Welcome, dear student! In this unit, we will learn about atoms — what they are made of, how scientists discovered their parts, and how the arrangement of electrons affects the properties of elements. This is one of the most important units in chemistry, so let’s take it step by step.
1.1 Dalton’s Atomic Theory and the Modern Atomic Theory
Historical Background
The ancient Greek philosopher Democritus (460–370 BC) first suggested that matter is made of tiny, indivisible particles called atomos. But this was just a philosophical idea — there was no experimental proof.
After nearly 2000 years, in 1808, John Dalton proposed a scientific atomic theory based on experimental evidence from the laws of chemistry.
Laws That Led to Dalton’s Theory
Before Dalton, two important laws were already known:
1. Law of Conservation of Mass (Lavoisier): Mass is neither created nor destroyed in a chemical reaction.
2. Law of Definite Proportions (Proust): A given compound always contains the same elements in the same proportion by mass.
For example, water is always 11.2% hydrogen and 88.8% oxygen by mass, no matter where the water comes from.
Solution:
Water is always 11.2% H and 88.8% O by mass.
Mass of H = 0.112 × 18.0 g = 2.02 g
Mass of O = 0.888 × 18.0 g = 15.98 g
Postulates of Dalton’s Atomic Theory
- All matter is made of atoms, which are indivisible and indestructible.
- All atoms of a given element are identical in mass and properties.
- Atoms of different elements have different masses and properties.
- Atoms combine in simple whole-number ratios to form compounds.
- Atoms cannot be created, destroyed, or subdivided in a chemical reaction.
Dalton also proposed the Law of Multiple Proportions: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
Compound A: 1.750 g N per 1 g O
Compound B: 0.8750 g N per 1 g O
Compound C: 0.4375 g N per 1 g O
Show this illustrates the law of multiple proportions.
Solution: Take ratios relative to the smallest value (C):
A/C = 1.750/0.4375 = 4
B/C = 0.8750/0.4375 = 2
C/C = 0.4375/0.4375 = 1
The ratios 4:2:1 are small whole numbers. ✓
Postulates of the Modern Atomic Theory
Later discoveries showed that some of Dalton’s postulates were not entirely correct. The modern atomic theory revised them:
- Atoms are divisible — they contain subatomic particles (protons, neutrons, electrons).
- Atoms of the same element can have different masses (isotopes).
- Atoms can be destroyed or created in nuclear reactions (but not in chemical reactions).
• Dalton’s theory explained the laws of conservation of mass, definite proportions, and multiple proportions.
• The parts that are still valid: matter is made of atoms; atoms combine in whole-number ratios; atoms are not created/destroyed in chemical reactions.
• The parts that are no longer valid: atoms are indivisible; all atoms of an element are identical.
1. Which postulate of Dalton’s theory is directly contradicted by the discovery of isotopes?
2. How does Dalton’s theory explain the law of definite proportions?
Answer 2: If all atoms of element X are identical and all atoms of element Y are identical, then any compound of X and Y must always have the same ratio of X atoms to Y atoms. Since each atom has a fixed mass, the mass ratio is also fixed.
1.2 Early Experiments to Characterize the Atom
1.2.1 Discovery of the Electron (J.J. Thomson, 1897)
Have you ever wondered how scientists discovered something as tiny as an electron? Let me walk you through it.
Thomson used a cathode ray tube — a glass tube with the air pumped out, containing two electrodes connected to a high-voltage source. When the current was turned on, a glowing beam appeared traveling from the negative electrode (cathode) to the positive electrode (anode).
Properties of cathode rays observed:
- They travel in straight lines.
- They bend toward a positive plate and away from a negative plate → they carry negative charge.
- They bend in a magnetic field.
- Their properties are the same regardless of the cathode material → electrons are a universal component of all matter.
Thomson measured the charge-to-mass ratio of the electron:
But he could not find the charge or mass separately. That was done by Robert Millikan (1909) using his famous oil drop experiment.
Millikan’s Oil Drop Experiment
Millikan sprayed tiny oil drops between two charged plates. By measuring how the drops fell under gravity and how they moved when the electric field was applied, he determined the charge on a single electron:
Using Thomson’s ratio, the electron mass was then calculated:
That is incredibly small — about 1/1840 of a proton’s mass!
• Thomson discovered the electron and measured e/m.
• Millikan measured the charge e on a single electron.
• Together, these gave the mass of the electron.
• The fact that cathode rays are the same regardless of material proves electrons are in ALL matter.
1. Why was Thomson’s cathode ray experiment important? State two conclusions.
2. If the charge-to-mass ratio of an electron is −5.686 × 10⁻¹² kg/C and the charge is −1.602 × 10⁻¹⁹ C, show how to calculate the mass of the electron.
(i) It proved that atoms are not indivisible — they contain smaller particles (electrons).
(ii) It showed that electrons are a universal constituent of all matter (since the results were independent of the cathode material).
Answer 2:
$$m_e = \left(\frac{m_e}{e}\right) \times e = 5.686 \times 10^{-12} \text{ kg/C} \times 1.602 \times 10^{-19} \text{ C} = 9.109 \times 10^{-31} \text{ kg}$$
1.2.2 Radioactivity and the Discovery of the Nucleus
Radioactivity is the spontaneous emission of particles and/or radiation from unstable nuclei of certain atoms (like uranium and radium).
Three types of radiation were discovered:
| Type | Symbol | Charge | Mass | Nature |
|---|---|---|---|---|
| Alpha (α) | ⁴₂He or α²⁺ | +2 | 4 amu | Helium nucleus |
| Beta (β) | ⁰₋₁e or β⁻ | −1 | ~0 amu | Electron from nucleus |
| Gamma (γ) | γ | 0 | 0 | High-energy EM radiation |
Think about this: alpha rays bend toward the negative plate (positive charge), beta rays bend toward the positive plate (negative charge), and gamma rays pass straight through (no charge).
Rutherford’s Gold Foil Experiment (1911)
This is one of the most famous experiments in science! Let me explain it carefully.
Rutherford and his team fired alpha particles at a very thin gold foil. They expected the particles to pass straight through (based on Thomson’s “plum pudding” model where positive charge is spread throughout the atom).
What actually happened:
- Most alpha particles passed straight through — the atom is mostly empty space.
- Some were deflected at small angles — there is something positive inside the atom.
- A very few bounced back at large angles (even 180°) — there is a tiny, dense, positively charged core!
The atom has a tiny, dense, positively charged nucleus at its center that contains nearly all the mass. Most of the atom is empty space, with electrons revolving around the nucleus.
Can you answer this? Why did MOST particles pass straight through? Think about it — the nucleus is extremely tiny compared to the whole atom, so most alpha particles simply miss it!
• It disproved Thomson’s plum-pudding model.
• Established the existence of a small, dense, positive nucleus.
• Showed that the atom is mostly empty space.
• Could NOT explain the stability of the atom or the arrangement of electrons.
1. Why didn’t ALL alpha particles bounce back in Rutherford’s experiment?
2. How did Rutherford’s experiment disprove Thomson’s plum-pudding model?
Answer 2: In Thomson’s model, positive charge is spread evenly throughout the atom. If this were true, alpha particles would experience only weak, uniform repulsion and would all pass through with at most very slight deflection. The observation that a few particles bounced back sharply proved that positive charge is concentrated in a tiny region (the nucleus), not spread out.
1.2.3 Discovery of the Neutron (Chadwick, 1932)
When beryllium metal was irradiated with alpha particles, a highly penetrating but neutral radiation was produced. James Chadwick showed that this radiation consisted of neutral particles with mass nearly equal to protons — the neutron.
1.3 Make-up of the Nucleus
1.3.1 Subatomic Particles
Now we know that atoms are made of three subatomic particles:
| Particle | Actual Mass (kg) | Relative Mass (amu) | Actual Charge (C) | Relative Charge | Location |
|---|---|---|---|---|---|
| Proton (p) | 1.672622 × 10⁻²⁷ | 1.007276 | +1.602 × 10⁻¹⁹ | +1 | Nucleus |
| Neutron (n) | 1.674927 × 10⁻²⁷ | 1.008665 | 0 | 0 | Nucleus |
| Electron (e⁻) | 9.109383 × 10⁻³¹ | 5.486 × 10⁻⁴ | −1.602 × 10⁻¹⁹ | −1 |
Notice that a proton is about 1836 times heavier than an electron, and a neutron is slightly heavier than a proton.
Atomic Number and Mass Number
Atomic number (Z) = number of protons in the nucleus. This defines the element.
Mass number (A) = number of protons + number of neutrons.
We write it as: AZX
For a neutral atom: number of electrons = number of protons = atomic number.
• Protons = 13
• Mass number = 27
• Neutrons = 27 − 13 = 14
• Electrons (neutral) = 13
1.3.2 Atomic Mass and Isotopes
Isotopes are atoms of the same element (same Z) that have different numbers of neutrons (different A).
For example, carbon has three isotopes: ¹²C, ¹³C, and ¹⁴C — all have 6 protons but 6, 7, or 8 neutrons respectively.
Most elements in nature are mixtures of isotopes. The atomic mass (average atomic mass) is calculated as a weighted average:
where $A_i$ = isotopic mass and $f_i$ = fractional abundance (not percentage — divide % by 100).
Solution:
Fractional abundances: f₁ = 0.5184, f₂ = 0.4816
$$\overline{A} = (106.90509)(0.5184) + (108.90476)(0.4816)$$
$$= 55.42 + 52.45 = 107.87 \text{ amu}$$
Solution: Let f = fractional abundance of ¹⁰B, then (1 − f) = abundance of ¹¹B.
$$10.811 = 10.013f + 11.009(1 – f)$$
$$10.811 = 10.013f + 11.009 – 11.009f$$
$$10.811 – 11.009 = -0.996f$$
$$-0.198 = -0.996f$$
$$f = 0.199 \text{ (19.9% for } ^{10}\text{B)}$$
$$1 – f = 0.801 \text{ (80.1% for } ^{11}\text{B)}$$
• Same Z, different A (different number of neutrons).
• Same chemical properties (same electron configuration).
• Different physical properties (different mass).
• Atomic mass on the periodic table is a weighted average, NOT a simple average.
• Convert percentage abundance to fractional abundance before calculating (divide by 100).
1. How many protons, neutrons, and electrons are in ⁶⁴₃₀Zn?
2. An element has two isotopes: ⁶Li (6.015 amu, 7.59%) and ⁷Li (7.016 amu, 92.41%). Calculate the atomic mass. Which isotope contributes more to the atomic mass and why?
Protons = 30, Neutrons = 64 − 30 = 34, Electrons = 30 (neutral atom).
Answer 2:
$$\overline{A} = (6.015)(0.0759) + (7.016)(0.9241) = 0.456 + 6.483 = 6.939 \text{ amu}$$
⁷Li contributes more because it has a much higher abundance (92.41%), so it has a greater weight in the weighted average.
1.4 Electromagnetic Radiation and Atomic Spectra
1.4.1 Electromagnetic Radiation (EMR)
In 1873, James Clerk Maxwell proposed that light consists of electromagnetic waves — waves with an electric field component and a magnetic field component that vibrate in perpendicular planes.
EMR has three primary characteristics:
- Wavelength (λ) — the distance between successive crests, measured in meters (or nm, pm, Å).
- Frequency (ν) — the number of cycles per second, measured in Hz (s⁻¹).
- Speed (c) — in a vacuum, $c = 3.00 \times 10^8$ m/s.
The relationship between them is fundamental:
Important: wavelength and frequency are inversely proportional — if one increases, the other decreases.
The electromagnetic spectrum arranges EMR by wavelength/frequency: radio waves → microwaves → infrared → visible → ultraviolet → X-rays → gamma rays. Visible light ranges from about 380 nm (violet) to 750 nm (red).
Solution:
$$\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 \text{ m/s}}{2.4 \times 10^6 \text{ s}^{-1}} = 125.0 \text{ m}$$
Solution:
$$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{10.0 \times 10^{-6} \text{ m}} = 3.00 \times 10^{13} \text{ Hz}$$
1.4.2 The Quantum Theory and Photon
Here is where physics gets interesting! In 1900, Max Planck proposed a revolutionary idea: energy is not continuous — it comes in small discrete packets called quanta.
The energy of one quantum of radiation:
where $h = 6.63 \times 10^{-34}$ J·s (Planck’s constant).
A photon is a particle of light — a single quantum of EMR.
Solution:
$$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{6.0 \times 10^{-7}} = 5.0 \times 10^{14} \text{ Hz}$$
$$E = h\nu = (6.63 \times 10^{-34})(5.0 \times 10^{14}) = 3.315 \times 10^{-19} \text{ J}$$
The Photoelectric Effect
In 1905, Einstein used Planck’s idea to explain the photoelectric effect — when light shines on certain metals, electrons are ejected. But only if the light frequency is above a minimum called the threshold frequency (ν₀).
Key equations:
If ν < ν₀, no electrons are ejected regardless of intensity. If ν > ν₀, higher intensity means MORE electrons (more photons), but each electron has the same maximum KE (determined by frequency, not intensity).
Einstein also showed (from his theory of relativity) that energy and mass are related:
This shows that even light has an “apparent mass” — supporting the idea that light has both wave AND particle properties. This is called the dual nature of light.
Solution:
Step 1: Find ν of incident light:
$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{7.5 \times 10^{-7}} = 4.0 \times 10^{14} \text{ Hz}$$
Step 2: Find ν₀:
$$\nu_0 = \nu – \frac{KE}{h} = 4.0 \times 10^{14} – \frac{1.5 \times 10^{-20}}{6.63 \times 10^{-34}}$$
$$= 4.0 \times 10^{14} – 2.26 \times 10^{13} = 3.77 \times 10^{14} \text{ Hz}$$
Step 3: Find λ₀:
$$\lambda_0 = \frac{c}{\nu_0} = \frac{3.0 \times 10^8}{3.77 \times 10^{14}} = 7.96 \times 10^{-7} \text{ m} = 796 \text{ nm}$$
• Energy is quantized — it comes in discrete packets (quanta/photons).
• E = hν = hc/λ — higher frequency means higher energy.
• Photoelectric effect: ν must exceed ν₀; KE depends on frequency, NOT intensity.
• Intensity affects the NUMBER of ejected electrons, not their energy.
• Dual nature: light behaves as both a wave (diffraction, interference) and a particle (photoelectric effect).
1. A photon has a wavelength of 1.0 × 10⁻¹⁰ m (gamma ray). Calculate its energy. How does this compare with a photon of λ = 1.0 × 10⁻⁶ m (infrared)?
2. The minimum energy to cause the photoelectric effect in potassium is 3.69 × 10⁻¹⁹ J. Will visible light of 520 nm eject electrons? If yes, calculate the KE of the ejected electrons.
For gamma ray: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{1.0 \times 10^{-10}} = 1.989 \times 10^{-15}$ J
For infrared: $E = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{1.0 \times 10^{-6}} = 1.989 \times 10^{-19}$ J
The gamma ray photon has about 10⁴ (ten thousand) times more energy than the infrared photon.
Answer 2:
Energy of 520 nm photon: $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{5.2 \times 10^{-7}} = 3.83 \times 10^{-19}$ J
Since $3.83 \times 10^{-19}$ J > $3.69 \times 10^{-19}$ J (E₀), YES, electrons will be ejected.
$KE = E – E_0 = 3.83 \times 10^{-19} – 3.69 \times 10^{-19} = 1.4 \times 10^{-20}$ J
1.4.3 Atomic Spectra
When an electric discharge passes through a gas, the atoms absorb energy and their electrons move to higher energy levels. When the electrons fall back to lower levels, they emit light. But this light is NOT a continuous rainbow — it consists of only certain discrete wavelengths, appearing as individual lines. This is called an atomic (line) spectrum.
Each element has a unique line spectrum — like a fingerprint! This is used to identify elements.
The energy change when an electron moves between levels:
1.4.4 The Bohr Model of the Hydrogen Atom (1913)
Niels Bohr combined Rutherford’s nuclear model with Planck’s quantum idea to explain the hydrogen spectrum. His key assumptions:
- The electron moves in circular orbits around the nucleus.
- Only certain orbits are allowed (quantized). Each has a fixed energy.
- While in an orbit, the electron does not radiate energy.
- The angular momentum of the electron is quantized: $mvr = \frac{nh}{2\pi}$ (n = 1, 2, 3, …).
- The electron absorbs energy to jump to a higher orbit and emits a photon when it falls to a lower orbit. The photon energy equals the energy difference: $\Delta E = E_{\text{higher}} – E_{\text{lower}}$.
From these assumptions, Bohr derived three important formulas:
1. Radius of the nth orbit:
where $a_0 = 0.53$ Å is the Bohr radius, and n is the principal quantum number.
2. Velocity of the electron in the nth orbit:
3. Energy of the electron in the nth orbit:
where $R_H = 2.18 \times 10^{-18}$ J is the Rydberg constant for hydrogen.
The negative sign means the electron is bound to the nucleus. The most negative value (most stable) is at n = 1 — this is the ground state. For n > 1, the atom is in an excited state.
Solution:
$$r_5 = 5^2 \times 0.53 \text{ Å} = 25 \times 0.53 = 13.25 \text{ Å}$$
$$v_5 = \frac{2.18 \times 10^6}{5} = 4.36 \times 10^5 \text{ m/s}$$
$$E_5 = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \text{ J}$$
Solution:
$$E_4 = \frac{-2.18 \times 10^{-18}}{16} = -1.3625 \times 10^{-19} \text{ J}$$
$$E_2 = \frac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19} \text{ J}$$
$$\Delta E = E_4 – E_2 = -1.3625 \times 10^{-19} – (-5.45 \times 10^{-19}) = 4.0875 \times 10^{-19} \text{ J}$$
$$\nu = \frac{\Delta E}{h} = \frac{4.0875 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.16 \times 10^{14} \text{ Hz}$$
$$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8}{6.16 \times 10^{14}} = 4.87 \times 10^{-7} \text{ m} = 487 \text{ nm (blue-green light)}$$
This transition (n = 4 → n = 2) is part of the Balmer series (transitions ending at n = 2), which falls in the visible region. Other series are Lyman (to n = 1, UV), Paschen (to n = 3, IR), Brackett (to n = 4, IR), and Pfund (to n = 5, IR).
• $r_n = n^2 \times 0.53$ Å — radius increases with n².
• $E_n = -2.18 \times 10^{-18}/n^2$ J — energy becomes less negative (higher) as n increases.
• $\Delta E = E_{\text{final}} – E_{\text{initial}}$ — for emission, final n < initial n, so ΔE is positive (photon emitted).
• Lyman series → UV; Balmer series → Visible; Paschen series → IR.
• Bohr’s model works ONLY for hydrogen and hydrogen-like ions (He⁺, Li²⁺).
1. Calculate the energy of an electron in the n = 3 state of hydrogen. How much energy is released when the electron falls from n = 3 to n = 1?
2. Which has greater energy: a photon emitted in a transition from n = 3 to n = 1, or from n = 2 to n = 1? Explain without calculating.
$$E_3 = \frac{-2.18 \times 10^{-18}}{9} = -2.42 \times 10^{-19} \text{ J}$$
$$E_1 = \frac{-2.18 \times 10^{-18}}{1} = -2.18 \times 10^{-18} \text{ J}$$
$$\Delta E = E_1 – E_3 = -2.18 \times 10^{-18} – (-2.42 \times 10^{-19}) = -1.94 \times 10^{-18} \text{ J}$$
The negative sign means energy is released. The photon carries $1.94 \times 10^{-18}$ J of energy.
Answer 2: The transition from n = 3 to n = 1 releases more energy. The energy difference is larger because n = 3 is further from the nucleus (higher energy) than n = 2, so the electron falls a greater distance. Since $\Delta E \propto (1/n_f^2 – 1/n_i^2)$, and $1/1^2 – 1/3^2 = 8/9$ while $1/1^2 – 1/2^2 = 3/4$, the n = 3 → 1 transition has a larger energy difference.
1.4.5 Limitations of the Bohr Model
- Works only for hydrogen and hydrogen-like ions — fails for multi-electron atoms.
- Cannot explain the fine structure of spectral lines (each line is actually a group of closely spaced lines).
- Does not explain the Zeeman effect (splitting of lines in a magnetic field).
- Combines classical physics with quantum ideas in an inconsistent way.
1.4.6 Wave-Particle Duality of Matter
In 1924, Louis de Broglie proposed that if light can behave as both wave and particle, then matter (like electrons) can also behave as a wave! He derived:
where m is mass and v is velocity. This is called the de Broglie wavelength.
This was experimentally confirmed by electron diffraction experiments — electrons passing through a crystal produced a diffraction pattern, just like waves!
Solution:
$$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(1.00 \times 10^7)} = 7.28 \times 10^{-11} \text{ m} = 0.728 \text{ Å}$$
This wavelength is comparable to the size of an atom — so wave effects are significant for electrons!
• De Broglie: $\lambda = h/mv$ — all moving matter has wave properties.
• Significant for very small particles (electrons) but negligible for large objects (a ball).
• This idea led to the development of quantum mechanics.
1.5 The Quantum Mechanical Model of the Atom
1.5.1 Heisenberg’s Uncertainty Principle
Werner Heisenberg (1927) stated that it is fundamentally impossible to know both the exact position AND exact momentum of an electron simultaneously:
where $\Delta x$ = uncertainty in position and $\Delta p$ = uncertainty in momentum.
This means if we know exactly where an electron is, we have no idea how fast it’s moving, and vice versa. This is why Bohr’s model (which assumed electrons have precise orbits) is incorrect!
1.5.2 Quantum Numbers
In the quantum mechanical model, electrons are described by four quantum numbers. Think of them as the “address” of each electron in an atom.
| Quantum Number | Symbol | Values | What it Describes |
|---|---|---|---|
| Principal | n | 1, 2, 3, … | Energy level / shell size |
| Azimuthal (Angular momentum) | l | 0 to (n−1) | Subshell shape (s, p, d, f) |
| Magnetic | mₗ | −l to +l (including 0) | Orbital orientation in space |
| Spin | mₛ | +½ or −½ | Spin direction of electron |
Subshell notation: l = 0 → s, l = 1 → p, l = 2 → d, l = 3 → f
Solution:
For 3p: n = 3, l = 1
mₗ can be: −1, 0, +1 (three orbitals)
mₛ can be: +½ or −½ for each orbital
So the six possible sets are:
(3, 1, −1, +½), (3, 1, −1, −½),
(3, 1, 0, +½), (3, 1, 0, −½),
(3, 1, +1, +½), (3, 1, +1, −½)
(a) (2, 0, 0, +½) (b) (3, 2, −3, +½) (c) (1, 1, 0, −½) (d) (4, 3, −2, +½)
Solution:
(a) Allowed: n=2, l=0 (2s), mₗ=0 ✓, mₛ=+½ ✓
(b) NOT allowed: n=3, l=2 means mₗ can only be −2, −1, 0, +1, +2. mₗ=−3 is outside this range.
(c) NOT allowed: n=1 means l can only be 0. l=1 is invalid.
(d) Allowed: n=4, l=3 (4f), mₗ=−2 is within −3 to +3 ✓
1.5.3 Shapes of Atomic Orbitals
An orbital is a region in space where there is a high probability of finding an electron. The shape depends on the value of l:
- s orbitals (l = 0): Spherical shape. Each shell has one s orbital.
- p orbitals (l = 1): Dumbbell shape. Three p orbitals (px, py, pz) oriented along x, y, z axes. Each shell from n = 2 has three p orbitals.
- d orbitals (l = 2): Cloverleaf/double-dumbbell shapes. Five d orbitals. Each shell from n = 3 has five d orbitals.
- f orbitals (l = 3): Complex shapes. Seven f orbitals. Each shell from n = 4 has seven f orbitals.
Maximum electrons per subshell: s → 2, p → 6, d → 10, f → 14
Maximum electrons per shell: 2n²
• n determines the shell; l determines the subshell; mₗ determines the orbital; mₛ determines the spin.
• No two electrons in an atom can have the same set of four quantum numbers (Pauli exclusion principle).
• Common invalid quantum number sets: l ≥ n, |mₗ| > l, mₛ ≠ ±½.
• Number of orbitals per subshell: s=1, p=3, d=5, f=7.
1. How many orbitals are in the n = 3 shell? What subshells do they belong to?
2. An electron has quantum numbers (4, 2, −1, −½). In which orbital is this electron?
n = 3 has subshells: 3s (l=0, 1 orbital), 3p (l=1, 3 orbitals), 3d (l=2, 5 orbitals).
Total orbitals = 1 + 3 + 5 = 9 orbitals (which matches n² = 9).
Answer 2:
n = 4, l = 2 → this is the 4d subshell.
mₗ = −1 → this is one specific d orbital (one of the five 4d orbitals).
So the electron is in a 4d orbital.
1.6 Electronic Configurations and Orbital Diagrams
1.6.1 Rules for Writing Electronic Configurations
To write the ground state electronic configuration of any element, we follow three rules:
1. Aufbau Principle (“building up”): Electrons fill orbitals starting from the lowest energy to the highest. The order is: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p
Remember: 4s fills BEFORE 3d, but 4s loses electrons BEFORE 3d when forming cations!
2. Pauli Exclusion Principle: No two electrons in an atom can have the same four quantum numbers. This means each orbital can hold at most 2 electrons with opposite spins.
3. Hund’s Rule: When filling degenerate orbitals (same energy, like the three 2p orbitals), electrons occupy separate orbitals with parallel spins before pairing up.
Solution:
(a) Carbon (Z=6): 1s² 2s² 2p²
Orbital diagram:
1s: ↑↓ 2s: ↑↓ 2p: ↑ ↑ _ (Hund’s rule — unpaired first!)
(b) Potassium (Z=19): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Note: 4s fills before 3d!
(c) Iron (Z=26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Orbital diagram for 3d: ↑ ↑ ↑ ↑ ↑↓ (five orbitals: four unpaired, one paired)
Solution:
(a) Fe²⁺: Remove 2 electrons from 4s first!
Fe: [Ar] 4s² 3d⁶ → Fe²⁺: [Ar] 3d⁶
(b) Fe³⁺: Remove 3 electrons (2 from 4s, 1 from 3d)
Fe: [Ar] 4s² 3d⁶ → Fe³⁺: [Ar] 3d⁵ (half-filled d subshell — extra stable!)
(c) Cu: Exception to Aufbau — [Ar] 4s¹ 3d¹⁰ (not 4s² 3d⁹)
This is because a completely filled d subshell (d¹⁰) is more stable. So Cu: [Ar] 4s¹ 3d¹⁰
• Always remove electrons from the HIGHEST n value first when forming cations (4s before 3d).
• Common exceptions: Cr = [Ar] 4s¹ 3d⁵ (half-filled), Cu = [Ar] 4s¹ 3d¹⁰ (fully filled).
• For ions: Fe²⁺ = [Ar] 3d⁶, Fe³⁺ = [Ar] 3d⁵, Cu⁺ = [Ar] 3d¹⁰, Cu²⁺ = [Ar] 3d⁹.
• Noble gas core notation: use [He], [Ne], [Ar] for inner electrons.
• Hund’s rule applies only to degenerate orbitals (same subshell, same energy).
1. Write the electronic configuration of sulfur (Z=16) and sulfide ion S²⁻.
2. Why is the electronic configuration of chromium [Ar] 4s¹ 3d⁵ instead of [Ar] 4s² 3d⁴?
S (Z=16): 1s² 2s² 2p⁶ 3s² 3p⁴ or [Ne] 3s² 3p⁴
S²⁻ (gains 2 electrons): [Ne] 3s² 3p⁶ = [Ar] configuration (same as argon!)
Answer 2:
A half-filled d subshell (d⁵) has extra stability due to symmetrical electron distribution and maximum exchange energy. By promoting one electron from 4s to 3d, chromium achieves the more stable [Ar] 4s¹ 3d⁵ configuration with five unpaired d electrons and one unpaired s electron, rather than [Ar] 4s² 3d⁴ which has only four unpaired d electrons.
1.7 Electronic Configurations and the Periodic Table
1.7.1 The Modern Periodic Table
The modern periodic table is arranged in order of increasing atomic number. The periodic law states: “Properties of elements are a periodic function of their atomic numbers.”
Each horizontal row is a period (1–7) and each vertical column is a group (1–18).
1.7.2 Classification of the Elements
| Block | Subshell Being Filled | Groups | Examples |
|---|---|---|---|
| s-block | s | 1, 2 (and He) | Li, Na, Mg, Ca |
| p-block | p | 13–18 | B, C, N, O, F, Ne |
| d-block | d | 3–12 | Sc, Ti, Fe, Cu, Zn |
| f-block | f | Lanthanides & Actinides | Ce, Th, U |
Representative elements = s-block + p-block elements. Transition elements = d-block. Inner transition elements = f-block.
1.7.3 Periodic Properties
Now let’s study the important periodic trends. These are very common exam questions!
A. Atomic Radius
The atomic radius is the distance from the nucleus to the outermost electrons.
Trends:
- Across a period (left → right): DECREASES. The nuclear charge increases, pulling electrons closer. Shielding stays roughly the same.
- Down a group: INCREASES. New shells are added, making the atom larger. Increased shielding also reduces the effective nuclear charge on outer electrons.
B. Ionization Energy (IE)
Ionization energy is the minimum energy required to remove the most loosely bound electron from a gaseous atom.
Trends:
- Across a period: INCREASES. Smaller radius + higher nuclear charge = harder to remove electron.
- Down a group: DECREASES. Larger radius + more shielding = easier to remove electron.
Exceptions to IE trend across a period:
- IE of Group 2 (Be, Mg) > IE of Group 13 (B, Al): Group 2 has a filled s subshell (stable), Group 13 starts filling p (easier to remove from higher energy p orbital).
- IE of Group 15 (N, P) > IE of Group 16 (O, S): Group 15 has half-filled p subshell (stable), Group 16 has one paired electron (electron-electron repulsion makes it easier to remove).
Successive IEs: IE₁ < IE₂ < IE₃ … with large jumps when removing from a new, inner shell.
Solution:
The large jump occurs between IE₃ and IE₄, meaning the first 3 electrons are in the outer shell and the 4th electron is being removed from an inner shell. An element with 3 valence electrons is in Group 13. Looking at IE₁ = 578 kJ/mol, this matches aluminum (Al, Z=13) with configuration [Ne] 3s² 3p¹.
C. Electron Affinity (EA)
Electron affinity is the energy change when an electron is added to a gaseous atom.
More negative EA = more energy released = greater tendency to accept an electron.
Trends: Generally becomes more negative (stronger attraction) across a period, and becomes less negative (weaker attraction) down a group. Exceptions: Group 2 and Group 15 have less negative EA (filled/half-filled subshells). Noble gases have positive EA (don’t want extra electron).
D. Electronegativity
Electronegativity is the ability of an atom to attract shared electrons in a chemical bond. It is a relative scale (Pauling scale, F = 4.0 is the most electronegative).
Trends: Same as IE — increases across a period, decreases down a group. Fluorine is the most electronegative element.
E. Metallic and Non-metallic Character
Metallic character increases down a group and decreases across a period (metals are on the left and bottom).
Non-metallic character increases across a period and decreases down a group (non-metals are on the right and top).
• Atomic radius: ← increases | ↓ increases
• Ionization energy: ← increases | ↓ decreases
• Electron affinity: ← more negative (generally) | ↓ less negative
• Electronegativity: ← increases | ↓ decreases
• Metallic character: ← decreases | ↓ increases
• Remember the exceptions! Group 2 vs 13, Group 15 vs 16 for IE and EA.
• Large jump in successive IE → change of shell → reveals group number.
1. Arrange the following in order of increasing atomic radius: Na, Mg, Al, K, Ca.
2. Explain why the first ionization energy of oxygen is less than that of nitrogen.
3. An element has successive IEs with a large jump between IE₂ and IE₃. Which group does it belong to?
Consider period and group: K (Period 4, Group 1) > Ca (Period 4, Group 2) > Na (Period 3, Group 1) > Mg (Period 3, Group 2) > Al (Period 3, Group 3).
Increasing order: Al < Mg < Na < Ca < K
Answer 2:
Nitrogen has the electronic configuration 1s² 2s² 2p³ — a half-filled p subshell which is extra stable. Removing an electron from this stable arrangement requires more energy. Oxygen has 1s² 2s² 2p⁴ — one of the p orbitals has a pair of electrons. The electron-electron repulsion in this paired arrangement makes it easier to remove one electron compared to nitrogen’s half-filled stable configuration.
Answer 3:
A large jump between IE₂ and IE₃ means the first 2 electrons are valence electrons and the 3rd comes from an inner shell. So the element has 2 valence electrons → Group 2.
1.7.4 Advantages of Periodic Classification
- Systematic study of elements — elements in the same group have similar properties.
- Prediction of properties of new/undiscovered elements.
- Helps understand the relationship between electronic configuration and properties.
- Useful for understanding trends in chemical behavior and bonding.
Quick Revision Notes — Exam Focus
1. Key Laws
• Definite Proportions: A compound always has the same elements in the same mass ratio.
• Multiple Proportions: When two elements form multiple compounds, mass ratios are small whole numbers.
2. Important Definitions
• Isotope: Atoms of the same element with different numbers of neutrons (same Z, different A).
• Atomic mass: Weighted average of isotopic masses based on natural abundances.
• Photon: A particle of light; a quantum of EMR.
• Ionization energy: Energy required to remove the outermost electron from a gaseous atom.
• Electron affinity: Energy change when an electron is added to a gaseous atom.
• Electronegativity: Ability of an atom to attract bonding electrons.
• Orbital: Region of high probability of finding an electron.
3. Key Formulas
4. Constants to Remember
5. Quantum Numbers Summary
| n | l | Subshell | mₗ values | # orbitals | # electrons |
|---|---|---|---|---|---|
| 1 | 0 | 1s | 0 | 1 | 2 |
| 2 | 0 | 2s | 0 | 1 | 2 |
| 2 | 1 | 2p | −1, 0, +1 | 3 | 6 |
| 3 | 0 | 3s | 0 | 1 | 2 |
| 3 | 1 | 3p | −1, 0, +1 | 3 | 6 |
| 3 | 2 | 3d | −2,−1, 0, +1, +2 | 5 | 10 |
| 4 | 0 | 4s | 0 | 1 | 2 |
| 4 | 1 | 4p | −1, 0, +1 | 3 | 6 |
| 4 | 2 | 4d | −2,−1, 0, +1, +2 | 5 | 10 |
| 4 | 3 | 4f | −3,−2,−1, 0, +1, +2, +3 | 7 | 14 |
6. Aufbau Order (Memorize!)
7. Exceptional Configurations
Copper (Cu, Z=29): [Ar] 4s¹ 3d¹⁰ (fully-filled d subshell — extra stable)
Molybdenum (Mo, Z=42): [Kr] 5s¹ 4d⁵ (same pattern as Cr)
Silver (Ag, Z=47): [Kr] 5s¹ 4d¹⁰ (same pattern as Cu)
8. Periodic Trends Summary
| Property | Across Period (→) | Down Group (↓) |
|---|---|---|
| Atomic Radius | Decreases | Increases |
| Ionization Energy | Increases | Decreases |
| Electron Affinity | More negative (generally) | Less negative |
| Electronegativity | Increases | Decreases |
| Metallic Character | Decreases | Increases |
| Non-metallic Character | Increases | Decreases |
9. Spectral Series of Hydrogen
Balmer series: Transitions to n = 2 → Visible region
Paschen series: Transitions to n = 3 → IR region
Brackett series: Transitions to n = 4 → IR region
Pfund series: Transitions to n = 5 → IR region
10. Common Mistakes to Avoid
❌ Forgetting the negative sign in Bohr energy formula.
❌ Writing 4s² 3d⁴ for Cr instead of 4s¹ 3d⁵.
❌ Removing electrons from 3d before 4s when writing ion configurations (ALWAYS remove from highest n first!).
❌ Confusing atomic number with mass number.
❌ Saying “IE increases continuously across a period” — remember the Group 2/13 and Group 15/16 exceptions!
❌ Using ΔE = E_initial − E_final instead of E_final − E_initial.
❌ Forgetting that frequency and wavelength are inversely proportional.
Challenge Exam Questions
Test yourself! Try each question before checking the answer.
Section A: Multiple Choice Questions
A) (3, 2, −1, +½) B) (2, 0, 0, −½) C) (3, 3, −2, +½) D) (4, 0, 0, −½)
For n = 3, the maximum value of l is n − 1 = 2. So l = 3 is not allowed. The set (3, 3, −2, +½) violates the rule that l must be less than n.
A) Planck’s constant B) The mass of the particle C) The velocity of the particle D) Both B and C
From λ = h/(mv), the wavelength is inversely proportional to both mass (m) and velocity (v). Increasing either m or v decreases λ.
A) Na B) Mg C) Al D) Ne
Neon is a noble gas with a completely filled outer shell (2s² 2p⁶), making it extremely stable. Removing an electron requires the most energy. Across Period 3: Na < Al < Mg < Ne (Ne is Period 2 but much higher IE than any Period 3 element).
A) 1.82 × 10⁻¹⁹ J B) 3.03 × 10⁻¹⁹ J C) 5.45 × 10⁻¹⁹ J D) 1.94 × 10⁻¹⁸ J
$$E_3 = \frac{-2.18 \times 10^{-18}}{9} = -2.42 \times 10^{-19} \text{ J}$$
$$E_2 = \frac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19} \text{ J}$$
$$\Delta E = E_2 – E_3 = -5.45 \times 10^{-19} – (-2.42 \times 10^{-19}) = -3.03 \times 10^{-19} \text{ J}$$
The negative sign indicates energy is emitted. The photon carries $3.03 \times 10^{-19}$ J.
A) [Ar] 3d³ B) [Ar] 4s¹ 3d² C) [Ar] 3d¹ D) [Ar] 4s² 3d¹
Cr: [Ar] 4s¹ 3d⁵. To form Cr³⁺, remove 1 electron from 4s and 2 from 3d (remove from highest n first).
Cr³⁺ = [Ar] 3d³. There are no 4s electrons left.
A) They have different atomic numbers.
B) They have the same number of neutrons.
C) They have different mass numbers.
D) They have different chemical properties.
Isotopes have the same atomic number (same protons) but different numbers of neutrons, giving them different mass numbers. They have the same chemical properties because chemical properties depend on electron configuration, which is the same for isotopes.
Section B: Fill in the Blanks
Section C: Short Answer Questions
Heisenberg’s uncertainty principle states that it is impossible to simultaneously determine the exact position (Δx) and exact momentum (Δp) of an electron: $\Delta x \cdot \Delta p \geq h/4\pi$.
Bohr’s model assumes that electrons travel in fixed circular orbits with precisely defined radii and velocities. This means both position and momentum are known exactly, which violates the uncertainty principle. Therefore, the concept of fixed orbits is fundamentally incorrect. In the quantum mechanical model, we can only describe the probability of finding an electron in a region (orbital), not its exact path.
Emission spectrum: Produced when excited atoms release energy as electrons fall from higher to lower energy levels. It appears as bright colored lines on a dark background. Each line corresponds to a specific wavelength of emitted photon.
Absorption spectrum: Produced when atoms absorb specific wavelengths of white light, causing electrons to jump from lower to higher energy levels. It appears as dark lines on a continuous bright background. The dark lines correspond to the same wavelengths as the bright lines in the emission spectrum of that element.
Both are characteristic of the element and can be used for identification.
Cation (e.g., Na → Na⁺): When an atom loses electrons, the number of protons stays the same but the number of electrons decreases. The effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus. Also, losing the outer shell sometimes occurs (e.g., Na: [Ne]3s¹ → Na⁺: [Ne]), significantly reducing the size.
Anion (e.g., Cl → Cl⁻): When an atom gains electrons, the nuclear charge stays the same but there are more electrons. This increases electron-electron repulsion and decreases the effective nuclear charge per electron. The electron cloud expands, making the anion larger than the parent atom.
Nitrogen (Z=7): 1s² 2s² 2p³
The three unpaired electrons are in the 2p subshell (n=2, l=1). By Hund’s rule, they occupy three separate p orbitals with parallel spins:
• Electron 1: (2, 1, −1, +½)
• Electron 2: (2, 1, 0, +½)
• Electron 3: (2, 1, +1, +½)
All three have the same spin (+½) — this is Hund’s rule in action!
Section D: Calculation Questions
$$\overline{A} = (34.969)(0.7577) + (36.966)(0.2423)$$
$$= 26.50 + 8.96 = 35.46 \text{ amu}$$
The calculated value (35.46 amu) is closer to 35 because ³⁵Cl has a much higher abundance (75.77%) than ³⁷Cl (24.23%). In a weighted average, the isotope with higher abundance pulls the average toward its mass.
(a) The energy of the incident photon
(b) The maximum kinetic energy of the ejected photoelectrons
(c) The velocity of the ejected photoelectrons (mₑ = 9.11 × 10⁻³¹ kg)
(a) $$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{4.0 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}$$
(b) $$KE_{\max} = E – E_0 = 4.97 \times 10^{-19} – 3.0 \times 10^{-19} = 1.97 \times 10^{-19} \text{ J}$$
(c) $$KE = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\frac{2 \times KE}{m_e}} = \sqrt{\frac{2 \times 1.97 \times 10^{-19}}{9.11 \times 10^{-31}}}$$
$$= \sqrt{4.33 \times 10^{11}} = 6.58 \times 10^5 \text{ m/s}$$
(a) The radius of the 3rd orbit of hydrogen
(b) The energy of an electron in the 3rd orbit
(c) The wavelength of the photon emitted when the electron falls from n = 3 to n = 1
(d) Identify the spectral series this transition belongs to
(a) $$r_3 = 3^2 \times 0.53 \text{ Å} = 9 \times 0.53 = 4.77 \text{ Å}$$
(b) $$E_3 = \frac{-2.18 \times 10^{-18}}{9} = -2.42 \times 10^{-19} \text{ J}$$
(c) $$E_1 = \frac{-2.18 \times 10^{-18}}{1} = -2.18 \times 10^{-18} \text{ J}$$
$$\Delta E = E_1 – E_3 = -2.18 \times 10^{-18} – (-2.42 \times 10^{-19}) = -1.94 \times 10^{-18} \text{ J}$$
$$\lambda = \frac{hc}{|\Delta E|} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{1.94 \times 10^{-18}} = 1.03 \times 10^{-7} \text{ m} = 103 \text{ nm}$$
(d) This transition ends at n = 1, so it belongs to the Lyman series (UV region). The wavelength of 103 nm confirms this is in the ultraviolet.
IE₁ = 496, IE₂ = 4562, IE₃ = 6912, IE₄ = 9544, IE₅ = 13354
(a) In which group of the periodic table does element X belong?
(b) Write the likely electronic configuration of X.
(c) Identify element X.
(a) Looking at the IEs: IE₁ = 496 (relatively low), then a very large jump to IE₂ = 4562. This means only ONE electron is easily removed (it’s in the outer shell), and the second electron is from an inner shell. Therefore, X belongs to Group 1.
(b) Group 1 elements have 1 valence electron. The configuration would be [noble gas] ns¹. Based on IE₁ = 496 kJ/mol matching sodium, the configuration is: [Ne] 3s¹
(c) Comparing IE₁ = 496 kJ/mol with known values, this matches sodium (Na, Z = 11).
(a) The velocity of the electron
(b) The kinetic energy of the electron
(c) How would the wavelength change if the velocity were doubled?
(a) Using $\lambda = \frac{h}{mv}$:
$$v = \frac{h}{m\lambda} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(1.0 \times 10^{-10})} = 7.28 \times 10^6 \text{ m/s}$$
(b) $$KE = \frac{1}{2}mv^2 = \frac{1}{2}(9.11 \times 10^{-31})(7.28 \times 10^6)^2 = 2.42 \times 10^{-17} \text{ J}$$
(c) Since $\lambda = h/(mv)$, if velocity is doubled, the wavelength is halved (wavelength is inversely proportional to velocity). New λ = 0.5 × 10⁻¹⁰ m = 5.0 × 10⁻¹¹ m.
(a) Phosphorus (Z=15) (b) Manganese (Z=25) (c) Zinc (Z=30) (d) S²⁻ ion
(a) P: [Ne] 3s² 3p³ → 3p has 3 unpaired electrons (↑ ↑ ↑ in three p orbitals). 3 unpaired.
(b) Mn: [Ar] 4s² 3d⁵ → 3d has 5 unpaired electrons (↑ ↑ ↑ ↑ ↑ in five d orbitals). 5 unpaired.
(c) Zn: [Ar] 4s² 3d¹⁰ → All orbitals are fully filled. 0 unpaired.
(d) S²⁻: S is [Ne] 3s² 3p⁴. S²⁻ gains 2 electrons: [Ne] 3s² 3p⁶ = [Ar]. All orbitals fully filled. 0 unpaired.
F, Ne, O, N, C, B, Be, Li
Li < B < Be < C < O < N < F < Ne
Explanation (all are Period 2, so IE generally increases left to right):
• Li (Group 1): lowest IE — only 1 valence electron, far from nucleus
• B (Group 13): lower than Be because its 2p electron is easier to remove than Be’s 2s electron
• Be (Group 2): higher than B — filled 2s² subshell is stable
• C (Group 14): normal trend resumes
• O (Group 16): lower than N — electron-electron repulsion in paired 2p electron
• N (Group 15): higher than O — half-filled 2p³ subshell is extra stable
• F (Group 17): high IE — nearly full shell
• Ne (Group 18): highest IE — completely filled shell, noble gas
Key: The two exceptions are B < Be and O < N.
(a) The threshold frequency
(b) The work function in joules
(c) The maximum kinetic energy of photoelectrons when light of 400 nm is used
(a) $$\nu_0 = \frac{c}{\lambda_0} = \frac{3.0 \times 10^8}{5.0 \times 10^{-7}} = 6.0 \times 10^{14} \text{ Hz}$$
(b) $$E_0 = h\nu_0 = (6.63 \times 10^{-34})(6.0 \times 10^{14}) = 3.98 \times 10^{-19} \text{ J}$$
(c) Energy of 400 nm photon: $$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.0 \times 10^8)}{4.0 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}$$
$$KE_{\max} = E – E_0 = 4.97 \times 10^{-19} – 3.98 \times 10^{-19} = 9.9 \times 10^{-20} \text{ J}$$
(a) The atomic radius of Na⁺ is smaller than that of Ne.
(b) The second ionization energy of sodium is much larger than the first.
(c) Fluorine has a smaller atomic radius than oxygen even though fluorine has more electrons.
(a) Na⁺ has the electron configuration [Ne] — the same as neon. Both have 10 electrons. However, Na⁺ has 11 protons while Ne has only 10 protons. The greater nuclear charge in Na⁺ pulls the 10 electrons closer, making Na⁺ smaller than Ne despite having the same number of electrons.
(b) Na: [Ne] 3s¹. The first electron is removed from the 3s orbital (outer shell, far from nucleus, well-shielded) — relatively easy. Na⁺: [Ne]. The second electron would have to be removed from the 2p subshell (inner shell, closer to nucleus, less shielding) — much more difficult. This is why IE₂ ≫ IE₁.
(c) Although F has one more electron than O, it also has one more proton (9 vs 8). The increased nuclear charge across the period has a stronger effect than the added electron-electron repulsion. The effective nuclear charge increases from O to F, pulling all electrons closer and reducing the radius.
From n = 4, the electron can fall to n = 3, 2, or 1. From n = 3, it can fall to n = 2 or 1. From n = 2, it can fall to n = 1. That gives 6 possible transitions:
Direct transitions from n = 4:
• 4 → 3: $\Delta E = \frac{-2.18 \times 10^{-18}}{9} – \frac{-2.18 \times 10^{-18}}{16} = 1.94 \times 10^{-19}$ J → $\lambda = 1026$ nm (IR, Paschen)
• 4 → 2: $\Delta E = 4.09 \times 10^{-19}$ J → $\lambda = 487$ nm (visible, Balmer)
• 4 → 1: $\Delta E = 2.04 \times 10^{-18}$ J → $\lambda = 97$ nm (UV, Lyman)
Cascade transitions:
• 3 → 2: $\Delta E = 3.03 \times 10^{-19}$ J → $\lambda = 657$ nm (visible, Balmer)
• 3 → 1: $\Delta E = 1.94 \times 10^{-18}$ J → $\lambda = 103$ nm (UV, Lyman)
• 2 → 1: $\Delta E = 1.64 \times 10^{-18}$ J → $\lambda = 122$ nm (UV, Lyman)
Shortest wavelength: The 4 → 1 transition (97 nm) because it has the largest energy change. Since $\lambda = hc/\Delta E$, larger ΔE means shorter λ.