Structure of Atom

This chapter is built as a chain of experiments, each breaking the previous model and demanding a better one, so read it in order rather than jumping around. It opens with the discovery of the three sub-atomic particles and their measured masses and charges, then moves to the two competing atomic models — Thomson's and Rutherford's — decided by the alpha-scattering experiment, followed by the bookkeeping of atomic number, mass number, isotopes and isobars. The middle shifts to light itself: the wave nature of electromagnetic radiation, Planck's quantum theory, the photoelectric effect, and the discrete line spectrum of hydrogen, all of which Bohr combined into his model of the hydrogen atom, its successes, and its fatal limitations. The final third is the modern quantum mechanical model: de Broglie's matter waves, Heisenberg's uncertainty principle, the four quantum numbers, the shapes and nodes of atomic orbitals, the energy ordering of orbitals in multi-electron atoms, and the three filling rules for electronic configuration. Four worked derivations rebuild the key formulas from scratch, and four worked examples extend NCERT's own solved problems to board and JEE/NEET level.

Discovery of the electron, proton and neutron

The story begins with electricity passed through a partially evacuated glass tube — a cathode ray discharge tube. At very low pressure and high voltage, a stream of particles flows from the cathode to the anode; these cathode rays travel in straight lines, make a zinc sulphide screen glow, and deflect in electric and magnetic fields exactly as negatively charged particles would, regardless of the electrode material or the gas used. This independence was decisive: the particles, called electrons, are a universal constituent of all atoms, not a property of any one element. In 1897 J.J. Thomson placed a cathode ray tube between perpendicular electric and magnetic fields and, by balancing the two deflections, measured the ratio of charge to mass of the electron:

where: $e$ is the magnitude of the electron's charge in coulombs and $m_e$ is its mass in kilograms. This ratio alone could not separate charge from mass, so in 1906–14 R.A. Millikan devised the oil-drop experiment: charged oil droplets, suspended between charged plates and observed through a telescope, fall or rise depending on the balance of gravity, viscous drag and the applied electric force. Millikan found that the charge on any droplet is always an integral multiple of a smallest unit, $q = ne$, and measured $e = 1.602176\times10^{-19}\ \text{C}$. Combining this with Thomson's ratio gives the electron's mass, $m_e = 9.109382\times10^{-31}\ \text{kg}$, about $1/1837$ the mass of a hydrogen atom. Modified cathode ray tubes, with a perforated cathode, also produced canal rays of positively charged particles whose charge-to-mass ratio depends on the gas used, unlike the electron's — proof that these are simply ionized atoms of the residual gas, not a universal particle. The lightest such positive ion, from hydrogen, was named the proton (characterised 1919). A neutral partner was still needed to balance atomic mass: in 1932 James Chadwick bombarded beryllium with alpha particles and detected neutral particles slightly heavier than the proton, naming them neutrons. Their measured properties are summarised below.

Thomson and Rutherford models, and the discovery of the nucleus

With three charged particles known, the question became: how are they arranged inside the atom? Thomson's 1898 answer, the plum-pudding or watermelon model, pictured the atom as a uniform sphere of positive charge, about $10^{-10}\ \text{m}$ in radius, with electrons embedded in it like seeds for maximum electrostatic stability. It correctly predicted overall electrical neutrality but could not survive the next experiment. In 1909 Ernest Rutherford, with Hans Geiger and Ernest Marsden, directed high-energy alpha particles at an extremely thin gold foil (about $100\ \text{nm}$ thick) surrounded by a circular fluorescent screen. If Thomson were right, the particles should pass through with only small deflections, since the positive charge and mass were spread evenly through the atom. Instead: most passed through completely undeflected; a small fraction deflected by small angles; and a very small fraction, about $1$ in $20{,}000$, bounced straight back by nearly $180^{\circ}$. Only a concentrated positive charge, occupying a minute fraction of the atom's volume, could repel a fast, massive alpha particle back the way it came. Rutherford put the nuclear radius at about $10^{-15}\ \text{m}$ against an atomic radius of about $10^{-10}\ \text{m}$ — if the nucleus were a cricket ball, the edge of the atom would lie roughly five kilometres away.

Rutherford's nuclear model followed: a tiny, dense, positively charged nucleus holds essentially all the mass; electrons revolve around it at high speed in circular orbits, held by electrostatic attraction, much like planets orbiting the sun. The number of protons in the nucleus is the atomic number, $Z$, equal to the number of electrons in a neutral atom:

The total count of protons and neutrons (together called nucleons) is the mass number, $A$:

where: $n$ is the number of neutrons. An atom is fully specified by the notation $^{A}_{Z}X$. Species with the same $Z$ but different $A$ are isotopes (for example protium $^1_1\text{H}$, deuterium $^2_1\text{H}$, tritium $^3_1\text{H}$, or $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$); since chemical behaviour depends on the number of electrons, all isotopes of an element share essentially the same chemistry. Species with the same $A$ but different $Z$ are isobars (for example $^{14}_6\text{C}$ and $^{14}_7\text{N}$). For an ion, remember that the number of protons stays fixed at $Z$, but the number of electrons differs from $Z$ by the magnitude of the charge — a cation has fewer electrons than protons, an anion has more.

Drawbacks of Rutherford's model

Rutherford's atom resembles the solar system, but that resemblance is fatal. Classical electromagnetic theory (Maxwell) demands that any accelerating charge radiates energy, and an orbiting electron is always accelerating, since its direction constantly changes even at constant speed. It should therefore continuously lose energy, spiral inward, and crash into the nucleus in roughly $10^{-8}\ \text{s}$ — yet real atoms are stable indefinitely. Making the electron stationary instead does not help, since it would simply be pulled straight into the nucleus, collapsing to a miniature Thomson model. Rutherford's model is also silent about how electrons are distributed or what energies they carry — it gives no account of chemical bonding or atomic spectra at all.

Electromagnetic radiation and Planck's quantum theory

Resolving this crisis required two breakthroughs about the nature of light itself. Electromagnetic radiation consists of oscillating electric and magnetic fields, mutually perpendicular and perpendicular to the direction of travel, requiring no medium and travelling at $c = 3.0\times10^8\ \text{m s}^{-1}$ in vacuum. Frequency $\nu$ and wavelength $\lambda$ are linked by

and the wavenumber $\bar\nu = 1/\lambda$ (commonly in $\text{cm}^{-1}$) counts wavelengths per unit length. Nineteenth-century wave theory explained diffraction and interference beautifully but failed completely to explain black-body radiation, the photoelectric effect, the temperature dependence of heat capacities, and atomic line spectra — all of which pointed to the same conclusion: energy exchange between matter and radiation is not continuous. In 1900 Max Planck proposed that the atoms in the walls of a black body absorb or emit energy only in discrete packets, called quanta, with energy directly proportional to frequency:

where: $h$ is Planck's constant, $6.626\times10^{-34}\ \text{J s}$, and $\nu$ is the frequency of the radiation. Energy exchange can only take one of the values $0, h\nu, 2h\nu, 3h\nu,\ldots, nh\nu$ — never anything in between, much as a person on a staircase can stand only on a step, never between two steps. This single postulate let Planck reproduce the observed intensity-versus-wavelength curve of black-body radiation exactly, and it planted the seed for everything that follows in this chapter.

The photoelectric effect

Heinrich Hertz observed in 1887 that shining light on certain metals (potassium, rubidium, caesium) ejects electrons instantly, with three puzzling features classical wave theory could not explain: there is no time lag between the light striking the surface and electron ejection; the number of electrons ejected rises with the intensity of the light, but their kinetic energy does not; and below a metal-specific threshold frequency $\nu_0$, no electrons are ejected at all, however intense the light. Classically, a brighter beam should eject faster electrons regardless of colour — the opposite of what was observed. Einstein resolved this in 1905 by treating light itself as a stream of particles, photons, each carrying energy $h\nu$. When a photon strikes an electron, it transfers its entire energy instantaneously. A minimum energy, the work function $h\nu_0 = W_0$, is needed just to free the electron from the metal; any energy beyond that becomes the photoelectron's kinetic energy:

where: $\nu$ is the frequency of the incident light, $\nu_0$ is the threshold frequency, and $v$ is the ejected electron's speed. A more intense beam simply carries more photons, ejecting more electrons of the same individual kinetic energy — resolving all three puzzles at once. Typical work functions (Table 2.2 of NCERT) are Li $2.42\ \text{eV}$, Na $2.3\ \text{eV}$, K $2.25\ \text{eV}$, Mg $3.7\ \text{eV}$, Cu $4.8\ \text{eV}$, Ag $4.3\ \text{eV}$.

Atomic spectra and the hydrogen line spectrum

White light passed through a prism spreads into a continuous rainbow because it contains every visible wavelength. Gas-phase atoms behave completely differently: when excited (by heat or an electric discharge) and made to emit, they radiate only at specific, sharply defined wavelengths, producing a line spectrum unique to each element, usable as a fingerprint (this is how helium was discovered, in the Sun, before it was found on Earth). Hydrogen, with a single electron, gives the simplest such spectrum, and its visible lines were fitted by Balmer in 1885 to a numerical pattern later generalised by Rydberg into

where: $\bar\nu$ is the wavenumber of the spectral line, $R_H = 1.09677\times10^{7}\ \text{m}^{-1}$ is the Rydberg constant, and $n_1, n_2$ are positive integers with $n_2 > n_1$. Fixing $n_1 = 1$ gives the Lyman series (ultraviolet); $n_1 = 2$ gives the Balmer series (the only one in the visible range); $n_1 = 3, 4, 5$ give the Paschen, Brackett and Pfund series (all infrared). This purely empirical formula, matching experiment to five significant figures, was the single biggest clue that electronic energy in an atom is quantized — and it is exactly what Bohr's model derives from first principles, as the second derivation below shows.

Bohr's model of the hydrogen atom

In 1913 Niels Bohr rescued the nuclear atom from its classical instability with four postulates built on Planck's quantum idea. First, the electron moves only in certain fixed circular orbits — called stationary states — without radiating energy, however much classical theory says it should. Second, each stationary state has a definite, unchanging energy; the electron changes energy only by jumping discontinuously between stationary states, absorbing energy to jump up or emitting it to jump down, with the frequency of absorbed or emitted radiation given by Bohr's frequency rule, $h\nu = |E_2 - E_1|$. Third, the electron's angular momentum is quantized in integer multiples of $h/2\pi$:

where: $m_e$ is the electron mass, $v$ its orbital speed, $r$ the orbital radius, and $n$ the principal quantum number, labelling the stationary states (and, for the atom as a whole, the electron shells $K, L, M, N,\ldots$ for $n = 1, 2, 3, 4,\ldots$). Working through these postulates (done in full in the first derivation below) gives quantized radii and energies:

for hydrogen, and, more generally, $r_n = 52.9(n^2/Z)\ \text{pm}$ and $E_n = -2.18\times10^{-18}(Z^2/n^2)\ \text{J}$ for any one-electron (hydrogen-like) species such as $\text{He}^+$, $\text{Li}^{2+}$ or $\text{Be}^{3+}$, with $Z$ the nuclear charge. The negative sign is a convention: energy is measured relative to a free, motionless electron infinitely far away (set to zero), so a bound electron at any finite $n$ necessarily has lower, more negative energy — the most negative value, at $n = 1$, is the most stable ground state. Substituting equation (10) into Bohr's frequency rule reproduces the Rydberg formula, equation (7), exactly (the second derivation below), which is why Bohr's model was accepted despite its unfamiliar postulates. It could not, however, explain the fine doublet splitting of spectral lines, the spectra of any atom with more than one electron, Zeeman and Stark splitting in magnetic and electric fields, or how atoms form chemical bonds.

Dual behaviour of matter and Heisenberg's uncertainty principle

Two further ideas dismantled Bohr's picture of a definite orbit even while explaining why hydrogen behaved as it did. In 1924 Louis de Broglie proposed that if radiation, ordinarily a wave, can behave as particles (photons), then matter, ordinarily particles, should equally show wave behaviour, with wavelength linked to momentum by

where: $m$ is the particle's mass, $v$ its speed, and $p = mv$ its momentum. This is not poetic language: electron beams were later shown to diffract exactly like waves, a fact now exploited in the electron microscope. Ordinary objects have wavelengths far too small ever to detect (a $0.1\ \text{kg}$ ball at $10\ \text{m s}^{-1}$ has $\lambda = 6.6\times10^{-34}\ \text{m}$), but electrons and other light sub-atomic particles have wavelengths of atomic dimensions, which is precisely why wave effects matter for them and not for cricket balls. In 1927 Werner Heisenberg added a second, independent limitation: it is fundamentally impossible to know an electron's exact position and exact momentum simultaneously. Quantitatively,

where: $\Delta x$ is the uncertainty in position and $\Delta p_x$ (or $m_e\Delta v_x$) is the uncertainty in momentum along the same direction. Pinning down position precisely forces momentum to become wildly uncertain, and vice versa — not a limitation of instruments but a statement about nature itself. Since a trajectory requires knowing both position and velocity at every instant, this rules out the very concept of a definite electron orbit; the fourth derivation below shows numerically just how badly the orbit picture fails once an electron is localised to nuclear dimensions, while a companion calculation for a macroscopic golf ball confirms the effect is negligible for anything you can hold in your hand.

Quantum numbers and the shapes of atomic orbitals

Schrödinger's wave equation (1926), which incorporates de Broglie's wave-particle duality and is consistent with Heisenberg's principle, replaces Bohr's orbits with orbitals — one-electron wave functions $\psi$ that carry no direct physical meaning themselves, but whose square, $|\psi|^2$, gives the probability density of finding the electron at a point. Solving the equation for hydrogen yields quantized energies and wave functions labelled by three quantum numbers that emerge naturally from the mathematics, plus a fourth describing electron spin, discovered experimentally by Uhlenbeck and Goudsmit in 1925. The principal quantum number, $n = 1, 2, 3,\ldots$, fixes the shell, the orbital's approximate size, and, for hydrogen-like species, its energy. The azimuthal quantum number, $l$, fixes the sub-shell and the orbital's shape; for a given $n$, $l$ ranges over $0, 1, 2,\ldots,(n-1)$, giving $s$ ($l=0$), $p$ ($l=1$), $d$ ($l=2$) and $f$ ($l=3$) sub-shells. The magnetic quantum number, $m_l$, fixes spatial orientation and ranges over $-l,\ldots,0,\ldots,+l$, giving $2l+1$ orbitals per sub-shell — one $s$, three $p$, five $d$, seven $f$. The spin quantum number, $m_s = +\tfrac12$ or $-\tfrac12$, describes intrinsic angular momentum orientation, drawn as an up or down arrow, and is essential because otherwise the line spectra of multi-electron atoms show doublets and triplets that three quantum numbers alone cannot explain. Boundary-surface diagrams — surfaces enclosing the region where the electron is found with about $90\%$ probability — give the conventional shapes: every $s$ orbital is a sphere centred on the nucleus, growing larger as $n$ increases ($4s > 3s > 2s > 1s$); each $p$ sub-shell has three dumbbell-shaped orbitals ($p_x, p_y, p_z$) of identical size, shape and energy, oriented along the three axes, with a nodal plane through the nucleus; each $d$ sub-shell has five orbitals, four cloverleaf-shaped and one ($d_{z^2}$) with a different shape but the same energy as the other four. Orbitals of the same sub-shell and identical energy are called degenerate. Every orbital also has nodes, split into radial nodes (spherical shells, $n - l - 1$ of them) and angular nodes (flat or conical surfaces through the nucleus, $l$ of them), giving $n - 1$ nodes in total.

Energies of orbitals, aufbau, Pauli and Hund's rule

In hydrogen, and only in hydrogen-like one-electron species, all orbitals of a given $n$ are degenerate: $1s < 2s = 2p < 3s = 3p = 3d < \ldots$. In every multi-electron atom this breaks down, because electrons repel one another besides being attracted by the nucleus, so energy depends on both $n$ and $l$: within a shell, $s < p < d < f$. Inner electrons partly screen outer electrons from the full nuclear charge $Ze$, so an outer electron feels a reduced effective nuclear charge $Z_{\text{eff}}e$; since $s$ electrons spend more time close to the nucleus than $p$, and $p$ more than $d$, shielding is strongest for $s$ and weakest for $d$ within the same shell — the microscopic reason $s$ lies below $p$ below $d$ in energy. The practical guide to ordering these energies is the $(n+l)$ rule: the lower the value of $n + l$, the lower the orbital's energy; if two orbitals tie on $n + l$, the one with lower $n$ has lower energy. This single rule generates the entire filling order used throughout chemistry:

Three rules govern how electrons fill these orbitals in the ground state. The aufbau ("building up") principle: orbitals fill in order of increasing energy, following equation (13) — the lowest-energy available orbital first. The Pauli exclusion principle (Wolfgang Pauli, 1926): no two electrons in an atom can share the same set of all four quantum numbers, equivalent to saying an orbital holds at most two electrons, with opposite spins; the maximum number of electrons in a shell of principal quantum number $n$ is therefore $2n^2$. Hund's rule of maximum multiplicity: electrons filling a set of degenerate orbitals occupy each singly, with parallel spins, before any pairs up — pairing begins only with the fourth electron in a $p$ sub-shell, the sixth in a $d$, and the eighth in an $f$. Electronic configurations, written as $1s^2 2s^2 2p^6\ldots$ or as orbital-diagram boxes with arrows, follow by applying all three rules in sequence, with the core often abbreviated by a noble gas (sodium, $[\text{Ne}]3s^1$). Two well-known exceptions arise because fully filled ($p^6, d^{10}, f^{14}$) and exactly half-filled ($p^3, d^5, f^7$) sub-shells carry extra stability, from symmetry and from exchange energy released when same-spin electrons in degenerate orbitals swap positions. Chromium therefore adopts $[\text{Ar}]3d^5 4s^1$ rather than $3d^4 4s^2$, and copper adopts $[\text{Ar}]3d^{10}4s^1$ rather than $3d^9 4s^2$.

Board level - protons, neutrons and electrons in a chloride ion (extends NCERT Examples 2.1 and 2.2)

Calculate the number of protons, neutrons and electrons in the $^{37}_{17}\text{Cl}^{-}$ ion.

Step 1. Read off $Z$ and $A$ directly from the given notation: $Z = 17$, $A = 37$.

Step 2. As an anion of charge $-1$, it has one more electron than the neutral atom. Protons $= Z = 17$ (unaffected by ionization). Electrons $= Z + 1 = 18$.

Step 3. Apply equation (3) to find the neutrons: number of neutrons $= A - Z = 37 - 17 = 20$. Answer: 17 protons, 20 neutrons, 18 electrons.

Competitive level (JEE / NEET) - wavelength and energy of the second Balmer line (extends NCERT Example 2.10)

Calculate the wavelength (in nm) and the photon energy (in eV) of the spectral line emitted when a hydrogen electron falls from $n = 4$ to $n = 2$. Use $R_H = 1.09677\times10^7\ \text{m}^{-1}$ and $h = 6.626\times10^{-34}\ \text{J s}$.

Step 1. This is a Balmer-series transition ($n_1 = 2$), so apply equation (7): $\bar\nu = R_H\left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right) = 1.09677\times10^7 \times (0.25 - 0.0625) = 1.09677\times10^7 \times 0.1875 = 2.0564\times10^6\ \text{m}^{-1}$.

Step 2. Invert the wavenumber: $\lambda = \dfrac{1}{\bar\nu} = \dfrac{1}{2.0564\times10^6} = 4.863\times10^{-7}\ \text{m} = 486.3\ \text{nm}$ — the blue-green $H_\beta$ line of the hydrogen spectrum.

Step 3. Find the photon energy using $E = hc\bar\nu$: $E = (6.626\times10^{-34})(3.0\times10^8)(2.0564\times10^6) = 4.09\times10^{-19}\ \text{J}$. Converting to electronvolts, $E = (4.09\times10^{-19})/(1.602\times10^{-19}) = 2.55\ \text{eV}$. Answer: $\lambda \approx 486\ \text{nm}$ (visible blue-green light), photon energy $\approx 2.55\ \text{eV}$. Timing note: about 90 seconds. Recognise the series from $n_1$ (here $2 \Rightarrow$ Balmer), plug both $n$ values into the Rydberg formula in one line, then invert for wavelength; compute the eV energy only if explicitly asked.

Board level - photoelectric effect on sodium (extends NCERT Example 2.9 and Table 2.2)

The work function of sodium is $2.3\ \text{eV}$. Light of wavelength $300\ \text{nm}$ falls on a sodium surface. Calculate (a) the threshold wavelength of sodium and (b) the kinetic energy and velocity of the ejected photoelectron. Use $hc = 1.988\times10^{-25}\ \text{J m}$ and $1\ \text{eV} = 1.602\times10^{-19}\ \text{J}$.

Step 1. Convert the work function to joules and find threshold wavelength from $W_0 = hc/\lambda_0$: $W_0 = 2.3\times1.602\times10^{-19} = 3.685\times10^{-19}\ \text{J}$, so $\lambda_0 = hc/W_0 = 5.394\times10^{-7}\ \text{m} = 539\ \text{nm}$ — light longer than this cannot eject an electron from sodium, however intense.

Step 2. Energy of the incident $300\ \text{nm}$ photon: $E_{\text{photon}} = hc/\lambda = (1.988\times10^{-25})/(3.0\times10^{-7}) = 6.626\times10^{-19}\ \text{J}$. Since $300\ \text{nm} < 539\ \text{nm}$, photoelectrons will indeed be ejected.

Step 3. Apply Einstein's photoelectric equation (6): $KE = E_{\text{photon}} - W_0 = 2.941\times10^{-19}\ \text{J} = 1.84\ \text{eV}$. Velocity from $KE=\tfrac12m_ev^2$: $v = \sqrt{2KE/m_e} = \sqrt{6.458\times10^{11}} = 8.04\times10^5\ \text{m s}^{-1}$. Answer: threshold wavelength $\approx 539\ \text{nm}$; kinetic energy $\approx 1.84\ \text{eV}$; ejected electron speed $\approx 8.04\times10^5\ \text{m s}^{-1}$.

Competitive level (JEE / NEET) - de Broglie wavelength and the uncertainty limit for an accelerated electron

An electron is accelerated from rest through a potential difference of $100\ \text{V}$. Calculate (a) its de Broglie wavelength, and (b) the minimum uncertainty in its velocity if its position is known to within $1\ \text{pm}$. Comment on whether the electron's orbit can be considered well defined.

Step 1. Kinetic energy gained: $KE = eV = (1.602\times10^{-19})(100) = 1.602\times10^{-17}\ \text{J}$. Speed from $KE = \tfrac12m_ev^2$: $v = \sqrt{2KE/m_e} = \sqrt{3.518\times10^{13}} = 5.93\times10^6\ \text{m s}^{-1}$.

Step 2. De Broglie relation, equation (11): $\lambda = \dfrac{h}{m_ev} = \dfrac{6.626\times10^{-34}}{(9.109\times10^{-31})(5.93\times10^6)} = 1.227\times10^{-10}\ \text{m} = 122.7\ \text{pm}$, matching the exam shortcut $\lambda(\text{nm}) = 1.227/\sqrt{V}$.

Step 3. Heisenberg's relation, equation (12), with $\Delta x = 1\ \text{pm} = 1\times10^{-12}\ \text{m}$: $\Delta v \geq \dfrac{h}{4\pi\,m_e\,\Delta x} = \dfrac{6.626\times10^{-34}}{1.1449\times10^{-41}} = 5.79\times10^7\ \text{m s}^{-1}$.

Step 4. Compare: this minimum velocity uncertainty is about ten times larger than the electron's own calculated speed. Answer: $\lambda \approx 122.7\ \text{pm}$; $\Delta v \geq 5.79\times10^7\ \text{m s}^{-1}$ — since this dwarfs the electron's actual speed, its velocity is essentially undefined once position is pinned down this precisely, so no meaningful orbit can be assigned to it. Timing note: about two minutes. Compute $v$ from $eV=\tfrac12m_ev^2$ first, since both parts need it; keep every quantity in SI units, and always finish with the size comparison ($\Delta v$ versus $v$), since that comparison is what the question really tests.

• Confusing atomic number ($Z$, protons) with mass number ($A$, protons plus neutrons) — check which one a question actually gives before writing $^A_ZX$.
• Forgetting to adjust electron count for ions: a cation has fewer electrons than $Z$, an anion has more; the proton count never changes with ionization.
• Treating "orbit" (a fixed classical path, Bohr's picture) and "orbital" (a quantum mechanical probability distribution, $|\psi|^2$) as interchangeable — they describe fundamentally different physics.
• Applying Bohr's formulas $r_n = n^2a_0/Z$ and $E_n = -2.18\times10^{-18}(Z^2/n^2)\ \text{J}$ to multi-electron atoms; they are exact only for one-electron, hydrogen-like species ($\text{H}$, $\text{He}^+$, $\text{Li}^{2+}$, $\text{Be}^{3+}$).
• Reading a negative orbital energy as a sign of an error; it simply means the electron is bound, with the zero reference at an infinitely distant, motionless electron.
• Mixing up the azimuthal quantum number $l$ (shape, sub-shell) with the magnetic quantum number $m_l$ (orientation) — $l$ is derived from $n$, and $m_l$ from $l$, never the reverse.
• Using a naive $n$-only filling order instead of the $(n+l)$ rule, which is why $4s$ ($n+l=4$) fills before $3d$ ($n+l=5$), and $6s$ before $4f$ and $5d$.
• Forgetting the exceptions chromium ($3d^54s^1$) and copper ($3d^{10}4s^1$), which trade one $4s$ electron for extra exchange-energy stability from a half-filled or fully-filled $d$ sub-shell.

• State the properties (relative charge, mass, discoverer) of the electron, proton and neutron, and reproduce Thomson's and Millikan's measured constants.
• Explain why Rutherford's alpha-scattering results disproved Thomson's model and led to the nuclear model of the atom.
• Compute atomic number, mass number, and the number of protons, neutrons and electrons in any neutral atom or ion, and correctly identify isotopes versus isobars.
• Explain, in one or two sentences, why the Rutherford model is classically unstable.
• State Planck's quantum hypothesis and Einstein's photoelectric equation, and use both to solve for photon energy, work function, threshold frequency or photoelectron kinetic energy.
• State Bohr's four postulates and derive the quantized radius and energy formulas for hydrogen and hydrogen-like species from them.
• Use the Rydberg formula to identify a spectral series and compute the wavelength, frequency or wavenumber of any hydrogen spectral line.
• State de Broglie's relation and Heisenberg's uncertainty principle, and explain why together they rule out the concept of a definite electron orbit.
• Assign the correct set of four quantum numbers to any electron, and sketch the boundary-surface shapes of $s$, $p$ and $d$ orbitals with their node counts.
• Write the ground-state electronic configuration of any element up to at least $Z=36$, correctly applying the aufbau principle, Pauli exclusion and Hund's rule, including the chromium and copper exceptions.

Ask the tutor: if you would like, I can walk through writing the full electronic configuration of any specific atom or ion you are struggling with, step by step using the aufbau, Pauli and Hund rules from this chapter.