Structure of Atom · TOPIC LESSON
Connect discrete atomic energies to photon wavelength, momentum and spectral lines.
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BEFORE YOU START
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CONNECT THE CONCEPT TO THE EXAM
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
Electromagnetic radiation has oscillating electric and magnetic fields and can propagate through vacuum. Its wavelength λ is the spatial repeat distance, while frequency ν counts cycles per second. In vacuum, c=λν, with c≈2.998×10⁸ m s⁻¹. Frequency and wavelength are inversely related when the propagation speed is fixed.
The electromagnetic spectrum runs, in increasing frequency, from radio through microwave, infrared, visible, ultraviolet, X-ray and gamma-ray regions. Boundaries are conventional and some source-based categories overlap. For calculations, the actual wavelength or frequency is more informative than the region name.
When light enters a medium, its frequency remains fixed at an ordinary stationary interface while its wavelength changes with phase speed. Use the vacuum wavelength in E=hc/λ unless the appropriate medium relation is specified. Inserting a shortened in-medium wavelength with vacuum c would incorrectly change the photon energy.
Intensity is energy delivered per area per time. At a fixed frequency, increasing intensity usually means more photons crossing a given area per second, not more energy in each photon. Wave amplitude and intensity are related, whereas photon energy depends on frequency. This distinction becomes essential in photoelectric and radiation-pressure questions.
c=λν; E_photon=hν=hc/λ_vac
h≈6.626×10⁻³⁴ J s; λ_vac must be in meters when using SI constants.
ORIGINAL WORKED EXAMPLE
A vacuum wavelength is 500 nm. Find frequency and photon energy using c=3.00×10⁸ m s⁻¹ and h=6.626×10⁻³⁴ J s.
Answer ν=6.00×10¹⁴ Hz; E≈3.98×10⁻¹⁹ J≈2.48 eV per photon.
02 / UNDERSTAND & APPLY
For a monochromatic pulse with energy U, photon number is N=U/(hν). For a beam of power P, photon rate is P/(hν). These relations assume the stated energy or power belongs to that monochromatic radiation. A broad spectrum requires adding contributions at different photon energies rather than assigning one arbitrary wavelength.
A mole of photons has energy N_Ahν, where N_A is Avogadro’s constant. Keep “per photon” and “per mole of photons” distinct: a few electronvolts per photon correspond to hundreds of kilojoules per mole. Mixing those scales is a common source of factors near 10²³.
In the elementary single-photon photoelectric model, emission requires hν at least equal to the work function φ. Above threshold, maximum electron kinetic energy is K_max=hν−φ. Raising intensity at fixed subthreshold frequency does not produce emission in this model; raising intensity above threshold can increase emitted-electron rate without increasing K_max.
The photoelectric equation describes a material surface and its work function, not an isolated hydrogen atom’s bound-state spectrum. An atomic ionization energy is a different threshold. Both illustrate quantized energy exchange, but the physical systems and allowed final states must be kept separate.
ORIGINAL WORKED EXAMPLE
A monochromatic pulse at 400 nm carries 1.00 mJ. Estimate photon number using hc=1.986×10⁻²⁵ J m.
Answer Approximately 2.01×10¹⁵ photons.
03 / UNDERSTAND & APPLY
A classical orbiting charged electron would radiate continuously and lose orbital energy, so a naive classical planetary atom is unstable. Bohr introduced stationary allowed states that do not radiate continuously and postulated that light is absorbed or emitted when the electron changes between them. This was a historically successful model for hydrogen-like one-electron systems.
For a nucleus of charge +Ze and one electron, the model combines Coulomb attraction as centripetal force with angular-momentum quantization m_evr=nħ. Here n is a positive integer and ħ=h/(2π). These assumptions yield discrete radii and energies rather than a continuum of arbitrary circular orbits.
The model applies directly to H, He⁺, Li²⁺ and other one-electron ions in its nonrelativistic approximation. Neutral helium has two electrons and is not hydrogen-like. Simply inserting Z=2 into the one-electron formula does not solve neutral helium because electron–electron interactions change the problem.
A modern orbital is not a literal Bohr orbit. Quantum mechanics describes stationary states through wavefunctions and probability distributions. The Bohr formulas retain useful energy predictions for ideal hydrogen-like species, but the circular-trajectory picture fails to capture orbital shapes, detailed selection rules, fine structure and many-electron spectra.
04 / UNDERSTAND & APPLY
Write Coulomb centripetal balance as m_ev²/r=k_eZe²/r² and quantized angular momentum as m_evr=nħ. From the second relation, v=nħ/(m_er). Substituting into the first gives r_n=n²ħ²/(m_ek_eZe²), or r_n=a₀n²/Z in the fixed-nucleus approximation.
The Bohr radius a₀ is approximately 0.0529 nm. The radius grows as n² and shrinks as 1/Z. The corresponding speed scales as Z/n. These are model orbit radii and speeds, not a statement that a quantum electron follows a measured classical trajectory at that radius.
Coulomb balance also gives kinetic energy K=k_eZe²/(2r), while electrostatic potential energy is V=−k_eZe²/r when zero is chosen at infinite separation. Hence V=−2K and total energy E=K+V=−K. Substitution gives E_n≈−13.6 Z²/n² eV.
The negative sign means the electron is bound relative to a separated stationary electron and nucleus. Larger n gives a less negative energy closer to zero, not a more strongly bound state. At n→∞ the bound-state sequence approaches the ionization limit. Finite nuclear mass produces a reduced-mass correction, and high-Z precision work requires additional corrections beyond this introductory formula.
r_n=a₀n²/Z; E_n=−13.6Z²/n² eV
Hydrogen-like, nonrelativistic fixed-nucleus approximation.
K=−E; V=2E; V=−2K
Coulombic Bohr stationary-state energy relations with zero at infinite separation.
ORIGINAL WORKED EXAMPLE
For He⁺ at n=2, find model radius, total energy, kinetic energy and potential energy using a₀=0.0529 nm.
Answer r=0.1058 nm; E=−13.6 eV; K=13.6 eV; V=−27.2 eV.
05 / UNDERSTAND & APPLY
For an emission transition from n_i to a lower n_f, the atom’s energy change E_f−E_i is negative, while the emitted photon energy E_i−E_f is positive. For absorption, the atom gains the positive energy needed to reach the higher state. Do not give a photon a negative energy merely because the atom loses energy.
For hydrogen-like emission, E_photon=13.6Z²(1/n_f²−1/n_i²) eV. Combining with E=hc/λ gives 1/λ=RZ²(1/n_f²−1/n_i²), where the appropriate Rydberg constant includes the intended approximation for nuclear mass. The bracket must be positive for the written emission ordering.
Frequency, photon energy and wavenumber 1/λ are proportional to the gap. Wavelength is inversely proportional. Therefore the largest gap produces the shortest wavelength. It is unsafe to rank wavelength merely by how many principal levels an arrow crosses because the levels are not equally spaced.
Atomic absorption from a specified initial state selects gaps to allowed final states, subject to selection rules in the fuller theory. An arbitrary photon energy smaller than ionization does not automatically get absorbed and leave the atom in an in-between stationary energy. Collisions and line broadening complicate real spectra, but the discrete-level model explains the basic line structure.
E_photon=E_i−E_f=13.6Z²(1/n_f²−1/n_i²) eV
Emission requires n_i>n_f; photon energy is positive.
λ(nm)≈1240/E_photon(eV)
Convenient rounded vacuum wavelength–energy conversion.
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ORIGINAL WORKED EXAMPLE
Estimate the photon energy and wavelength for H, n=4→2.
Answer E_photon=2.55 eV; λ≈486 nm, a Balmer line.
06 / UNDERSTAND & APPLY
Transitions ending at n_f=1 form the Lyman series, n_f=2 the Balmer series, and n_f=3 the Paschen series. In hydrogen these lie respectively in ultraviolet, with prominent visible Balmer lines and an ultraviolet limit, and infrared. Higher named series have still larger fixed lower levels.
For a fixed lower level, the smallest emission gap comes from n_i=n_f+1 and gives the longest series wavelength. The largest gap is approached as n_i→∞ and gives the shortest wavelength, the series limit. The lines crowd toward that limit because successive high-n levels become closer together in energy.
For hydrogen, the series-limit energy is 13.6/n_f² eV. The Balmer limit is therefore 3.40 eV, corresponding to about 365 nm. Calling the whole Balmer series visible is an overstatement: its high-frequency limit lies in ultraviolet, and visibility ranges themselves have no perfectly sharp physiological boundary.
For the same n_i→n_f transition in hydrogen-like ions, gaps scale as Z² and wavelengths scale as 1/Z² in the simple fixed-nucleus approximation. This comparison is powerful because the principal-level bracket cancels. It applies to corresponding one-electron species, not arbitrary neutral atoms of increasing atomic number.
ORIGINAL WORKED EXAMPLE
If a particular hydrogen transition has wavelength 486 nm in the simple model, what is the corresponding He⁺ wavelength?
Answer Approximately 121.5 nm, neglecting reduced-mass and other small corrections.
07 / UNDERSTAND & APPLY
Ionization from a hydrogen-like level n requires at least 13.6Z²/n² eV to reach the zero-energy continuum with zero final kinetic energy. If an absorbed photon supplies more than that threshold, the excess can appear as kinetic energy of the outgoing electron, neglecting nuclear recoil in the basic model.
Excitation from one bound level to another needs only their difference, which is smaller than ionization from the same initial state. An excited atom is therefore easier to ionize than its ground-state counterpart. “Ionization energy of hydrogen” normally refers to ground-state hydrogen unless another initial level is specified.
For an ensemble initially excited to a maximum principal level N, the maximum number of distinct pairwise transition energies among levels 1 through N is N(N−1)/2 in the elementary unrestricted-level counting model. This counts possible pairs across an ensemble, not photons emitted by one atom in one descent.
A single atom beginning at level N can emit at most N−1 photons in a strictly downward cascade through adjacent principal levels, and it can emit fewer if it skips levels. Detailed quantum selection rules and sublevel populations can restrict observable lines. State the ideal counting convention before applying the pair formula.
ORIGINAL WORKED EXAMPLE
A hydrogen atom begins at n=3 and absorbs a 2.00 eV photon that ionizes it. Find the approximate outgoing electron kinetic energy, neglecting recoil.
Answer Approximately 0.489 eV.
08 / UNDERSTAND & APPLY
A photon in vacuum has momentum magnitude p=E/c=h/λ, despite having zero rest mass. The Newtonian expression p=mv with a photon rest mass inserted is not the appropriate relation. Photon momentum is a vector directed along propagation; absorption or reflection transfers momentum to a material surface.
For normal incidence on a perfectly absorbing stationary surface, each absorbed photon transfers its incident normal momentum p. For perfect specular reflection back along the incident direction, photon momentum reverses from +p to −p, so the surface receives 2p. The factor two comes from momentum change, not from doubling the incident photon energy.
For a normally incident beam of intensity I, the energy arriving per area per time is I. Dividing by c gives momentum arriving per area per time, so radiation pressure is I/c for complete absorption and 2I/c for perfect reflection. Force is pressure times illuminated area, or P/c and 2P/c for a fully intercepted beam of power P.
For an opaque surface with reflected fraction r and absorbed fraction 1−r under the same normal-incidence idealization, pressure is (1+r)I/c. Transmission, oblique incidence, diffuse reflection and moving surfaces require a more careful momentum balance. Never use the perfect-mirror factor without identifying how the outgoing radiation is directed.
p_photon=E/c=h/λ
Vacuum photon momentum magnitude.
P_rad=I/c (absorber); P_rad=2I/c (perfect mirror)
Normal incidence; stationary surface; complete beam interception for F=P_rad A.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A normally incident 3.00 W beam is completely intercepted by a perfect stationary mirror. Find force using c=3.00×10⁸ m s⁻¹.
Answer 2.00×10⁻⁸ N normal to the mirror.
09 / UNDERSTAND & APPLY
Bohr quantization explains hydrogen-like energy gaps remarkably well at introductory precision, but it does not supply the modern probability interpretation. A diagram of levels should place higher n states progressively closer together near zero; equally spaced rungs would misrepresent the inverse-square energies even if the labels were correct.
The sign of total bound-state energy is not the sign of kinetic energy. Kinetic energy remains positive, potential energy is more negative, and the total is negative. In a transition calculation, subtract the two total energies; do not subtract orbital radii and expect an energy gap.
Constants such as 13.6 eV and 1240 eV nm are rounded. Small disagreements at the final digit can result from mixing rounded constants, reduced-mass conventions or reported vacuum versus air wavelengths. Use the constants supplied in the problem consistently rather than presenting more precision than the model supports.
An efficient workflow identifies the species and electron count, the initial and final states or continuum, the quantity requested, and the appropriate conversion. Use Z² scaling for energies, n²/Z scaling for model radii, and inverse energy scaling for wavelength. Check that emission decreases atomic energy and that every photon energy remains positive.
10 / UNDERSTAND & APPLY
Use the rounded constants stated in the teaching examples. Each answer includes the model or sign distinction that makes the arithmetic meaningful.
ORIGINAL PRACTICE · 1
Both halve.
ν=c/λ and E=hc/λ are each inversely proportional to wavelength.
ORIGINAL PRACTICE · 2
−3.40 eV.
E=−13.6/2²=−3.40 eV. The ionization threshold from that level is +3.40 eV.
ORIGINAL PRACTICE · 3
The n=3 radius is nine times larger.
For fixed Z, r_n is proportional to n², so the ratio is 3²/1²=9.
ORIGINAL PRACTICE · 4
122.4 eV.
Li²⁺ has one electron and Z=3. The threshold is 13.6×3²=122.4 eV.
ORIGINAL PRACTICE · 5
n=3→1.
Its energy bracket 1−1/9 is greater than 1−1/4, so the photon energy is larger and wavelength shorter.
ORIGINAL PRACTICE · 6
Six pair transitions; at most three photons in one downward cascade.
The ensemble count is 4×3/2=6. One atom can descend 4→3→2→1, producing three photons; skipping levels gives fewer.
ORIGINAL PRACTICE · 7
2.00×10⁻²⁷ kg m s⁻¹.
p=E/c=6.00×10⁻¹⁹/(3.00×10⁸). Zero rest mass does not imply zero momentum.
ORIGINAL PRACTICE · 8
3.00×10⁻⁶ Pa.
P_rad=I/c=900/(3.00×10⁸). A perfect reflector would give twice this pressure under the same normal-incidence conditions.
KEEP THE BIG IDEAS
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