Structure of Atom
Welcome to HSLC Guru. In this article, we present a comprehensive English-medium study guide for ASSEB Class 11 Chemistry Chapter 2 — Structure of Atom. This chapter takes you on a fascinating journey from the discovery of subatomic particles to the modern quantum mechanical model of the atom. You will learn about cathode rays, Thomson’s plum pudding model, Rutherford’s gold foil experiment, Bohr’s atomic model, the hydrogen spectrum, dual nature of matter, Heisenberg’s uncertainty principle, quantum numbers, orbitals, and rules for filling electrons. The article includes detailed theory, solved numerical problems on Bohr’s energy, wavelength and wavenumber, MCQs, fill-in-the-blanks, true/false questions, glossary and a formula table — all aligned strictly with the ASSEB syllabus.
Chapter Summary
The journey into the structure of atom began with the study of cathode rays in discharge tubes at very low pressure and high voltage. These rays travel in straight lines, get deflected by electric and magnetic fields, and consist of negatively charged particles called electrons. J. J. Thomson determined the charge-to-mass ratio (e/m) of the electron and proposed the plum pudding model, in which the atom was visualised as a sphere of positive charge with electrons embedded in it. Robert Millikan’s oil-drop experiment then provided the precise charge of an electron (1.602 × 10⁻¹⁹ C). Goldstein’s discovery of anode rays (canal rays) led to the identification of the proton, while Chadwick later discovered the neutral neutron.
Rutherford’s famous α-particle scattering experiment, in which alpha particles were bombarded on a thin gold foil, revealed that most of the atom is empty space, the entire positive charge and nearly all the mass are concentrated in a tiny nucleus, and electrons revolve around this nucleus. The atomic number (Z) equals the number of protons, and the mass number (A) equals protons + neutrons. Isotopes have the same Z but different A (e.g., ¹H, ²H, ³H), isobars have the same A but different Z (e.g., ⁴⁰Ar, ⁴⁰K, ⁴⁰Ca), and isotones have the same number of neutrons. Rutherford’s model, however, could not explain the stability of the atom or the line spectrum of hydrogen.
To overcome these limitations, scientists studied electromagnetic radiation characterised by wavelength (λ), frequency (ν), wavenumber (ν̄) and velocity c = νλ. Max Planck’s quantum theory proposed that energy is emitted or absorbed in discrete packets called quanta, with E = hν. Einstein extended this to explain the photoelectric effect, where light striking a metal surface ejects electrons only if its frequency exceeds a threshold (ν₀); the kinetic energy of ejected electrons is ½mv² = hν − hν₀. Niels Bohr combined these ideas to propose his model: electrons revolve in fixed circular orbits of definite energy (stationary states), angular momentum is quantised (mvr = nh/2π), and energy is emitted or absorbed only when an electron jumps between orbits.
Bohr’s model successfully explained the hydrogen spectrum, which consists of distinct series: Lyman (UV, n₁=1), Balmer (visible, n₁=2), Paschen (IR, n₁=3), Brackett (IR, n₁=4) and Pfund (far-IR, n₁=5). Louis de Broglie proposed the dual nature of matter with λ = h/mv. Heisenberg’s uncertainty principle states that the position and momentum of a particle cannot be simultaneously measured exactly. The quantum mechanical model uses four quantum numbers: principal (n), azimuthal (l), magnetic (m) and spin (s). Orbitals are designated s, p, d, f with shapes spherical, dumb-bell, double dumb-bell and complex. Electrons fill orbitals according to the Aufbau principle, Hund’s rule of maximum multiplicity and the Pauli exclusion principle, giving the electronic configurations of atoms (e.g., Cr: [Ar]3d⁵4s¹, Cu: [Ar]3d¹⁰4s¹).
Key Formulae (KaTeX)
$E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\text{J}$
$r_n = \frac{n^2 h^2}{4\pi^2 m e^2 k}$
$\bar{\nu} = R_H\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)$
$\lambda = \frac{h}{mv}$
$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$
$E = h\nu$
$\hat{H}\psi = E\psi$
1-Mark Questions
Q1. What is a cathode ray?
Answer: Cathode rays are streams of negatively charged particles (electrons) emitted from the cathode of a discharge tube operated at low pressure and high voltage.
Q2. Who discovered the proton?
Answer: The proton was discovered by E. Goldstein during his experiments on anode (canal) rays.
Q3. Define atomic number.
Answer: Atomic number (Z) is the number of protons present in the nucleus of an atom; in a neutral atom it also equals the number of electrons.
Q4. Define mass number.
Answer: Mass number (A) is the total number of protons and neutrons (collectively called nucleons) present in the nucleus of an atom.
Q5. What are isotopes?
Answer: Isotopes are atoms of the same element that have the same atomic number but different mass numbers, e.g., ¹H, ²H, ³H.
Q6. What are isobars?
Answer: Isobars are atoms of different elements that have the same mass number but different atomic numbers, e.g., ⁴⁰Ar, ⁴⁰K, ⁴⁰Ca.
Q7. State Planck’s quantum theory in one line.
Answer: Energy is emitted or absorbed not continuously but in small discrete packets called quanta, with energy E = hν.
Q8. What is the value of Planck’s constant?
Answer: Planck’s constant h = 6.626 × 10⁻³⁴ J·s.
Q9. Name the spectral series of hydrogen lying in the visible region.
Answer: The Balmer series of hydrogen lies in the visible region.
Q10. Write the maximum number of electrons in a shell with principal quantum number n.
Answer: The maximum number of electrons in a shell is given by 2n².
2-3 Mark Questions
Q1. Distinguish between cathode rays and anode rays.
Answer: (i) Cathode rays consist of negatively charged electrons; anode rays consist of positively charged ions of the gas. (ii) The e/m ratio of cathode rays is the same for all gases, but for anode rays it depends on the gas used. (iii) Cathode rays are deflected towards the positive plate; anode rays are deflected towards the negative plate.
Q2. State three observations of Rutherford’s α-scattering experiment and the conclusion drawn.
Answer: (i) Most α-particles passed straight through the gold foil — atom is mostly empty space. (ii) A few were deflected through small angles — there is a positively charged centre. (iii) Very few (1 in 20,000) bounced back — the entire positive charge and almost all mass are concentrated in a tiny dense nucleus.
Q3. Define isotopes, isobars and isotones with one example each.
Answer: Isotopes — same Z, different A (¹H, ²H, ³H). Isobars — same A, different Z (⁴⁰Ar, ⁴⁰K, ⁴⁰Ca). Isotones — same number of neutrons (¹⁴C and ¹⁶O both have 8 neutrons).
Q4. Explain the photoelectric effect.
Answer: When light of frequency greater than a threshold frequency (ν₀) strikes a metal surface, electrons are ejected. The kinetic energy of the ejected electrons is ½mv² = hν − hν₀. The number of electrons ejected depends on the intensity of light, but their kinetic energy depends only on its frequency.
Q5. State the postulates of Bohr’s model of the atom.
Answer: (i) Electrons revolve around the nucleus only in certain permitted circular orbits called stationary states. (ii) The angular momentum of the electron is quantised: mvr = nh/2π. (iii) Energy is emitted or absorbed only when an electron jumps from one orbit to another, ΔE = E₂ − E₁ = hν.
Q6. Write any three limitations of Bohr’s model.
Answer: (i) It applies only to one-electron systems (H, He⁺, Li²⁺); fails for multi-electron atoms. (ii) It cannot explain fine structure of spectral lines or the Zeeman and Stark effects. (iii) It violates Heisenberg’s uncertainty principle by assigning fixed paths to electrons.
5-7 Mark Questions (with Numericals)
Q1. Calculate the energy of an electron in the second Bohr orbit of a hydrogen atom.
Answer: Using $E_n = -\frac{2.18 \times 10^{-18}}{n^2}\,\text{J}$, for n = 2: E₂ = −2.18 × 10⁻¹⁸ / 4 = −5.45 × 10⁻¹⁹ J. The negative sign indicates that the electron is bound to the nucleus. The energy required to remove the electron from this orbit is +5.45 × 10⁻¹⁹ J.
Q2. Calculate the wavenumber and wavelength of the spectral line emitted when an electron in a hydrogen atom jumps from n = 3 to n = 2 (R_H = 1.097 × 10⁷ m⁻¹).
Answer: Using $\bar{\nu} = R_H\left(\frac{1}{n_1^2} – \frac{1}{n_2^2}\right)$ with n₁ = 2, n₂ = 3: ν̄ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × (5/36) = 1.524 × 10⁶ m⁻¹. Wavelength λ = 1/ν̄ = 6.56 × 10⁻⁷ m = 656 nm. This is the H-α line of the Balmer series in the visible region (red).
Q3. Calculate the de Broglie wavelength associated with an electron moving with a velocity of 1.0 × 10⁶ m/s. (mass of electron = 9.11 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J·s)
Answer: Using $\lambda = \frac{h}{mv}$: λ = (6.626 × 10⁻³⁴) / (9.11 × 10⁻³¹ × 1.0 × 10⁶) = 7.27 × 10⁻¹⁰ m = 7.27 Å. This wavelength is comparable to atomic dimensions, which is why electrons are diffracted by crystal lattices, confirming their wave nature.
Q4. Explain the four quantum numbers used to describe an electron in an atom.
Answer: (i) Principal quantum number (n) — n = 1, 2, 3 …; gives main energy level/shell, size and energy of orbital. (ii) Azimuthal/Subsidiary (l) — l = 0 to (n−1); gives shape of orbital and subshell (s, p, d, f). (iii) Magnetic (m) — m = −l … 0 … +l; gives spatial orientation of orbital. (iv) Spin (s) — s = +½ or −½; describes the spin direction of the electron. Together they specify the complete state of an electron.
Q5. State and explain Aufbau principle, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity. Hence write the electronic configuration of Cr (Z = 24) and Cu (Z = 29).
Answer: Aufbau: Electrons fill orbitals in order of increasing (n + l) energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d… Pauli: No two electrons in an atom can have all four quantum numbers identical; hence an orbital can hold a maximum of 2 electrons with opposite spins. Hund’s rule: Electrons fill degenerate orbitals singly with parallel spins before pairing. Cr (24): [Ar] 3d⁵ 4s¹ (half-filled stability); Cu (29): [Ar] 3d¹⁰ 4s¹ (fully-filled stability).
Multiple Choice Questions (MCQs)
Q1. The discoverer of the electron is —
(a) Rutherford (b) Goldstein (c) J. J. Thomson (d) Chadwick
Answer: (c) J. J. Thomson.
Q2. The charge on an electron is —
(a) 1.6 × 10⁻¹⁹ C (b) 9.11 × 10⁻³¹ C (c) 1.67 × 10⁻²⁷ C (d) zero
Answer: (a) 1.6 × 10⁻¹⁹ C.
Q3. Which series of the H-spectrum lies in the ultraviolet region?
(a) Balmer (b) Paschen (c) Lyman (d) Brackett
Answer: (c) Lyman series.
Q4. The shape of a p-orbital is —
(a) spherical (b) dumb-bell (c) double dumb-bell (d) complex
Answer: (b) dumb-bell.
Q5. The maximum number of electrons in the M-shell is —
(a) 8 (b) 18 (c) 32 (d) 2
Answer: (b) 18 (using 2n² with n = 3).
Q6. The principle that no two electrons in an atom have the same set of four quantum numbers is —
(a) Aufbau (b) Hund (c) Pauli (d) Heisenberg
Answer: (c) Pauli’s exclusion principle.
Q7. The de Broglie equation is —
(a) E = mc² (b) λ = h/mv (c) E = hν (d) Δx·Δp ≥ h/4π
Answer: (b) λ = h/mv.
Q8. Number of orbitals in n = 3 is —
(a) 3 (b) 6 (c) 9 (d) 18
Answer: (c) 9 (= n²).
Q9. The electronic configuration of Cr (Z = 24) is —
(a) [Ar]3d⁴4s² (b) [Ar]3d⁵4s¹ (c) [Ar]3d⁶ (d) [Ar]4s²4p⁴
Answer: (b) [Ar]3d⁵4s¹.
Q10. The energy of an electron in the first Bohr orbit of H atom is —
(a) −13.6 eV (b) +13.6 eV (c) −3.4 eV (d) zero
Answer: (a) −13.6 eV.
Fill in the Blanks
Q1. The neutron was discovered by ________.
Answer: James Chadwick.
Q2. The energy of a quantum is given by E = ________.
Answer: hν.
Q3. The Balmer series of hydrogen lies in the ________ region.
Answer: visible.
Q4. The maximum number of electrons in an orbital is ________.
Answer: two (with opposite spins).
Q5. The azimuthal quantum number for a d-orbital is ________.
Answer: l = 2.
True / False
Q1. Cathode rays travel in straight lines.
Answer: True.
Q2. Isobars have the same number of protons.
Answer: False — isobars have the same mass number but different atomic numbers.
Q3. The angular momentum of an electron in a Bohr orbit is quantised.
Answer: True.
Q4. Heisenberg’s uncertainty principle is significant for macroscopic objects.
Answer: False — it is significant only for microscopic particles like electrons.
Q5. An s-orbital is spherically symmetrical.
Answer: True.
Glossary
| Term | Meaning |
|---|---|
| Cathode ray | Stream of electrons emitted from cathode in a discharge tube |
| Anode ray | Stream of positive ions formed in a discharge tube |
| Atomic number (Z) | Number of protons in the nucleus |
| Mass number (A) | Total number of protons and neutrons |
| Isotopes | Same Z, different A |
| Isobars | Same A, different Z |
| Isotones | Same number of neutrons |
| Quantum | Smallest discrete packet of energy |
| Photoelectric effect | Emission of electrons by metals on illumination by light of suitable frequency |
| Threshold frequency | Minimum frequency required to eject electrons (ν₀) |
| Stationary state | Permitted Bohr orbit of definite energy |
| Spectral series | Group of spectral lines obtained by transitions to a fixed lower orbit |
| de Broglie wavelength | λ = h/mv associated with a moving particle |
| Uncertainty principle | Δx·Δp ≥ h/4π |
| Orbital | Region of space with high probability of finding an electron |
| Aufbau principle | Electrons fill orbitals in order of increasing (n + l) energy |
| Hund’s rule | Pair electrons in degenerate orbitals only after each is singly occupied |
| Pauli exclusion principle | No two electrons in an atom have identical four quantum numbers |
Formula Table
| Quantity | Formula |
|---|---|
| Energy of n-th Bohr orbit (H) | E_n = −2.18 × 10⁻¹⁸ / n² J = −13.6/n² eV |
| Radius of n-th Bohr orbit | r_n = n²h² / (4π²me²k); r₁ = 0.529 Å |
| Wavenumber (Rydberg) | ν̄ = R_H (1/n₁² − 1/n₂²); R_H = 1.097 × 10⁷ m⁻¹ |
| Energy of photon | E = hν = hc/λ |
| Photoelectric KE | ½mv² = hν − hν₀ |
| de Broglie wavelength | λ = h/mv = h/p |
| Heisenberg uncertainty | Δx·Δp ≥ h/4π |
| Schrödinger equation | Ĥψ = Eψ |
| Maximum electrons in shell | 2n² |
| Number of orbitals in shell | n² |
| Angular momentum | mvr = nh/2π |
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