Atoms — Concepts, Formulas & Solved Numericals

What Are Atoms, Really?

Rutherford’s gold foil experiment in 1909 shattered the “plum pudding” model and gave us something far more elegant — the nuclear atom. Most of the atom is empty space, with a tiny, dense nucleus carrying all the positive charge, surrounded by electrons orbiting at relatively vast distances.

Bohr took this further in 1913. He asked: if electrons orbit freely, they’d radiate energy continuously (Maxwell’s equations say so), spiral inward, and atoms would collapse in nanoseconds. Clearly they don’t. So Bohr proposed that electrons exist only in certain allowed orbits where they don’t radiate. This is the quantum restriction that makes atoms stable.

For Class 12 and JEE Main, the Bohr model is your workhorse. It’s not perfectly accurate (it fails for multi-electron atoms), but it gives exact answers for hydrogen-like atoms (H, He⁺, Li²⁺) and those are what exams test. Understand the assumptions, the derivations, and you’ll handle 90% of questions in this chapter.

Key Terms & Definitions

Atomic number (Z): Number of protons in the nucleus. For hydrogen, Z = 1; for helium, Z = 2. In Bohr’s model, Z appears in every formula — it scales the energy and orbit size.

Principal quantum number (n): Labels the allowed orbits. n = 1 is the ground state (lowest energy, most stable). n = 2, 3, 4… are excited states.

Ground state: The lowest energy state of an atom. Hydrogen’s ground state energy is −13.6 eV.

Excited state: When an electron absorbs energy and jumps to a higher orbit (n > 1).

Ionisation energy: The energy required to remove the electron completely from the atom (take it to n = ∞). For hydrogen ground state, this is 13.6 eV.

Emission spectrum: When an electron falls from higher to lower orbit, it emits a photon. The collection of all such photon frequencies forms the emission spectrum.

Absorption spectrum: When white light passes through hydrogen gas, photons matching exact transition energies get absorbed. Same frequencies as emission, but now seen as dark lines.

Hydrogen-like atoms: Single-electron systems — H, He⁺, Li²⁺, Be³⁺. The Bohr model works exactly for these. For multi-electron atoms, it breaks down due to electron-electron repulsion.

Core Concepts & Derivations

Bohr’s Postulates

Bohr’s model rests on three postulates:

Allowed orbits: Electrons move in circular orbits where the angular momentum is quantised: $mvr = \frac{nh}{2\pi} = n\hbar$
No radiation in stable orbits: An electron in an allowed orbit doesn’t emit radiation, despite being in circular motion (which classically would require radiation).
Photon emission/absorption: When an electron transitions from orbit $n_i$ to $n_f$ , the energy of the emitted/absorbed photon equals the energy difference: $h\nu = E_i - E_f$

Radius of nth Orbit

Balancing centripetal force with Coulomb attraction:

\frac{mv^2}{r} = \frac{kZe^2}{r^2}

Combined with the quantisation condition $mvr = n\hbar$ , we get:

r_n = \frac{n^2 a_0}{Z}

where $a_0 = 0.529 \text{ Å} = 0.529 \times 10^{-10} \text{ m}$ is the Bohr radius (radius of H atom in ground state).

Key point: radius scales as $n^2$ and inversely as $Z$ . So He⁺ (Z=2) has orbits half the size of hydrogen for the same n.

Energy of nth Orbit

E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV}

The negative sign means the electron is bound — you need to add energy to free it. At n = ∞, E = 0 (electron just barely free).

Why does energy scale as $Z^2$ ? Higher nuclear charge pulls the electron tighter, lowering the energy more deeply. And $1/n^2$ because higher orbits are farther away, less bound.

Velocity of Electron

v_n = \frac{Z e^2}{2\epsilon_0 h n} = \frac{Z}{n} \times 2.18 \times 10^6 \text{ m/s}

For hydrogen ground state: $v_1 \approx 2.18 \times 10^6$ m/s — about 0.7% of the speed of light. Fast, but non-relativistic, which is why Bohr’s model (ignoring relativity) works reasonably well.

Spectral Series

When electrons transition between orbits, they emit photons of specific frequencies. These group into series:

Series	Final orbit ( $n_f$ )	Region	Exam Relevance
Lyman	1	Ultraviolet	JEE asks wavelength calculations
Balmer	2	Visible	CBSE diagrams; visible lines
Paschen	3	Infrared	Less frequently tested
Brackett	4	Infrared	Rarely tested
Pfund	5	Far IR	Almost never tested

JEE Main has repeatedly asked: “What is the wavelength of the first line of the Lyman series?” and “How many spectral lines are emitted when electrons fall from n=4 to n=1?” Balmer series questions appear in CBSE boards almost every year.

Rydberg Formula

\frac{1}{\lambda} = RZ^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

where $R = 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant.

Number of Spectral Lines

If electrons are distributed across multiple levels up to n, the maximum number of spectral lines possible:

\text{Number of lines} = \frac{n(n-1)}{2}

For n = 4: lines = 4×3/2 = 6. These are transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.

Solved Examples

Example 1 — CBSE Level

Q: Find the radius of the third orbit of the hydrogen atom.

Solution:

Using $r_n = n^2 a_0 / Z$ , with n = 3, Z = 1:

r_3 = \frac{9 \times 0.529}{1} = 4.761 \text{ Å}

That’s it. Just plug n² × 0.529 Å. For hydrogen, Z = 1 always drops out.

Example 2 — CBSE Level

Q: An electron in a hydrogen atom transitions from n = 3 to n = 1. Find the energy of the emitted photon.

Solution:

E_3 = \frac{-13.6}{9} = -1.51 \text{ eV}

E_1 = \frac{-13.6}{1} = -13.6 \text{ eV}

Energy of photon = $E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV}$

The photon has energy 12.09 eV, falling in the Lyman series (UV region).

Example 3 — JEE Main Level

Q: Find the wavelength of the first line of the Balmer series for hydrogen. (R = 1.097 × 10⁷ m⁻¹)

Solution:

First line of Balmer means transition from n = 3 to n = 2.

\frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \times \frac{5}{36}

\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} = 1.524 \times 10^6 \text{ m}^{-1}

\lambda = \frac{1}{1.524 \times 10^6} = 6563 \text{ Å} = 656.3 \text{ nm}

This is the red H-alpha line — the most famous line in astrophysics and a favourite in both CBSE and JEE.

Example 4 — JEE Main Level

Q: A hydrogen atom is in its second excited state. Find (a) the energy, (b) the orbital radius, (c) the velocity of the electron.

Solution:

Second excited state means n = 3 (ground = n=1, first excited = n=2, second excited = n=3).

(a) Energy:

E_3 = \frac{-13.6}{9} = -1.51 \text{ eV}

(b) Radius:

r_3 = 9 \times 0.529 = 4.76 \text{ Å}

(c) Velocity:

v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5 \text{ m/s}

Students confuse “second excited state” with n = 2. Count carefully: ground state is n = 1. First excited is n = 2. Second excited is n = 3. This slip costs marks every year.

Example 5 — JEE Advanced Level

Q: A photon of energy 12.5 eV is incident on a hydrogen atom in the ground state. Will it be absorbed? If not, what happens?

Solution:

Hydrogen can only absorb photons whose energy exactly matches a transition energy. Let’s check transitions from n = 1:

n = 1 → 2: ΔE = 13.6 − 3.4 = 10.2 eV
n = 1 → 3: ΔE = 13.6 − 1.51 = 12.09 eV
n = 1 → 4: ΔE = 13.6 − 0.85 = 12.75 eV
Ionisation: 13.6 eV

12.5 eV matches none of these exactly. The photon will not be absorbed. It passes through the hydrogen atom without interaction.

This is why atomic absorption spectra show sharp lines, not broad bands — only exact energies are absorbed. This appeared as a conceptual question in JEE Advanced 2019.

Example 6 — JEE Advanced Level

Q: For a hydrogen-like ion He⁺, find the minimum energy required to excite it from the ground state to the first excited state.

Solution:

He⁺ has Z = 2.

E_1(\text{He}^+) = -\frac{13.6 \times 4}{1} = -54.4 \text{ eV}

E_2(\text{He}^+) = -\frac{13.6 \times 4}{4} = -13.6 \text{ eV}

Energy needed = $E_2 - E_1 = -13.6 - (-54.4) = 40.8 \text{ eV}$

Notice: He⁺ ground state energy (−54.4 eV) equals hydrogen’s ionisation energy times Z² = 13.6 × 4. The Z² scaling is the core of hydrogen-like atom problems.

Exam-Specific Tips

CBSE Board Exam Strategy

CBSE typically awards 3-5 marks to this chapter. Focus areas:

Derivation of Bohr radius and energy — frequently asked as a 5-mark derivation. Know it cold.
Spectral series diagram — draw it with series names, final orbits, and wavelength regions.
Numerical on wavelength using Rydberg formula — 1-2 numericals guaranteed.

CBSE marking scheme gives separate marks for formula, substitution, and final answer. Even if your arithmetic is wrong, you get 2/3 marks for correct formula and setup. Never leave a numerical blank.

JEE Main Weightage

Atoms + Nuclei together form one chapter in JEE Main pattern. Expect 1-2 questions per paper. High-yield topics:

Energy level calculations for H and He⁺
Number of spectral lines from a given n
Rydberg formula wavelength calculations
Ionisation energy and excitation energy

JEE Main 2023 Shift 2 asked: “The ratio of the shortest wavelength of the Lyman series to the shortest wavelength of the Balmer series is ___.” Answer: the shortest wavelength (series limit) corresponds to $n_i \to \infty$ , giving $1/\lambda = RZ^2/n_f^2$ . For Lyman: $n_f=1$ , Balmer: $n_f=2$ . Ratio = 1:4. Recognise this pattern.

Important Numerical Values to Memorise

Bohr radius: $a_0 = 0.529$ Å
Ground state energy of H: −13.6 eV
Rydberg constant: $R = 1.097 \times 10^7$ m⁻¹
First Balmer line wavelength: 6563 Å (red)
Series limit of Lyman series: 912 Å

Common Mistakes to Avoid

Mistake 1: Wrong excited state numbering “Second excited state” = n = 3, not n = 2. Count from ground state (n=1). This is the most common error in CBSE papers.

Mistake 2: Forgetting Z² in hydrogen-like atoms For He⁺, Li²⁺, energy scales as Z². Students use −13.6/n² for all atoms. Always check: is this hydrogen (Z=1) or a hydrogen-like ion?

Mistake 3: Getting the Rydberg formula direction wrong It must be $1/n_f^2 - 1/n_i^2$ with $n_f < n_i$ for emission. If you write it backwards, you get a negative wavelength — a clear sign something’s wrong.

Mistake 4: Confusing “remove from ground state” with “remove from excited state” Ionisation energy from ground state = 13.6 eV. From n = 2, it’s only 3.4 eV. Read the question: which state is the electron starting from?

Mistake 5: Applying Bohr model to multi-electron atoms Bohr gives exact results only for H, He⁺, Li²⁺ etc. (one electron). Never use −13.6Z²/n² for sodium, oxygen, or any atom with more than one electron.

Practice Questions

Q1. What is the energy of the electron in the third orbit of He⁺?

He⁺ has Z = 2.

E_3 = -\frac{13.6 \times 4}{9} = -\frac{54.4}{9} = -6.04 \text{ eV}

Q2. How many spectral lines are possible when electrons in hydrogen atoms jump from n = 5 to n = 1?

Total lines = $\frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10$

The transitions are: 5→4, 5→3, 5→2, 5→1, 4→3, 4→2, 4→1, 3→2, 3→1, 2→1.

Q3. Find the wavelength of the series limit of the Lyman series for hydrogen.

Series limit means $n_i \to \infty$ , so $1/n_i^2 \to 0$ .

\frac{1}{\lambda} = R\left(\frac{1}{1^2} - 0\right) = R = 1.097 \times 10^7 \text{ m}^{-1}

\lambda = 912 \text{ Å}

Q4. The ionisation energy of He⁺ is 54.4 eV. What is the energy of the photon emitted when He⁺ transitions from n = 3 to n = 2?

Ionisation energy of He⁺ = 54.4 eV = ground state binding energy. So $E_1 = -54.4$ eV.

E_n = \frac{E_1}{n^2}

E_3 = \frac{-54.4}{9} = -6.04 \text{ eV}

E_2 = \frac{-54.4}{4} = -13.6 \text{ eV}

Photon energy = $E_3 - E_2 = -6.04 - (-13.6) = 7.56 \text{ eV}$

Q5. An electron in hydrogen is in the n = 4 state. It falls to the ground state in two steps. Which paths give the maximum and minimum energy first photon?

For maximum energy first photon: go directly as far down as possible in the first step, then finish. The biggest first jump is 4→2 (energy = 13.6(1/4 − 1/16) = 13.6 × 3/16 = 2.55 eV) vs 4→3 (0.66 eV). Actually, 4→1 directly gives the most energy (12.75 eV) in one step, but if forced into two steps:

Maximum first photon: 4→2 (ΔE = 2.55 eV), then 2→1 (ΔE = 10.2 eV) Minimum first photon: 4→3 (ΔE = 0.66 eV), then 3→1 (ΔE = 12.09 eV)

Total energy is always the same (12.75 eV) regardless of path — energy conservation.

Q6. The velocity of an electron in the ground state of hydrogen is $v$ . What is the velocity in the first excited state of Li²⁺?

v_{n,Z} = \frac{Z}{n} \times v_1

For H ground state (Z=1, n=1): $v = v_1$

For Li²⁺ first excited state (Z=3, n=2):

v = \frac{3}{2} v_1 = 1.5v

Q7. A hydrogen atom in the ground state is hit by an electron with kinetic energy 12.1 eV. What state does it go to?

Unlike photons, electrons can give part of their energy. But the atom can only absorb specific amounts. Check transition energies:

Ground → n=2: 10.2 eV ✓ (electron has 12.1 eV, can give 10.2 eV)
Ground → n=3: 12.09 eV ✓ (electron can give 12.09 eV, keeping 0.01 eV as KE)
Ground → n=4: 12.75 eV ✗ (exceeds available energy)

The atom goes to n = 3 (highest accessible level). The electron exits with KE = 12.1 − 12.09 = 0.01 eV.

Q8. In a hydrogen sample, all atoms are in n = 3 state. Find the number of different photon frequencies that can be emitted.

Possible transitions from n = 3: 3→2, 3→1, 2→1.

That’s 3 different frequencies.

Using the formula: $\frac{n(n-1)}{2} = \frac{3 \times 2}{2} = 3$ . ✓

FAQs

Why do electrons not fall into the nucleus?

In classical physics, an accelerating charge radiates energy. An orbiting electron is always accelerating (centripetal), so it should spiral inward and crash into the nucleus in ~10⁻¹¹ seconds. Bohr’s postulate simply forbids radiation in allowed orbits. The deeper explanation comes from quantum mechanics — the electron is not really a particle in a fixed orbit, but a standing wave pattern around the nucleus. The ground state wavefunction has no lower energy configuration to fall into.

Why does the Bohr model fail for helium?

Helium has two electrons. The Coulomb repulsion between these two electrons cannot be handled by Bohr’s simple model, which only accounts for nucleus-electron attraction. You’d need perturbation theory or variational methods from quantum mechanics. Bohr’s model is exact only for single-electron systems.

What is the difference between ionisation energy and excitation energy?

Excitation energy is the energy needed to move the electron from its current state to a higher allowed orbit (not free). Ionisation energy takes the electron all the way to n = ∞ (completely free). For hydrogen ground state: first excitation energy = 10.2 eV; ionisation energy = 13.6 eV.

Is the Balmer series visible to the naked eye?

Yes — the Balmer series falls in the visible region (roughly 400–700 nm). The four visible lines are H-alpha (656 nm, red), H-beta (486 nm, blue-green), H-gamma (434 nm, violet), H-delta (410 nm, violet). The rest of the Balmer series falls in the UV.

What does the negative sign in the energy formula mean?

It means the electron is bound to the nucleus. We set the reference: at n = ∞, the electron is completely free, with E = 0. Any bound state (finite n) has lower energy, hence negative. The more negative the energy, the more tightly bound the electron. Ground state (−13.6 eV) is more tightly bound than n = 2 (−3.4 eV).

Can the Rydberg formula be used for He⁺?

Yes, for any hydrogen-like ion. Just include Z²: $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ . For He⁺, Z = 2, so all wavelengths are 1/4 of the corresponding hydrogen wavelengths.

What is the significance of the Lyman series limit?

The series limit (912 Å for Lyman) is the shortest wavelength in that series, corresponding to transitions from $n = \infty$ to $n = 1$ . It marks the onset of ionisation from the ground state — any photon with λ < 912 Å can ionise ground-state hydrogen. This is why the interstellar medium strongly absorbs UV radiation below this wavelength.

How many orbits does hydrogen have?

Technically, infinitely many. There are allowed orbits for every positive integer n = 1, 2, 3, … stretching to infinity (though for large n they merge into the continuum). In practice, for problems, you only work with n = 1 through n = 6 or so.