: $E_n = -\frac{13.6}{n^2}\ \text{eV}$
- Ground state: $E_1 = -13.6\ \text{eV} = -2.178\times 10^{-18}\ \text{J}$
- As n → ∞, $E_n \to 0^-$ (levels approach zero from below)
Ionization energy (to remove the electron completely): $E{\text{ion}} = 13.6\ \text{eV}$ per atom; per mole: $E{\text{ion}}^{\text{mol}} \approx 1.3\times 10^{3}\ \text{kJ/mol}$
Energy change for a transition: $\Delta E = Ef - Ei$
- If \Delta E > 0: absorption (electron goes to higher n)
- If \Delta E < 0: emission (photon emitted; electron goes to lower n)
Photon energy relations: $E_{\text{ph}} = h\nu = \frac{hc}{\lambda}$
Wavelength from transition: $\lambda = \frac{hc}{|\Delta E|}$
Example: ground to n=4
- $\Delta E = E4 - E1 = -\frac{13.6}{16} - (-13.6) = 12.75\ \text{eV}$
- $E_{\text{ph}} = 12.75\ \text{eV} = 2.04\times 10^{-18}\ \text{J}$
- $\lambda \approx \frac{hc}{2.04\times 10^{-18}} \approx 97\ \text{nm}$
Balmer series (transitions to $n_f=2$ ) yield visible lines (red ~656 nm, blue ~486 nm, etc.).
Absorption vs emission visual cue: absorption raises energy level; emission lowers it; energy gaps smaller for some transitions than others.
Wave-particle duality (brief context): light can behave as waves or particles depending on the experiment; matter also shows wave-like behavior (De Broglie).

Photon energy: $E_{\text{ph}} = h\nu = \frac{hc}{\lambda}$
Wavelength from energy: $\lambda = \frac{hc}{E}$
Common scale: $1\ \text{eV} \approx 1.602\times 10^{-19}\ \text{J}\Rightarrow \lambda \approx 1240\ \text{nm} \text{ per eV}$ (conversion examples vary with E in eV or J)
Units reminder: for exam, give wavelength in nanometers (nm); 1 m = $10^{9}\ \text{nm}$

Absorption: \Delta E = Ef - Ei > 0 (photon absorbed)
Emission: \Delta E = Ef - Ei < 0 (photon emitted; take absolute value for photon energy)
Photon energy magnitude: $|\Delta E| = E_{\text{ph}} = h\nu = \frac{hc}{\lambda}$
Practical note: on exam, compute final minus initial energy in joules, take absolute value for wavelength or frequency

De Broglie relation: $\lambda = \frac{h}{mv}$
Electron example: $m_e = 9.11\times 10^{-31}\ \text{kg},\ v \approx 1\times 10^{6}\ \text{m/s}$
- $\lambda \approx \frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 10^{6}} \approx 7\times 10^{-10}\ \text{m} \approx 0.7\ \text{nm}$
Macroscopic object (e.g., baseball): wavelength ~ $10^{-34}\ \text{m}$ (undetectable)
TEM and diffraction demonstrate electron waves; orbitals arise from wavefunctions solving Schrödinger equation

Core equation: $\hat{H}\psi = E\psi$
- $\hat{H} = T + V$ (kinetic + potential energy)
- $\psi$ is the wavefunction; |\psi|^2 gives probability density
Principal quantum number: $n$ (and angular quantum number $l$ , magnetic $m_l$ ) label orbitals
Orbitals (solutions):
- s orbital: spherical probability cloud
- p orbitals: dumbbell shape (3 orientations)
Key conceptual point: orbitals arise from solving the Schrödinger equation; electrons described by wavefunctions rather than precise orbits; nodes are regions with zero probability
Hydrogen energy depends primarily on $n$ (non-rel, single-electron model)

Core task: given a transition, find the wavelength (nm) or energy (J) using
- $\Delta E = Ef - Ei = -\frac{13.6}{nf^2} + \frac{13.6}{ni^2} \quad(\text{in eV})$
- $|\Delta E| = h\nu = \frac{hc}{\lambda}$
- $\lambda = \frac{hc}{|\Delta E|}$
Use the correct unit: report wavelength in nanometers (nm)
Example reminder: ionization energy per atom = $13.6\ \text{eV}$ ; per mole ≈ $1.3\times 10^{3}\ \text{kJ/mol}$
Remember conversions: 1 m = $10^{9}\ \text{nm}$ ; 1 eV = $1.602\times 10^{-19}\ \text{J}$
On the exam, you may be asked to apply the Balmer-serving idea or interpret an orbital diagram in energy terms
Tools: calculator is allowed; phones or laptops prohibited; periodic table reference may be allowed depending on instructor