Class 11 Chemistry Chapter 2 notes, in-text problems (2.12.18), NCERT exercises (2.1 to 2.67), revision formula sheets, and chapter tests.
Introduction: From Dalton to Quantum Mechanics
Atom (400 B.C., Indian & Greek philosophers): from Greek 'a-tomio' = 'uncut-able' — the indivisible building block of matter (then only a speculation).
Dalton's atomic theory (1808): atom = ultimate, indivisible particle of matter. Explained laws of conservation of mass, constant composition & multiple proportions.
Failure: could not explain why glass/ebonite get charged on rubbing.
Late 19th century: experiments proved atoms are divisible → made of sub-atomic particles: electrons, protons, neutrons.
2.1 Discovery of Sub-atomic Particles
Discovered from electrical discharge through gases (cathode ray tubes). Basic rule: "Like charges repel; unlike charges attract."
2.1.1 Discovery of Electron
Discovery: electrical discharge through a gas at very low pressure and very high voltage (cathode ray tube, mid-1850s) produces cathode rays — a stream of particles moving from cathode (−) to anode (+). (Faraday, 1830, had shown electricity has a particulate nature.)
Cathode Ray Discharge Tube — 3D View
Results of the Cathode Ray Experiments
(i) The cathode rays start from the cathode and move towards the anode.
(ii) The rays themselves are not visible, but their behaviour can be observed with certain materials (fluorescent or phosphorescent) which glow when hit by them. Television picture tubes are cathode ray tubes.
(iii) In the absence of an electrical or magnetic field, these rays travel in straight lines.
(iv) In the presence of an electrical or magnetic field, they behave like negatively charged particles, called electrons.
(v) The characteristics of cathode rays (electrons) do not depend upon the material of the electrodes or the nature of the gas present in the tube.
Key Conclusion
Since the characteristics of electrons are independent of the electrode material and the gas taken, electrons are the basic constituents of all atoms.
Mnemonic: Cathode Ray Properties
"S.V.N.D. — Straight, Visible, Negative, Don't-care" → rays travel in Straight lines, are Visible via ZnS glow, are Negative (deflect in fields), and Don't depend on gas or electrode material. ⚡ In one line: electron = universal building block of every atom.
2.1.2 Charge to Mass Ratio of Electron
Measurement: Thomson (1897) applied electric and magnetic fields perpendicular to the electron beam and balanced them to restore the original path. Deflection depends on:
Charge — greater charge → greater deflection.
Mass — lighter particle → greater deflection.
Field strength — higher voltage / stronger field → greater deflection.
Result:
$$\frac{e}{m_e} = 1.758820 \times 10^{11}\text{ C kg}^{-1}$$
Charge on the electron = $-e$ (negative).
2.1.3 Charge on the Electron (Millikan's Oil Drop Experiment)
Oil drop experiment (190614): X-rays ionise the air; oil droplets from an atomiser pick up charge by colliding with gaseous ions; their fall between charged condenser plates is watched through a microscope and controlled by the electric field.
Result: charge on every droplet is an integral multiple of e — q = n·e (n = 1, 2, 3...).
Millikan's Oil Drop Apparatus — 3D View
Charge on the electron: $-1.602176 \times 10^{-19}$ C (Millikan found $-1.6 \times 10^{-19}$ C)
Mass of the electron: $m_e = \dfrac{e}{e/m_e} = 9.1094 \times 10^{-31}$ kg
2.1.4 Discovery of Protons and Neutrons
Electrical discharge carried out in a modified cathode ray tube led to the discovery of canal rays carrying positively charged particles. Their characteristics:
Unlike cathode rays, the mass of the positively charged particles depends upon the nature of the gas in the tube — they are simply positively charged gaseous ions.
The charge to mass ratio of the particles depends on the gas from which these originate.
Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
Their behaviour in magnetic or electrical fields is opposite to that observed for electrons (cathode rays).
Proton: smallest & lightest positive ion, obtained from hydrogen (characterised 1919). Neutron: electrically neutral particle, discovered by Chadwick (1932) by bombarding beryllium with α-particles; mass slightly greater than proton.
Table 2.1: Properties of Fundamental Particles
Name
Symbol
Absolute charge (C)
Relative charge
Mass (kg)
Mass (u)
Electron
$e$
−1.602176×10⁻¹⁹
−1
9.109382×10⁻³¹
0.00054
Proton
$p$
+1.602176×10⁻¹⁹
+1
1.6726216×10⁻²⁷
1.00727
Neutron
$n$
0
0
1.674927×10⁻²⁷
1.00867
Mnemonic: Who Found What
"E-Tom, P-Gold, N-Chad" → ElectronTomson (1897), ProtonGoldstein's canal rays (1919), NeutronChadwick (1932, Be + α). Masses: p & n are ~1 u "twins"; e is a 0.00054 u feather. ⚡ In one line: A = Z + n (protons + neutrons = nucleons).
2.2 Atomic Models
After sub-atomic particles were discovered, scientists needed to explain: the stability of the atom • comparison of elements' physical & chemical properties • molecule formation • origin and nature of electromagnetic radiation absorbed/emitted by atoms. → led to atomic models.
2.2.1 Thomson Model of Atom
Postulate (1898): atom = a sphere of uniform positive charge (radius ~10⁻¹⁰ m) with electrons embedded in it in the most stable electrostatic arrangement — the "plum pudding / watermelon model".
Thomson's 'Plum Pudding' Model — 3D Sphere of Positive Charge
Mass: assumed uniformly distributed over the atom.
Success: explains overall neutrality of the atom.
Failure: contradicted by Rutherford's α-scattering results.
X-rays & Radioactivity — key facts
X-rays (Röentgen, 1895): produced when fast electrons strike a dense metal target; not deflected by fields; λ ~ 0.1 nm; very high penetrating power.
Radioactivity (Becquerel): spontaneous emission of radiation by heavy elements — developed by the Curies, Rutherford, Soddy.
α-rays = He²⁺ nuclei (2+ charge, mass 4 u); β-rays = electrons; γ-rays = neutral EM radiation.
Penetrating power: α < β (100×) < γ (1000×).
2.2.2 Rutherford's Nuclear Model of Atom
Experiment (1909): Rutherford, Geiger & Marsden bombarded a thin gold foil (~100 nm) with high-energy α-particles; a ZnS screen around the foil flashed at each impact.
Rutherford's α-particle Scattering Experiment
Interactive 3D Simulation: Rutherford α-Particle Scattering
Observations (quite unexpected)
(i) Most of the α-particles passed through the gold foil undeflected.
(ii) A small fraction of the α-particles was deflected by small angles.
(iii) A very few α-particles (~1 in 20,000) bounced back, i.e., were deflected by nearly 180°.
Conclusions
(i) Most of the atom is empty space (most α-particles passed straight through).
(ii) Positive charge is concentrated in a tiny volume (nucleus) — only that could repel and deflect α-particles.
(iii)ratom ≈ 10⁻¹⁰ m; rnucleus ≈ 10⁻¹⁵ m — if a cricket ball were a nucleus, the atom would be ~5 km across!
Rutherford's Nuclear Model of Atom
All positive charge and nearly all mass → densely packed in a tiny central nucleus.
Electrons revolve around the nucleus in circular orbits (solar-system model: nucleus = sun, electrons = planets).
Electrostatic attraction holds electrons and nucleus together.
Rutherford's Nuclear Atom — 3D View (Solar-System Model)
2.1
Calculate the number of protons, neutrons and electrons in ⁸⁰₃₅Br.
Step 1: Identify Z and A
For ⁸⁰₃₅Br: $Z = 35$, $A = 80$, and the species is neutral.
Step 2: Apply the Relations
Number of protons = number of electrons = $Z = \mathbf{35}$
Number of neutrons $= A - Z = 80 - 35 = \mathbf{45}$
Protons = 35, Neutrons = 45, Electrons = 35
2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Step 1: Identify the Element
Atomic number = number of protons = 16 → the element is sulphur (S).
Step 2: Find the Mass Number
Atomic mass number = protons + neutrons = 16 + 16 = 32
Step 3: Determine the Charge
The species is not neutral (protons ≠ electrons). It is an anion with charge = excess electrons = 18 − 16 = 2.
Symbol is $^{32}_{16}\text{S}^{2-}$
Note: Before using the notation $^{A}_{Z}X$, find out whether the species is a neutral atom, a cation or an anion. The number of neutrons is always given by $A - Z$, whether the species is neutral or an ion.
2.2.3 Atomic Number and Mass Number
Atomic number (Z): number of protons in the nucleus (= number of electrons in a neutral atom). Protons + neutrons = nucleons; their total = mass number (A).
Atomic number (Z) = number of protons = number of electrons (neutral atom)
Mass number (A) = number of protons (Z) + number of neutrons (n)
2.2.4 Isobars and Isotopes
The composition of any atom is represented using the element symbol (X) with the atomic mass number (A) as superscript and the atomic number (Z) as subscript on the left hand side ($^{A}_{Z}X$).
Isobars
Atoms with the same mass number (A) but different atomic numbers (Z). Example: $^{14}_{6}\text{C}$ and $^{14}_{7}\text{N}$; $^{40}_{18}\text{Ar}$, $^{40}_{19}\text{K}$, $^{40}_{20}\text{Ca}$.
Isotopes
Atoms with identical atomic number (Z) but different mass numbers (A) — difference is due to neutrons. Example: $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$; $^{12}_{6}\text{C}, ^{13}_{6}\text{C}, ^{14}_{6}\text{C}$.
Isotones
Species having the same number of neutrons ($A - Z$) but different atomic and mass numbers.
• $^{14}_{6}\text{C}$ ($n = 14 - 6 = 8$) and $^{16}_{8}\text{O}$ ($n = 16 - 8 = 8$)
• $^{31}_{15}\text{P}$ and $^{32}_{16}\text{S}$ ($n = 16$ each).
Isoelectronic Species
Atoms, molecules, or ions having the same total number of electrons.
• 10-electron series: $\text{N}^{3-}, \text{O}^{2-}, \text{F}^{-}, \text{Ne}, \text{Na}^{+}, \text{Mg}^{2+}, \text{Al}^{3+}$
• Molecules: $\text{CH}_4, \text{NH}_3, \text{H}_2\text{O}, \text{HF}$ (all have 10 $e^-$); $\text{CO}, \text{N}_2, \text{CN}^{-}$ (14 $e^-$).
Isodiaphers
Species having the same neutron excess ($n - p$ or $A - 2Z$).
• $^{238}_{92}\text{U}$ ($n - p = 146 - 92 = 54$) and $^{234}_{90}\text{Th}$ ($n - p = 144 - 90 = 54$).
Chemical properties of atoms are controlled by the number of electrons, which is determined by the number of protons in the nucleus. Neutrons have very little effect on chemical properties — therefore, all the isotopes of a given element show the same chemical behaviour.
Mnemonic: Isotope vs Isobar vs Isotone
"P.A.N." → IsotoPe = same Protons • IsobAr = same A (mass number) • IsotoNe = same Neutrons. ⚡ In one line: isotopes of an element → identical chemistry, different mass.
2.2.5 Drawbacks of Rutherford Model
Instability: a revolving electron is an accelerated charge → must continuously radiate energy (Maxwell's theory) and spiral into the nucleus in ~10⁻⁸ s. But atoms are stable!
Stationary electrons? Would simply be pulled into the nucleus by attraction.
Silent about how electrons are distributed around the nucleus and what energies they have.
2.3 Developments Leading to Bohr's Model of Atom
Historically, results observed from the studies of interactions of radiations with matter have provided immense information regarding the structure of atoms and molecules. Neils Bohr utilised these results to improve upon the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr's model:
(i) Dual character of the electromagnetic radiation — radiations possess both wave-like and particle-like properties, and
(ii) Experimental results regarding atomic spectra (Section 2.3.3).
2.3.1 Wave Nature of Electromagnetic Radiation
Electromagnetic radiation (Maxwell, 1870; confirmed by Hertz): an accelerating charged particle produces alternating electric & magnetic fields that travel as waves — no medium needed.
Important Properties of Electromagnetic Waves
The oscillating electric and magnetic fields are perpendicular to each other, and both are perpendicular to the direction of propagation of the wave.
Unlike sound waves or waves on water, electromagnetic waves do not require a medium and can move through vacuum.
Types differ in wavelength (or frequency) → together form the electromagnetic spectrum: radio (~10⁶ Hz), microwave (~10¹⁰ Hz), IR (~10¹³ Hz), visible (~10¹⁵ Hz — only region our eyes detect), UV (~10¹⁶ Hz).
Electromagnetic Wave — Perpendicular E & B Fields (3D View)
Characterisation of Electromagnetic Radiation
Frequency (ν): the number of waves that pass a given point in one second. SI unit: hertz (Hz, s⁻¹), after Heinrich Hertz.
Wavelength (λ): SI unit metre (m); smaller units like cm, nm (10⁻⁹ m), Å (10⁻¹⁰ m) and pm (10⁻¹² m) are used for the much smaller wavelengths of electromagnetic radiation.
In vacuum, all types of electromagnetic radiation, regardless of wavelength, travel at the same speed — the speed of light, $c = 3.0 \times 10^{8}$ m s⁻¹ (precisely $2.997925 \times 10^{8}$ m s⁻¹).
$$c = \nu\lambda$$
The other commonly used quantity, especially in spectroscopy, is the wavenumber (ν̄) — the number of wavelengths per unit length. Its SI unit is m⁻¹, but the commonly used unit is cm⁻¹ (not SI).
Mnemonic: EM Spectrum (increasing frequency)
"Raging Martians Invaded Venus Using X-ray Guns" → Radio, Microwave, Infrared, Visible, Ultraviolet, X-rays, Gamma. ⚡ In one line: c = νλ; visible light ≈ 10¹⁵ Hz; higher ν → higher E = hν.
2.3
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to?
Step 1: Formula
The wavelength $\lambda = \dfrac{c}{\nu}$, where $c$ is the speed of electromagnetic radiation in vacuum.
λ = 219.3 m — a characteristic radiowave wavelength.
2.4
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1 nm = 10⁻⁹ m)
Using equation $c = \nu\lambda$:
Frequency of Violet Light (400 nm)
$$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}\text{ m s}^{-1}}{400 \times 10^{-9}\text{ m}} = \mathbf{7.50 \times 10^{14}\text{ Hz}}$$ Frequency of Red Light (750 nm)
$$\nu = \frac{3.0 \times 10^{8}\text{ m s}^{-1}}{750 \times 10^{-9}\text{ m}} = \mathbf{4.00 \times 10^{14}\text{ Hz}}$$
The range of the visible spectrum is from 4.0×10¹⁴ to 7.5×10¹⁴ Hz in frequency units.
2.5
Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.
(a) Calculation of Wavenumber (ν̄)
$\lambda = 5800\text{ Å} = 5800 \times 10^{-8}$ cm
$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5800 \times 10^{-8}\text{ cm}} = \mathbf{1.724 \times 10^{4}\text{ cm}^{-1}}$$ (b) Calculation of Frequency (ν)
$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8}\text{ m s}^{-1}}{5800 \times 10^{-10}\text{ m}} = \mathbf{5.172 \times 10^{14}\text{ s}^{-1}}$$
2.3.2 Particle Nature of Electromagnetic Radiation: Planck's Quantum Theory
Wave theory (classical physics) failed to explain:
(i) the nature of emission of radiation from hot bodies (black-body radiation)
(ii) ejection of electrons from a metal surface when radiation strikes it (photoelectric effect)
(iii) variation of heat capacity of solids as a function of temperature
(iv) line spectra of atoms with special reference to hydrogen
* Diffraction = bending of a wave around an obstacle. ** Interference = combination of two waves.
→ Energy is absorbed or emitted only in discrete amounts (quantised), never continuously.
Black Body Radiation
Black body: an ideal body that absorbs and emits radiation of all frequencies uniformly — approximated by a cavity with a tiny hole (carbon black is close). It radiates the same energy per unit area as it absorbs.
Observation: intensity vs wavelength rises to a maximum then falls; as temperature rises, the maximum shifts to shorter λ (hot iron: dull red → red → white → blue). Wave theory could not explain this — Max Planck (1900) did:
Planck's Quantum Theory
Postulate: atoms/molecules emit or absorb energy only in discrete quantities — the smallest packet of EM energy is a quantum, whose energy is proportional to its frequency:
$$E = h\nu$$
where Planck's constant $h = 6.626 \times 10^{-34}$ J s.
Quantisation: The Staircase Analogy
Quantisation has been compared to standing on a staircase — a person can stand on any step but not in between two steps. The energy can take any one of the values $E = 0, h\nu, 2h\nu, 3h\nu...nh\nu...$, but cannot take on any values between them.
Photoelectric Effect
In 1887, H. Hertz performed an experiment in which electrons (electric current) were ejected when certain metals (for example potassium, rubidium, caesium) were exposed to a beam of light. This phenomenon is called the Photoelectric Effect. The observed results were:
(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface — there is no time lag between the striking of the light beam and the ejection of electrons.
(ii) The number of electrons ejected is proportional to the intensity (brightness) of light.
(iii) For each metal, there is a characteristic minimum frequency, ν₀ (threshold frequency), below which the photoelectric effect is not observed. At ν > ν₀, the ejected electrons come out with certain kinetic energy, which increases with the increase of frequency of the light used.
Classical physics fails: it predicted both the number and K.E. of ejected electrons to depend on brightness. Actually, K.E. depends only on frequency — red light (ν = 4.34.6 × 10¹⁴ Hz) never ejects electrons from potassium (ν₀ = 5.0 × 10¹⁴ Hz), but even weak yellow light does.
Einstein (1905): light = stream of photons; each photon transfers its energy hν instantaneously to one electron. Minimum energy needed to eject an electron = hν₀ (work function, W₀):
More intense light → more photons → more electrons ejected (same K.E. each). W₀ values: Li 2.42 eV, Na 2.3 eV, K 2.25 eV, Mg 3.7 eV, Cu 4.8 eV, Ag 4.3 eV.
Mnemonic: Photoelectric Effect
"Income = Rent + Savings" → hν (photon income) = hν₀ (rent to escape = work function) + ½mv² (savings = K.E.). "I→N, F→E" → Intensity decides the Number of electrons; Frequency decides their Energy. ⚡ In one line: ν < ν₀ → nothing happens, however bright; ejection has zero time lag.
Dual Behaviour of Electromagnetic Radiation
Light has dual nature:wave-like while propagating (interference, diffraction) + particle-like while interacting with matter (black body, photoelectric effect). Electrons and other microscopic particles also show this wave-particle duality.
2.6
Calculate the energy of one mole of photons of radiation whose frequency is 5×10¹⁴ Hz.
Step 1: Energy of One Photon
$E = h\nu = (6.626 \times 10^{-34}\text{ J s}) \times (5 \times 10^{14}\text{ s}^{-1}) = 3.313 \times 10^{-19}$ J
Step 2: Energy of One Mole of Photons
$$E = (3.313 \times 10^{-19}\text{ J}) \times (6.022 \times 10^{23}\text{ mol}^{-1})$$
E = 199.51 kJ mol⁻¹
2.7
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Step 1: Power of the Bulb
Power = 100 watt = 100 J s⁻¹
Step 2: Energy of One Photon
$$E = h\nu = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}\text{ J s})(3 \times 10^{8}\text{ m s}^{-1})}{400 \times 10^{-9}\text{ m}} = 4.969 \times 10^{-19}\text{ J}$$ Step 3: Number of Photons per Second
$$N = \frac{100\text{ J s}^{-1}}{4.969 \times 10^{-19}\text{ J}}$$
N = 2.012 × 10²⁰ photons per second
2.8
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×10⁵ J mol⁻¹. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Step 1: Energy of a 300 nm Photon
$$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}\text{ J s})(3.0 \times 10^{8}\text{ m s}^{-1})}{300 \times 10^{-9}\text{ m}} = 6.626 \times 10^{-19}\text{ J}$$ Step 2: Energy of One Mole of Photons
$= 6.626 \times 10^{-19} \times 6.022 \times 10^{23} = 3.99 \times 10^{5}$ J mol⁻¹
Step 3: Work Function (Minimum Energy)
Minimum energy to remove one mole of electrons $= (3.99 - 1.68) \times 10^{5} = \mathbf{2.31 \times 10^{5}}$ J mol⁻¹
Minimum energy for one electron $= \dfrac{2.31 \times 10^{5}}{6.022 \times 10^{23}} = \mathbf{3.84 \times 10^{-19}}$ J
The threshold frequency ν₀ for a metal is 7.0×10¹⁴ s⁻¹. Calculate the kinetic energy of an electron emitted when radiation of frequency ν = 1.0×10¹⁵ s⁻¹ hits the metal.
2.3.3 Evidence for the Quantized Electronic Energy Levels: Atomic Spectra
Spectrum: a prism splits white light into coloured bands (violet deviated most, red least) → continuous spectrum (violet 7.5×10¹⁴ Hz → red 4×10¹⁴ Hz; e.g., a rainbow). Excited atoms/molecules emit radiation while returning to lower energy states.
Emission Spectrum
The spectrum of radiation emitted by a substance that has absorbed energy. Atoms, molecules or ions that have absorbed radiation are said to be "excited". Energy is supplied to the sample by heating it or irradiating it, and the wavelength (or frequency) of the emitted radiation is recorded as the sample gives up the absorbed energy.
Absorption Spectrum
Like the photographic negative of an emission spectrum. A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelengths leave dark spaces in the bright continuous spectrum.
Spectroscopy: study of emission/absorption spectra. Gas-phase atoms emit only at specific wavelengths → line (atomic) spectra — bright lines separated by dark spaces.
Spectral "Fingerprints"
Each element has a unique line emission spectrum — a "fingerprint" for identifying elements (Bunsen pioneered this; Rb, Cs, Tl, In, Ga, Sc were discovered spectroscopically; helium was found in the sun).
Line Spectrum of Hydrogen
Hydrogen spectrum: electric discharge through H₂ → excited H atoms emit radiation of discrete frequencies → several series of lines. Balmer (1885) formula (visible lines); Rydberg generalised it for all series:
The series of lines described by this formula are called the Balmer series — the only lines in the hydrogen spectrum which appear in the visible region. The Swedish spectroscopist Johannes Rydberg noted that all series of lines in the hydrogen spectrum could be described by the expression:
Rydberg constant for hydrogen = 109,677 cm⁻¹. Common features of all line spectra: (i) each element's line spectrum is unique, and (ii) there is regularity within each spectrum.
Series
n₁
n₂
Spectral Region
Lyman
1
2, 3, ...
Ultraviolet
Balmer
2
3, 4, ...
Visible
Paschen
3
4, 5, ...
Infrared
Brackett
4
5, 6, ...
Infrared
Pfund
5
6, 7, ...
Infrared
The first five series of lines corresponding to n₁ = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Brackett and Pfund series, respectively.
Mnemonic: Hydrogen Spectral Series
"Learn Better, Play Better, Perform Fine" → Lyman, Balmer (Learn Better: UV → Visible), Paschen, Brackett (Play Better: IR), Pfund (Perform Fine: IR). ⚡ In one line: only Balmer is visible; ν̄ = 109,677 (1/n₁² − 1/n₂²) cm⁻¹.
2.4 Bohr's Model for Hydrogen Atom
Neils Bohr (1913) — first quantitative explanation of the H-atom structure & its spectrum, using Planck's quantum idea. Postulates:
(i) Electron moves in fixed circular orbits (stationary states) of definite radius & energy around the nucleus.
(ii) Energy of an orbit is constant; transition between states requires absorption/emission of exactly ΔE (never gradual).
(iv) Angular momentum is quantised:
$$m_e v r = n\frac{h}{2\pi} \quad n = 1, 2, 3...$$
Angular Momentum
Angular momentum = I·ω; for an electron in a circular orbit, I = mer², ω = v/r → angular momentum = mevr.
Bohr's Model — Allowed Orbits & Photon Emission (3D View)
Interactive 3D Model: Bohr's Hydrogen Atom & Spectral Transitions
Mnemonic: Bohr's Postulates
"O.E.F.A. — Old Elephants Fly Always" → Orbits fixed (allowed states), Energy changes only in jumps, Frequency rule ν = ΔE/h, Angular momentum mvr = nh/2π. ⚡ In one line: Eₙ = −2.18×10⁻¹⁸/n² J and rₙ = n² × 52.9 pm (H); for He⁺/Li²⁺ multiply by Z², divide r by Z.
What Does Bohr's Theory Tell Us?
For the hydrogen atom, Bohr's theory gives:
a) Stationary states are numbered by the principal quantum number n = 1, 2, 3... b) Orbital radii:
$$r_n = n^2 a_0 \quad (a_0 = 52.9\text{ pm})$$
n = 1 is the Bohr orbit (52.9 pm); r grows as n². c) Orbital energies:
$$E_n = -\frac{R_H}{n^2} = -\frac{2.18 \times 10^{-18}}{n^2}\text{ J}$$
Ground state (n = 1): E₁ = −2.18×10⁻¹⁸ J; n = 2: E₂ = −0.545×10⁻¹⁸ J.
Why is the Electronic Energy (Eₙ) Negative?
Eₙ is negative because the electron's energy in the atom is lower than a free electron at rest (E = 0, at n = ∞). Decreasing n makes Eₙ more negative → more stable. n = 1 = ground state (most stable); n = ∞ = ionised atom.
Bohr's Theory for Hydrogen-Like Species
d) Applies to one-electron species (He⁺, Li²⁺, Be³⁺ ...):
$$E_n = -\frac{2.18 \times 10^{-18}Z^2}{n^2}\text{ J}, \qquad r_n = \frac{52.9\,n^2}{Z}\text{ pm}$$
Higher Z → more negative E and smaller r (electron more tightly bound). e) Electron speed: increases with Z, decreases with n.
2.10
What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Since $n_i = 5$ and $n_f = 2$, this transition gives a spectral line in the visible region (Balmer series).
Step 1: Energy Difference
$$\Delta E = 2.18 \times 10^{-18}\left(\frac{1}{5^2} - \frac{1}{2^2}\right)\text{ J} = 2.18 \times 10^{-18} \times (-0.21) = -4.58 \times 10^{-19}\text{ J}$$
(It is an emission energy.)
Absorption: nf > ni (ΔE positive) • Emission: ni > nf (ΔE negative, energy released). Many atoms → all possible transitions → many lines; line intensity ∝ number of photons of that wavelength.
2.4.2 Limitations of Bohr's Model
Limitations of Bohr's model:
(i) Cannot explain finer details (doublets), spectra of multi-electron atoms (even He), or line splitting in a magnetic field (Zeeman effect) / electric field (Stark effect).
(ii) Cannot explain chemical bonding (molecule formation).
2.5 Towards Quantum Mechanical Model of the Atom
Two key ideas led from Bohr's model to the quantum mechanical model:
1. Dual behaviour of matter
2. Heisenberg uncertainty principle
2.5.1 Dual Behaviour of Matter (de Broglie Relation)
de Broglie (1924): matter, like radiation, has dual behaviour (particle + wave). For any material particle:
$$\lambda = \frac{h}{m v} = \frac{h}{p}$$
Confirmed by electron diffraction → basis of the electron microscope (≈15 million × magnification). Every moving object has a wave character, but for large masses λ is far too small to detect.
Mnemonic: de Broglie Relation
"λ = h/mv — Feather waves, Rock doesn't" → tiny mass (electron) → big detectable wavelength; large mass (ball) → immeasurably small λ. ⚡ In one line: every moving object has a wave character — only small ones show it.
2.12
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s⁻¹?
According to the de Broglie equation:
$$\lambda = \frac{h}{mv} = \frac{(6.626 \times 10^{-34}\text{ J s})}{(0.1\text{ kg})(10\text{ m s}^{-1})}$$
λ = 6.626 × 10⁻³⁴ m (J = kg m² s⁻²) — too small to be observed.
2.13
The mass of an electron is 9.1×10⁻³¹ kg. If its K.E. is 3.0×10⁻²⁵ J, calculate its wavelength.
Step 1: Find the Velocity
Since $\text{K.E.} = \frac{1}{2}mv^2$:
$$v = \sqrt{\frac{2 \times 3.0 \times 10^{-25}\text{ J}}{9.1 \times 10^{-31}\text{ kg}}} = 812\text{ m s}^{-1}$$ Step 2: Apply de Broglie Equation
$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}\text{ J s}}{(9.1 \times 10^{-31}\text{ kg})(812\text{ m s}^{-1})} = 8.967 \times 10^{-10}\text{ m}$$
λ = 896.7 nm
2.14
Calculate the mass of a photon with wavelength 3.6 Å.
$\lambda = 3.6\text{ Å} = 3.6 \times 10^{-10}$ m; velocity of photon = velocity of light.
$$m = \frac{h}{\lambda v} = \frac{6.626 \times 10^{-34}\text{ J s}}{(3.6 \times 10^{-10}\text{ m})(3 \times 10^{8}\text{ m s}^{-1})}$$
m = 6.135 × 10⁻²⁹ kg
2.5.2 Heisenberg's Uncertainty Principle
Heisenberg (1927): it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron:
$$\Delta x \cdot \Delta p_x \geq \frac{h}{4\pi} \quad\text{or}\quad \Delta x \cdot \Delta v_x \geq \frac{h}{4\pi m}$$
Analogy: to "see" an electron you need light of very small wavelength, but such high-momentum photons disturb the electron — you get position, but lose velocity.
Significance
Rules out fixed trajectories → we speak only of the probability of finding the electron (|ψ|²), not its path.
Significant only for microscopic objects (electron); negligible for macroscopic ones (a 1 mg object has negligible Δv·Δx).
Mnemonic: Uncertainty Principle
"WHERE + HOW FAST? Pick ONE!" → Δx·Δp ≥ h/4π: pin down position and velocity blurs. ⚡ In one line: no fixed trajectories — probability (|ψ|²) replaces the orbit.
2.15
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?
$\Delta x \Delta p = \dfrac{h}{4\pi}$ or $\Delta x \cdot m\Delta v = \dfrac{h}{4\pi}$
$\Delta x = 0.1\text{ Å} = 10^{-11}$ m
$$\Delta v = \frac{h}{4\pi m \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31} \times 10^{-11}}$$
Δv = 0.579 × 10⁷ m s⁻¹ = 5.79 × 10⁶ m s⁻¹ (1 J = 1 kg m² s⁻²)
2.16
A golf ball has a mass of 40 g and a speed of 45 m/s. If the speed can be measured within an accuracy of 2%, calculate the uncertainty in the position.
The uncertainty in speed is 2%, i.e., $\Delta v = \dfrac{2}{100} \times 45 = 0.9\text{ m s}^{-1}$
Using $\Delta x \cdot \Delta v = \dfrac{h}{4\pi m}$:
$$\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 0.04 \times 0.9}$$
Δx = 1.46 × 10⁻³³ m
This is nearly ~10¹⁸ times smaller than the diameter of a typical atomic nucleus — for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.
Reasons for the Failure of the Bohr Model
In the Bohr model, an electron is regarded as a charged particle moving in well-defined circular orbits about the nucleus — the wave character of the electron is not considered. Further, an orbit is a clearly defined path, and this path can be completely defined only if both the position and the velocity of the electron are known exactly at the same time — not possible according to the Heisenberg uncertainty principle. The Bohr model therefore not only ignores the dual behaviour of matter but also contradicts Heisenberg's uncertainty principle. An insight into the structure of the atom which accounts for wave-particle duality and is consistent with the uncertainty principle came with the advent of quantum mechanics.
2.6 Quantum Mechanical Model of Atom
Quantum mechanics (Heisenberg & Schrödinger, 1926): a theory for microscopic objects that accounts for wave-particle duality + the uncertainty principle. Its fundamental equation (Schrödinger wave equation) is written as Ĥψ = Eψ, solved for a system to give its energy E and wave function ψ. Classical (Newtonian) mechanics applies only to macroscopic objects.
Hydrogen Atom and the Schrödinger Equation
Solving the Schrödinger equation for H: gives quantized energy levels and their wave functions ψ, each characterised by three quantum numbers (n, l, ml).
Atomic orbital: the wave function ψ of a one-electron species — it is a mathematical function (no physical meaning by itself). The probability density |ψ|² at a point gives the chance of finding the electron there.
Multi-electron atoms: the equation can't be solved exactly → approximate methods. Orbitals are similar to hydrogen's but contracted by the higher nuclear charge; their energies depend on both n and l.
Important Features of the Quantum Mechanical Model of Atom
1. Electron energies in atoms are quantized (specific values only).
2. Quantized energy levels arise from the wave-like nature of electrons (solutions of the wave equation).
3. Position & velocity cannot both be known (uncertainty principle) → we speak only of probability of finding the electron.
4. An atomic orbital = wave function ψ of an electron; each orbital has a definite energy and holds at most 2 electrons.
5. Probability of finding the electron ∝ |ψ|² (probability density, always positive).
2.6.1 Orbitals and Quantum Numbers
Atomic orbitals differ in size, shape and orientation — each is fixed by three quantum numbers (n, l, ml).
Principal Quantum Number (n)
Positive integer (n = 1, 2, 3...); determines the size and largely the energy of the orbital and identifies the shell — n² orbitals per shell:
n
1
2
3
4
...
Shell
K
L
M
N
...
Size and energy of the orbital increase with increase of n — the electron is located further from the nucleus.
Azimuthal Quantum Number (l)
Defines the shape of the orbital. For a given n, l = 0 to (n − 1); a shell has n sub-shells:
Value of l
0
1
2
3
4
5
Sub-shell notation
s
p
d
f
g
h
Mnemonic: Quantum Numbers
"Nice Little Monkeys Sing" → Number (n = shell/size), Little (l = shape), Monkeys (ml = orientation), Sing (ms = spin ±½). "Smart People Don't Fail Geometry" → l = 0,1,2,3,4 → s, p, d, f, g. Shells: K L M N = 1 2 3 4 ("KeLvin Meets Newton"). ⚡ In one line: orbitals in shell = n²; max electrons = 2n²; orbitals per subshell = 2l + 1.
Magnetic Orbital Quantum Number (ml)
Gives the orientation of the orbital; for a given l, 2l + 1 values of ml = −l ... 0 ... +l
Value of l
0
1
2
3
4
5
Sub-shell notation
s
p
d
f
g
h
Number of orbitals
1
3
5
7
9
11
Electron Spin 's' (Fourth Quantum Number, ms)
Spin quantum number (ms) — Uhlenbeck & Goudsmit (1925): explains doublets/triplets in spectra. Electron spins about its own axis — ms = +½ (↑) or −½ (↓). Two electrons in one orbital must have opposite spins.
• For s-orbital ($l=0$): $L = 0$ (no orbital angular momentum)
• For p-orbital ($l=1$): $L = \sqrt{2}\,\frac{h}{2\pi}$
• For d-orbital ($l=2$): $L = \sqrt{6}\,\frac{h}{2\pi}$
Spin Magnetic Moment ($\mu_s$)
Determined by the number of unpaired electrons ($n$):
Orbit (Bohr): a definite circular path — has no real meaning (uncertainty principle). Orbital: the one-electron wave function ψ; its |ψ|² = probability density of finding the electron in a small volume.
2.17
What is the total number of orbitals associated with the principal quantum number n = 3?
For n = 3, the possible values of l are 0, 1 and 2:
• one 3s orbital (n = 3, l = 0, ml = 0)
• three 3p orbitals (n = 3, l = 1, ml = −1, 0, +1)
• five 3d orbitals (n = 3, l = 2, ml = −2, −1, 0, +1, +2)
Total = 1 + 3 + 5 = 9 — also obtained from the relation number of orbitals = n² = 3² = 9.
2.18
Using s, p, d, f notations, describe the orbital with the following quantum numbers: (a) n = 2, l = 1; (b) n = 4, l = 0; (c) n = 5, l = 3; (d) n = 3, l = 2.
n
l
Orbital
(a)
2
1
2p
(b)
4
0
4s
(c)
5
3
5f
(d)
3
2
3d
2.6.2 Shapes of Atomic Orbitals
Orbital shape = boundary surface enclosing ~90% probability of finding the electron.
s-orbitals (l = 0):spherically symmetric sphere around the nucleus; size grows with n (4s > 3s > 2s > 1s); ns has (n − 1) nodes (1s: 0, 2s: 1, 3s: 2).
p-orbitals (l = 1):dumbbell of two lobes with a nodal plane through the nucleus; three mutually perpendicular ones — px, py, pz (same size/energy, different orientation); radial nodes = n − 2; size 4p > 3p > 2p.
d-orbitals (l = 2): five orbitals — dxy, dyz, dxz, dx²−y², dz² (first four clover-shaped, dz² has a ring); all five 3d have equal energy; minimum n = 3.
Interactive 3D Atomic Orbitals & Nodal Surfaces Explorer
s-Orbitals — 3D Boundary Surfaces (90% probability)
2p-Orbitals — Three Mutually Perpendicular Dumbbells (3D Lobes)
Same size, shape & energy — only orientation differs • 1 angular node (l = 1) • size: 4p > 3p > 2p
3d-Orbitals — Five Equivalent Shapes (3D Lobes, n ≥ 3)
Nodes
Besides radial nodes (where the probability density function is zero), the probability density functions for np and nd orbitals are zero at the plane(s) passing through the nucleus (origin) — e.g., for a pz orbital the xy-plane is a nodal plane; these are called angular nodes. Number of angular nodes = l (one for p, two for d orbitals). Total number of nodes = n − 1 = l angular nodes + (n − l − 1) radial nodes. 🧠 "A.T.R." → Angular = l, Total = n − 1, Radial = n − l − 1.
2.6.3 Energies of Orbitals
Hydrogen (1-electron): energy depends only on n → orbitals of the same shell are degenerate:
Multi-electron atoms: energy depends on n AND l → within a shell, s < p < d < f, so staggering occurs (4s < 3d, 6s < 5d, 4f < 6p). Cause: electronelectron repulsion + shielding → outer electron feels effective nuclear charge Zeffe (s orbitals shield most; energy rises s → p → d → f).
The (n + l) Rule
The lower the value of (n + l) for an orbital, the lower is its energy. If two orbitals have the same value of (n + l), the orbital with the lower value of n has the lower energy. For example, 4s (n + l = 4) has lower energy than 3d (n + l = 5). Also, energies of orbitals in the same subshell decrease with increase in atomic number: E2s(H) > E2s(Li) > E2s(Na) > E2s(K).
2.6.4 Filling of Orbitals in Atom
The filling of electrons into the orbitals of different atoms takes place according to the aufbau principle, which is based on the Pauli exclusion principle, the Hund's rule of maximum multiplicity and the relative energies of the orbitals.
Aufbau Principle
'Aufbau' = building up: in the ground state, orbitals fill in order of increasing energy:
Order of Filling of Orbitals (Aufbau Diagonal Rule)
Pauli Exclusion Principle
Pauli (1926):no two electrons in an atom can have the same set of four quantum numbers → an orbital holds at most 2 electrons with opposite spins. Maximum electrons in a shell = 2n².
Hund's Rule of Maximum Multiplicity
Electrons fill degenerate orbitals singly first; pairing starts only after each orbital has one electron (p: 4th, d: 6th, f: 8th electron starts pairing). Half-filled / fully-filled sets gain extra stability.
Mnemonic: The Three Filling Rules
"A-P-H: Address, Pairing, House-full" → Aufbau: grab the lowest-energy Address first (diagonal rule; (n+l) rule → 4s fills before 3d); Pauli: max Pair = 2 electrons per orbital, opposite spins; Hund: like a bus — every seat gets one passenger before pairing starts (pairing begins with the 4th, 6th, 8th electron in p, d, f). ⚡ In one line: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s.
2.6.5 Electronic Configuration of Atoms
Electronic configuration = distribution of electrons among orbitals. Two ways to write it:
(i) Notation: subshell + electron count as superscript, e.g., 1s² 2s² 2p⁶ (write shell number before subshell).
(ii) Orbital diagram: boxes = orbitals, arrows ↑↓ = electrons (shows all four quantum numbers).
First 20 elements: H 1s¹; He 1s²; Li 1s²2s¹; Be 1s²2s²; B→Ne: 2p fills → Ne 1s²2s²2p⁶; Na→Ar: [Ne]3s¹ → [Ne]3s²3p⁶. K, Ca: 4s fills before 3d ([Ar]4s¹, [Ar]4s²). Sc→Zn: 3d fills (with exceptions: Cr = [Ar]3d⁵4s¹, Cu = [Ar]3d¹⁰4s¹). Ga→Kr: 4p fills. Core electrons = filled inner shells (e.g., [Ar]); valence electrons = outermost-shell electrons.
2.6.6 Stability of Completely Filled and Half-Filled Subshells
Ground state = lowest-energy configuration. When 4s and 3d differ only slightly in energy, an electron jumps 4s → 3d to give a half-filled (d⁵, f⁷) or completely filled (d¹⁰, f¹⁴) subshell — hence Cr = 3d⁵4s¹, Cu = 3d¹⁰4s¹ (not 3d⁴4s², 3d⁹4s²).
Causes of Stability of Completely Filled and Half-Filled Subshells
1. Symmetrical distribution of electrons → stability; less mutual shielding → electrons more strongly attracted by the nucleus.
2. Exchange energy: electrons of the same spin can exchange positions in degenerate orbitals; the released energy is maximum for half-filled/fully-filled configurations.
Mnemonic: Extra Stability (Cr & Cu)
"SEE = Symmetry + Exchange energy = Extra stability" → half-filled (d⁵, f⁷) and fully-filled (d¹⁰, f¹⁴) subshells beat the normal order: Cr = 3d⁵4s¹, Cu = 3d¹⁰4s¹ ("Cr likes 5, Cu likes 10"). ⚡ In one line: a fully or half-filled subshell is worth an electron jump from 4s → 3d.
Practice Quiz (25 Questions)
Q01 Which of the following statements about cathode rays is incorrect?
Incorrect: This is a correct property of cathode rays.
Correct Answer: The characteristics of cathode rays (electrons) do NOT depend on the electrode material or the nature of the gas — this is why electrons are basic constituents of all atoms.
Incorrect: They are negatively charged, so this behaviour is correct.
Incorrect: This is how the bright spot was observed in the discharge tube.
Q02 The charge to mass ratio (e/mₑ) of the electron measured by J.J. Thomson is:
Incorrect: This is the magnitude of the charge on the electron in coulombs.
Correct Answer: Thomson measured e/mₑ = 1.758820 × 10¹¹ C kg⁻¹ by balancing electric and magnetic fields perpendicular to the electron path.
Incorrect: This is the mass of the electron in kg.
Incorrect: This is Planck's constant in J s.
Q03 Millikan's oil drop experiment helped determine:
Incorrect: e/mₑ was measured by Thomson using cathode ray tubes.
Incorrect: The neutron was discovered by Chadwick in 1932.
Incorrect: The proton was characterised in 1919 from the lightest positive ion (hydrogen).
Correct Answer: Millikan found q = n·e (n = 1, 2, 3...), giving the charge on the electron as −1.602176 × 10⁻¹⁹ C.
Q04 Chadwick discovered the neutron in 1932 by bombarding a thin sheet of:
Correct Answer: Bombarding beryllium with α-particles emitted electrically neutral particles with mass slightly greater than protons — named neutrons.
Incorrect: Gold foil bombardment was Rutherford's scattering experiment.
Incorrect: Hydrogen discharge produced the lightest positive ion, the proton.
Incorrect: β-particles are electrons, not used in Chadwick's experiment.
Q05 In Rutherford's α-particle scattering experiment, approximately how many α-particles bounced back (deflected by nearly 180°)?
Incorrect: Deflections by 180° were far rarer than this.
Incorrect: Still far more frequent than observed.
Correct Answer: Only about 1 in 20,000 α-particles bounced back, showing the positive charge is concentrated in a very small volume (the nucleus).
Incorrect: Most α-particles passed straight through the foil.
Q06 If a cricket ball (radius ≈ 5 cm) represents a nucleus, the radius of the atom would be about:
Incorrect: The ratio of nuclear to atomic radius is 10⁻⁵, so the atom would be much larger.
Correct Answer: Since the atom (10⁻¹⁰ m) is 10⁵ times larger than the nucleus (10⁻¹⁵ m), a 5 cm nucleus corresponds to an atomic radius of about 5 km.
Incorrect: This corresponds to a factor of 10⁴, not 10⁵.
Incorrect: This corresponds to a factor of 10⁶, which is too large.
Q07 Which of the following pairs represents isobars?
Incorrect: Protium and deuterium have the same atomic number — they are isotopes.
Incorrect: These are isotopes of chlorine (same Z = 17).
Incorrect: Same atomic number (6), different mass numbers — isotopes.
Correct Answer: ¹⁴₆C and ¹⁴₇N have the same mass number (14) but different atomic numbers — isobars.
Q08 According to Maxwell's electromagnetic theory, the Rutherford model fails to explain atomic stability because:
Correct Answer: An electron in an orbit undergoes acceleration; Maxwell's theory requires it to emit radiation, lose energy and spiral into the nucleus — but atoms are stable.
Incorrect: Electron mass is not the issue in the classical instability argument.
Incorrect: Electrons and the nucleus attract each other electrostatically.
Incorrect: Circular motion involves continuous change of direction, hence acceleration.
Q09 Electromagnetic radiation of frequency 1368 kHz has a wavelength of:
Incorrect: This is ten times too large — check the kHz conversion.
Incorrect: This is ten times too small.
Correct Answer: λ = c/ν = (3.0 × 10⁸ m s⁻¹)/(1368 × 10³ s⁻¹) = 219.3 m — a radiowave.
Incorrect: This confuses the frequency value with wavelength.
Q10 The wavenumber of yellow radiation of wavelength 5800 Å is:
Incorrect: That is the frequency in Hz, not the wavenumber.
Q16 Which phenomenon could Bohr's model NOT explain?
Correct Answer: Bohr's model could not explain the Zeeman effect (magnetic field), the Stark effect (electric field), fine structure/doublets, or spectra of multi-electron atoms.
Incorrect: Explaining the hydrogen spectrum was Bohr's greatest success.
Incorrect: Bohr's theory works for one-electron (hydrogen-like) species.
Incorrect: Bohr's postulates accounted for the stability of the hydrogen atom.
Q17 The radius of the second Bohr orbit (n = 2) of the hydrogen atom is:
Incorrect: This is the radius of the first orbit (n = 1), the Bohr orbit a₀.
Incorrect: The radius is proportional to n², not n.
Incorrect: This corresponds to n = 3 (9 × 52.9 pm).
Q18 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Incorrect: This counts only direct transitions to the ground state.
Correct Answer: Number of lines = n(n−1)/2 = 6 × 5/2 = 15, from all possible downward transitions among levels 1 to 6.
Incorrect: This counts only transitions from n = 6 to each lower level.
Incorrect: The correct formula n(n−1)/2 gives 15, not 12.
Q19 The de Broglie wavelength associated with a moving cricket ball cannot be detected because:
Incorrect: According to de Broglie, every object in motion has a wave character.
Incorrect: Slower speed increases wavelength, but mass dominates.
Incorrect: λ = h/mv — wavelength is inversely proportional to mass.
Correct Answer: For a ball of mass 0.1 kg at 10 m/s, λ ≈ 6.6 × 10⁻³⁴ m — far too small to observe. Only subatomic particles have detectable de Broglie wavelengths.
Q20 An important implication of the Heisenberg uncertainty principle is that:
Correct Answer: Since position and velocity cannot be determined simultaneously to arbitrary precision, the trajectory of an electron cannot be defined — probability replaces the orbit concept.
Incorrect: Stationary electrons would be pulled into the nucleus by electrostatic attraction.
Incorrect: The uncertainty principle is precisely what invalidates fixed orbits.
Incorrect: The effect is significant only for microscopic objects; for macroscopic objects the uncertainties are negligible.
Q21 Which set of quantum numbers is NOT possible?
Incorrect: This is the valid set for a 1s electron.
Incorrect: This is a valid set for a 2p electron.
Correct Answer: For n = 3, l can only be 0, 1 or 2 (l ≤ n − 1). The value l = 3 is not allowed for n = 3.
Incorrect: This is a valid set for a 3p electron.
Q22 The total number of orbitals associated with the principal quantum number n = 3 is:
Incorrect: This counts only the subshells (3s, 3p, 3d), not the orbitals.
Correct Answer: Number of orbitals = n² = 3² = 9 (one 3s + three 3p + five 3d orbitals).
Incorrect: This counts only the three 3p orbitals plus something else — the correct total is 9.
Incorrect: This is the maximum number of electrons (2n²), not orbitals.
Q23 The ground state electronic configuration of chromium (Z = 24) is:
Incorrect: This follows the simple filling order, but Cr gains extra stability with a half-filled d subshell.
Incorrect: A completely empty 4s with d⁶ is not the observed configuration.
Incorrect: This gives d⁵, but leaves 4s² — the actual shift empties 4s to half-fill... note the observed configuration has only one 4s electron.
Correct Answer: Cr adopts the half-filled 3d⁵ configuration with 4s¹ due to the extra stability of symmetrical distribution and maximum exchange energy.
Q24 The maximum number of electrons that can be present in the shell with n = 4 is:
Correct Answer: Maximum electrons in a shell = 2n² = 2 × 16 = 32 (4s² 4p⁶ 4d¹⁰ 4f¹⁴).
Incorrect: This is n² (the number of orbitals); each orbital holds 2 electrons.
Incorrect: 18 is the capacity of the n = 3 shell.
Incorrect: 8 is the capacity of the n = 2 shell.
Q25 Which of the following orbitals is NOT possible?
Incorrect: 2s is possible (n = 2, l = 0).
Incorrect: 2p is possible (n = 2, l = 1).
Correct Answer: For n = 3, l can only be 0, 1 or 2 (s, p, d). The f subshell requires l = 3, so the minimum n for an f orbital is 4 (4f). Similarly, 1p is impossible.
Incorrect: 4d is possible (n = 4, l = 2).
NCERT Exercises (2.1 2.67)
2.1
(i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.
(i) Mass of one electron = $9.109 \times 10^{-31}$ kg $= 9.109 \times 10^{-28}$ g Number of electrons weighing 1 g $= \frac{1}{9.109 \times 10^{-28}} = \mathbf{1.098 \times 10^{27}}$ electrons
(ii) Mass of 1 mole of electrons $= 9.109 \times 10^{-31} \times 6.022 \times 10^{23} = \mathbf{5.486 \times 10^{-7}}$ kg (≈ 0.55 mg) Charge of 1 mole of electrons $= 1.6022 \times 10^{-19} \times 6.022 \times 10^{23} = \mathbf{9.65 \times 10^{4}}$ C
2.2
(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP. Will the answer change if the temperature and pressure are changed?
(i) CH₄ has 10 electrons per molecule (6 C + 4 H). Total electrons $= 10 \times 6.022 \times 10^{23} = \mathbf{6.022 \times 10^{24}}$
(ii) Moles of $^{14}$C $= \frac{7 \times 10^{-3}}{14} = 5 \times 10^{-4}$ mol → atoms $= 5 \times 10^{-4} \times 6.022 \times 10^{23} = 3.011 \times 10^{20}$ (a) Neutrons per $^{14}$C atom = 8 → total neutrons $= 8 \times 3.011 \times 10^{20} = \mathbf{2.409 \times 10^{21}}$ (b) Mass $= 2.409 \times 10^{21} \times 1.675 \times 10^{-27}$ kg $= \mathbf{4.035 \times 10^{-6}}$ kg
(iii) Moles of NH₃ $= \frac{0.034}{17} = 2 \times 10^{-3}$ mol → molecules $= 1.2044 \times 10^{21}$ (a) Protons per NH₃ = 10 → total protons $= \mathbf{1.2044 \times 10^{22}}$ (b) Mass $= 1.2044 \times 10^{22} \times 1.6726 \times 10^{-27}$ kg $= \mathbf{2.014 \times 10^{-5}}$ kg
No — the number of atoms/molecules in a given mass is independent of temperature and pressure, so the answer will not change.
2.3
How many neutrons and protons are there in the following nuclei? ¹³₆C, ¹⁶₈O, ²⁴₁₂Mg, ⁵⁶₂₆Fe, ⁸⁸₃₈Sr
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A): (i) Z = 17, A = 35. (ii) Z = 92, A = 233. (iii) Z = 4, A = 9.
(i) Z = 17 → chlorine: $\mathbf{^{35}_{17}Cl}$ (ii) Z = 92 → uranium: $\mathbf{^{233}_{92}U}$ (iii) Z = 4 → beryllium: $\mathbf{^{9}_{4}Be}$
2.5
Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wavenumber (ν̄) of the yellow light.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
$\lambda = 4000$ pm $= 4000 \times 10^{-12}$ m $= 4 \times 10^{-9}$ m Energy of one photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4 \times 10^{-9}} = 4.969 \times 10^{-17}$ J Number of photons $= \frac{1\text{ J}}{4.969 \times 10^{-17}\text{ J}} = \mathbf{2.012 \times 10^{16}}$ photons
2.9
A photon of wavelength 4 × 10⁻⁷ m strikes a metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10⁻¹⁹ J).
(i) Energy of the photon $$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{4 \times 10^{-7}} = 4.97 \times 10^{-19}\text{ J} = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} = \mathbf{3.10\text{ eV}}$$ (ii) Kinetic energy $\text{K.E.} = E - W_0 = 3.10 - 2.13 = \mathbf{0.97\text{ eV}} = 1.554 \times 10^{-19}$ J (iii) Velocity of the photoelectron $$v = \sqrt{\frac{2 \times \text{K.E.}}{m}} = \sqrt{\frac{2 \times 1.554 \times 10^{-19}}{9.109 \times 10^{-31}}}$$
v = 5.84 × 10⁵ m s⁻¹
2.10
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol⁻¹.
Energy per photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{242 \times 10^{-9}} = 8.214 \times 10^{-19}$ J Ionisation energy per mole $= 8.214 \times 10^{-19} \times 6.022 \times 10^{23} = 4.947 \times 10^{5}$ J mol⁻¹
IE = 494.7 kJ mol⁻¹ ≈ 495 kJ mol⁻¹
2.11
A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 µm. Calculate the rate of emission of quanta per second.
Energy of one quantum $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19}$ J Rate of emission $= \frac{25\text{ J s}^{-1}}{3.487 \times 10^{-19}\text{ J}}$
= 7.17 × 10¹⁹ quanta per second
2.12
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency (ν₀) and work function (W₀) of the metal.
Since the electrons are emitted with zero velocity, the incident light is at the threshold. $$\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^{8}\text{ m s}^{-1}}{6800 \times 10^{-10}\text{ m}} = \mathbf{4.41 \times 10^{14}\text{ s}^{-1}}$$ $W_0 = h\nu_0 = (6.626 \times 10^{-34}\text{ J s})(4.41 \times 10^{14}\text{ s}^{-1}) = \mathbf{2.924 \times 10^{-19}}$ J
2.13
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionisation enthalpy of H atom (energy required to remove the electron from n = 1 orbit).
$E_5 = \dfrac{-2.18 \times 10^{-18}}{5^2} = -8.72 \times 10^{-20}$ J Energy required for ionisation from n = 5 $= \mathbf{8.72 \times 10^{-20}}$ J Ionisation enthalpy from n = 1 $= 2.18 \times 10^{-18}$ J — which is 25 times larger ($\frac{2.18 \times 10^{-18}}{8.72 \times 10^{-20}} = 25$).
2.15
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Number of emission lines $= \dfrac{n(n-1)}{2} = \dfrac{6 \times 5}{2}$
= 15 lines
2.16
(i) The energy associated with the first orbit in the hydrogen atom is −2.18 × 10⁻¹⁸ J atom⁻¹. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr's fifth orbit for hydrogen atom.
Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
The longest wavelength corresponds to the smallest energy transition in the Balmer series, i.e., n = 3 → n = 2. $$\bar{\nu} = 109{,}677\left(\frac{1}{2^2} - \frac{1}{3^2}\right)\text{ cm}^{-1} = 109{,}677 \times \frac{5}{36}$$
ν̄ = 1.5233 × 10⁴ cm⁻¹
2.18
What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit, and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is −2.18 × 10⁻¹¹ ergs.
The electron energy in a hydrogen atom is given by Eₙ = (−2.18 × 10⁻¹⁸)/n² J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
$E_2 = \dfrac{-2.18 \times 10^{-18}}{4} = -5.45 \times 10^{-19}$ J Energy required for ionisation from n = 2 $= \mathbf{5.45 \times 10^{-19}}$ J Longest wavelength $$\bar{\nu} = \frac{R_H}{2^2} = \frac{109{,}677}{4} = 27{,}419.25\text{ cm}^{-1}$$ $\lambda = \dfrac{1}{27{,}419.25}$ cm
λ = 3.647 × 10⁻⁵ cm
2.20
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s⁻¹.
By the de Broglie equation ($m_e = 9.11 \times 10^{-31}$ kg): $$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(2.05 \times 10^{7})} = \frac{6.626 \times 10^{-34}}{1.868 \times 10^{-23}}$$
λ = 3.548 × 10⁻¹¹ m ≈ 35.5 pm
2.21
The mass of an electron is 9.1 × 10⁻³¹ kg. If its K.E. is 3.0 × 10⁻²⁵ J, calculate its wavelength.
Isoelectronic groups: (Na⁺, Mg²⁺) with 10 electrons; (K⁺, Ca²⁺, S²⁻, Ar) with 18 electrons.
2.23
(i) Write the electronic configurations of the following ions: (a) H⁻ (b) Na⁺ (c) O²⁻ (d) F⁻. (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s¹ (b) 2p³ (c) 3p⁵? (iii) Which atoms are indicated by the following configurations? (a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹.
(i) (a) H⁻: 1s² (b) Na⁺: 1s²2s²2p⁶ (c) O²⁻: 1s²2s²2p⁶ (d) F⁻: 1s²2s²2p⁶ (ii) (a) 3s¹ → configuration ends at Na → Z = 11 (b) 2p³ → configuration 1s²2s²2p³ → N → Z = 7 (c) 3p⁵ → configuration ends at Cl → Z = 17 (iii) (a) [He]2s¹ → Li (b) [Ne]3s²3p³ → P (c) [Ar]4s²3d¹ → Sc
2.24
What is the lowest value of n that allows g orbitals to exist?
For g orbitals, $l = 4$. Since $l \le n - 1$, we need $n \ge l + 1 = 5$.
Lowest value of n = 5
2.25
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
For a 3d orbital: • Principal quantum number: n = 3 • Azimuthal quantum number (d): l = 2 • Magnetic quantum number: ml = −2, −1, 0, +1, +2
2.26
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
(i) For a neutral atom, number of protons = number of electrons = 29 (Z = 29, the element is copper). (ii) Electronic configuration: $\mathbf{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1}$ (i.e., [Ar]3d¹⁰4s¹)
2.27
Give the number of electrons in the species H⁺, H₂ and O₂⁺.
• H⁺: hydrogen atom has 1 electron; the positive ion has 0 electrons. • H₂: two H atoms together have 2 electrons. • O₂⁺: neutral O₂ has 16 electrons; the cation has 15 electrons.
2.28
(i) An atomic orbital has n = 3. What are the possible values of l and ml? (ii) List the quantum numbers (ml and l) of electrons for a 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.
(i) For n = 3: l = 0, 1, 2. • l = 0 → ml = 0 • l = 1 → ml = −1, 0, +1 • l = 2 → ml = −2, −1, 0, +1, +2 (ii) For 3d: l = 2 and ml = −2, −1, 0, +1, +2. (iii) 1p — not possible (n = 1 allows only l = 0). 2s — possible (n = 2, l = 0). 2p — possible (n = 2, l = 1). 3f — not possible (n = 3 allows l up to 2 only).
2.29
Using s, p, d notations, describe the orbital with the following quantum numbers: (a) n = 1, l = 0; (b) n = 3, l = 1; (c) n = 4, l = 2; (d) n = 4, l = 3.
(a) n = 1, l = 0 → 1s (b) n = 3, l = 1 → 3p (c) n = 4, l = 2 → 4d (d) n = 4, l = 3 → 4f
2.30
Explain, giving reasons, which of the following sets of quantum numbers are not possible: (a) n = 0, l = 0, ml = 0, ms = +½ (b) n = 1, l = 0, ml = 0, ms = −½ (c) n = 1, l = 1, ml = 0, ms = +½ (d) n = 2, l = 1, ml = 0, ms = −½ (e) n = 3, l = 3, ml = −3, ms = +½ (f) n = 3, l = 1, ml = 0, ms = +½
(a) Not possible — n cannot be zero; the minimum value of n is 1. (b) Possible — valid set for a 1s electron. (c) Not possible — for n = 1, l cannot be 1 (l ≤ n − 1 = 0). (d) Possible — valid set for a 2p electron. (e) Not possible — for n = 3, l cannot be 3 (l ≤ n − 1 = 2). (f) Possible — valid set for a 3p electron.
2.31
How many electrons in an atom may have the following quantum numbers? (a) n = 4, ms = −½ (b) n = 3, l = 0.
(a) Total electrons in the n = 4 shell = 2n² = 32. Half of them have ms = −½.
(a) 16 electrons
(b) n = 3, l = 0 corresponds to the 3s subshell (one orbital).
(b) 2 electrons
2.32
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
According to Bohr's postulate, $m_e v r = n\frac{h}{2\pi}$ Multiplying both sides by $\frac{2\pi}{m_e v}$: $$2\pi r = n\frac{h}{m_e v} = n\lambda$$ where $\lambda = \frac{h}{m_e v}$ is the de Broglie wavelength of the electron. Thus the circumference of the Bohr orbit (2πr) is an integral multiple of the electron's de Broglie wavelength — the electron wave is in phase around the orbit (a standing wave), which justifies the quantisation of orbits.
2.33
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?
For He⁺ (Z = 2), n = 4 → n = 2: $$\bar{\nu}_{He^+} = R_H Z^2\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R_H\left(\frac{1}{4} - \frac{1}{16}\right) = 4R_H \times \frac{3}{16} = \frac{3R_H}{4}$$ For hydrogen, we need $R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = \frac{3R_H}{4}$, i.e., $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$. This holds for $n_1 = 1, n_2 = 2$: $\frac{1}{1} - \frac{1}{4} = \frac{3}{4}$.
The transition n = 2 → n = 1 (Lyman series) of hydrogen.
2.34
Calculate the energy required for the process He⁺(g) → He²⁺(g) + e⁻. The ionisation energy for the H atom in the ground state is 2.18 × 10⁻¹⁸ J atom⁻¹.
The energy required is the ionisation energy of He⁺ from its ground state (n = 1, Z = 2): $$E = \frac{2.18 \times 10^{-18} \times Z^2}{n^2} = 2.18 \times 10^{-18} \times \frac{(2)^2}{(1)^2}$$
E = 8.72 × 10⁻¹⁸ J atom⁻¹
2.35
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across the length of a scale of length 20 cm long.
Diameter $= 0.15$ nm $= 0.15 \times 10^{-9}$ m; scale length = 20 cm $= 0.2$ m $$\text{Number of atoms} = \frac{0.2\text{ m}}{0.15 \times 10^{-9}\text{ m}} = \mathbf{1.33 \times 10^{9}}\text{ atoms}$$
2.36
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of a carbon atom if the length of this arrangement is 2.4 cm.
Diameter of one atom $= \dfrac{2.4 \times 10^{-2}\text{ m}}{2 \times 10^{8}} = 1.2 \times 10^{-10}$ m Radius $= \dfrac{1.2 \times 10^{-10}}{2} = 6 \times 10^{-11}$ m
Radius = 0.6 Å = 60 pm
2.37
The diameter of a zinc atom is 2.6 Å. Calculate (a) the radius of a zinc atom in pm and (b) the number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
(a) Radius = diameter/2 = 1.3 Å $= \mathbf{130}$ pm (b) Number of atoms $= \dfrac{1.6 \times 10^{-2}\text{ m}}{2.6 \times 10^{-10}\text{ m}}$
= 6.154 × 10⁷ atoms ≈ 6.15 × 10⁷ atoms
2.38
A certain particle carries 2.5 × 10⁻¹⁶ C of static electric charge. Calculate the number of electrons present in it.
In Millikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is −1.282 × 10⁻¹⁸ C, calculate the number of electrons present on it.
In Rutherford's experiment, generally the thin foil of heavy atoms like gold, platinum etc. has been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
Light atoms like aluminium have nuclei with a smaller positive charge and less mass. Therefore: • The repulsive force on the α-particles would be much weaker. • Most α-particles would pass through with only very small deflections. • Almost no α-particles would bounce back (deflection by nearly 180° would practically not be observed), unlike with heavy nuclei such as gold.
2.41
Symbols ⁷⁹₃₅Br and ⁷⁹Br can be written, whereas symbols ³⁵₇₉Br and ³⁵Br are not acceptable. Answer briefly.
In atomic notation the superscript on the left must be the mass number (A) and the subscript must be the atomic number (Z). For bromine: A = 79 and Z = 35, so ⁷⁹₃₅Br is the full correct symbol and ⁷⁹Br (mass number only) is also acceptable. Writing ³⁵₇₉Br would wrongly assign A = 35 (superscript) and Z = 79 (subscript), and ³⁵Br would wrongly imply A = 35 — both are incorrect because the mass number of bromine is 79, not 35.
2.42
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Let the number of protons = p. Then neutrons $= p + 0.317p = 1.317p$ $$p + 1.317p = 81 \implies 2.317p = 81 \implies p = 34.95 \approx 35$$ So protons = 35 (Z = 35 → bromine) and neutrons = 81 − 35 = 46.
Symbol: ⁸¹₃₅Br
2.43
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
Let electrons = e. Then neutrons $= 1.111e$; and since the ion carries one unit of negative charge, protons $= e - 1$. $$A = p + n \implies 37 = (e - 1) + 1.111e = 2.111e - 1$$ $$e = \frac{38}{2.111} = 18; \quad p = 17; \quad n = 20$$ Z = 17 → chlorine.
Symbol: ³⁷₁₇Cl⁻
2.44
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.
Let electrons = e. Then neutrons $= 1.304e$ and protons $= e + 3$ (3 units of positive charge). $$56 = (e + 3) + 1.304e = 2.304e + 3 \implies e = \frac{53}{2.304} = 23$$ So electrons = 23, protons = 26 (Z = 26 → iron), neutrons = 30.
Symbol: ⁵⁶₂₆Fe³⁺
2.45
Arrange the following types of radiations in increasing order of frequency: (a) radiation from a microwave oven (b) amber light from a traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
(c) FM radio < (a) microwave oven < (b) amber light < (e) X-rays < (d) cosmic rays
2.46
Nitrogen laser produces radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 10²⁴, calculate the power of this laser.
Energy of one photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{337.1 \times 10^{-9}} = 5.897 \times 10^{-19}$ J Total energy $= 5.897 \times 10^{-19} \times 5.6 \times 10^{24} = 3.30 \times 10^{6}$ J
Power = 3.30 × 10⁶ W ≈ 3.3 MW
2.47
Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) the distance travelled by this radiation in 30 s, (c) the energy of a quantum, and (d) the number of quanta present if it produces 2 J of energy.
In astronomical observations, signals observed from distant stars are generally weak. If the photon detector receives a total of 3.15 × 10⁻¹⁸ J from the radiation of 600 nm, calculate the number of photons received by the detector.
Energy of one photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{600 \times 10^{-9}} = 3.313 \times 10^{-19}$ J $$N = \frac{3.15 \times 10^{-18}}{3.313 \times 10^{-19}} = 9.5$$
≈ 9.5, i.e., about 910 photons received by the detector.
2.49
Lifetimes of the molecules in the excited states are often measured by using a pulsed radiation source of duration nearly in the nanosecond range. If the radiation source has a duration of 2 ns and the number of photons emitted during the pulse is 2.5 × 10¹⁵, calculate the energy of the source.
Taking the typical pulsed radiation wavelength as λ = 500 nm (ν = 6 × 10¹⁴ Hz): Energy of one photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{500 \times 10^{-9}} = 3.98 \times 10^{-19}$ J Total energy of the source $= 2.5 \times 10^{15} \times 3.98 \times 10^{-19}$
E = 9.95 × 10⁻⁴ J ≈ 1 mJ (power ≈ 5 × 10⁵ W for a 2 ns pulse)
2.50
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and the energy difference between two excited states.
The work function for the caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) the threshold wavelength and (b) Planck's constant. λ (nm): 500, 450, 400; v × 10⁻⁵ (m s⁻¹): 2.55, 4.35, 5.35
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Energy of the photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{256.7 \times 10^{-9}} = 7.746 \times 10^{-19}$ J Maximum kinetic energy (stopping potential) $= eV = (1.602 \times 10^{-19})(0.35) = 5.607 \times 10^{-20}$ J $W_0 = E - \text{K.E.}_{max} = 7.746 \times 10^{-19} - 0.561 \times 10^{-19} = 7.185 \times 10^{-19}$ J
W₀ = 4.49 eV
2.54
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 10⁷ m s⁻¹, calculate the energy with which it is bound to the nucleus.
Energy of the photon $= \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{150 \times 10^{-12}} = 1.325 \times 10^{-15}$ J Kinetic energy of the ejected electron $= \frac{1}{2}mv^2 = \frac{1}{2}(9.11 \times 10^{-31})(1.5 \times 10^{7})^2 = 1.025 \times 10^{-16}$ J Binding energy $= E_{photon} - \text{K.E.} = 1.325 \times 10^{-15} - 0.1025 \times 10^{-15}$
= 1.223 × 10⁻¹⁵ J
2.55
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as ν = 3.29 × 10¹⁵ (Hz) [1/3² − 1/n²]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
n = 5; the transition (5 → 3) lies in the infrared region (Paschen series).
2.56
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Dual behaviour of matter proposed by de Broglie led to the discovery of the electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is 1.6 × 10⁶ m s⁻¹, calculate the de Broglie wavelength associated with this electron.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Mass of neutron $= 1.675 \times 10^{-27}$ kg; $\lambda = 800$ pm $= 8 \times 10^{-10}$ m $$v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{(1.675 \times 10^{-27})(8 \times 10^{-10})}$$
v = 4.94 × 10² m s⁻¹
2.59
If the velocity of the electron in Bohr's first orbit is 2.19 × 10⁶ m s⁻¹, calculate the de Broglie wavelength associated with it.
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10⁵ m s⁻¹. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
λ = 1.516 × 10⁻³⁸ m — far too small to be observed.
2.61
If the position of the electron is measured within an accuracy of ±0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πₘ × 0.05 nm, is there any problem in defining this value?
$\Delta x = 0.002$ nm $= 2 \times 10^{-12}$ m $$\Delta p = \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 2 \times 10^{-12}} = \mathbf{2.637 \times 10^{-23}\text{ kg m s}^{-1}}$$ The stated momentum $= \dfrac{h}{4\pi \times (0.05 \times 10^{-9})} = 1.05 \times 10^{-24}$ kg m s⁻¹. Yes, there is a problem — the minimum uncertainty in momentum (2.637 × 10⁻²³ kg m s⁻¹) is much larger than the stated momentum value itself (1.05 × 10⁻²⁴ kg m s⁻¹), so defining the momentum with such precision is physically meaningless.
2.62
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy, list them: 1. n = 4, l = 2, ml = −2, ms = −½ 2. n = 3, l = 2, ml = 1, ms = +½ 3. n = 4, l = 1, ml = 0, ms = +½ 4. n = 3, l = 2, ml = −2, ms = −½ 5. n = 3, l = 1, ml = −1, ms = +½ 6. n = 4, l = 1, ml = 0, ms = +½
Apply the (n + l) rule — lower (n + l) means lower energy; if (n + l) is equal, lower n has lower energy: • 1: 4d (n + l = 6) • 2: 3d (n + l = 5) • 3: 4p (n + l = 5) • 4: 3d (n + l = 5) • 5: 3p (n + l = 4) • 6: 4p (n + l = 5)
The bromine atom possesses 35 electrons. It contains 6 electrons in the 2p orbital, 6 electrons in the 3p orbital and 5 electrons in the 4p orbital. Which of these electrons experiences the lowest effective nuclear charge?
The effective nuclear charge decreases with increasing principal quantum number n due to greater shielding by inner electrons. Among 2p, 3p and 4p electrons, the 4p electrons (n = 4) are farthest from the nucleus and most shielded.
The 4p electrons experience the lowest effective nuclear charge.
2.64
Among the following pairs of orbitals, which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
(i) 2s — smaller n means it is closer to the nucleus and less shielded, experiencing larger Zeff. (ii) 4d — for the same n, Zeff decreases with increasing l; 4d (l = 2) penetrates more than 4f (l = 3). (iii) 3p — within the same shell, s < p < d in penetration; the 3p electron is more strongly attracted than the 3d electron.
2.65
The unpaired electrons in Al and Si are present in the 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Silicon (Z = 14) has a higher nuclear charge than aluminium (Z = 13), while the shielding by inner electrons is similar.
The 3p unpaired electron in Si experiences the larger effective nuclear charge.
2.66
Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
(a) P — [Ne]3s²3p³: three half-filled 3p orbitals → 3 unpaired electrons. (b) Si — [Ne]3s²3p²: 2 unpaired electrons. (c) Cr — [Ar]3d⁵4s¹: six half-filled orbitals → 6 unpaired electrons. (d) Fe — [Ar]3d⁶4s²: 3d⁶ has one paired and four unpaired → 4 unpaired electrons. (e) Kr — [Ar]3d¹⁰4s²4p⁶: completely filled configuration → 0 unpaired electrons.
2.67
(a) How many subshells are associated with n = 4? (b) How many electrons will be present in the subshells having ms value of −½ for n = 4?
(a) Number of subshells in the nth shell = n. For n = 4, the subshells are 4s (l = 0), 4p (l = 1), 4d (l = 2), 4f (l = 3).
(a) 4 subshells
(b) Total electrons in the n = 4 shell = 2n² = 32. Half of them have ms = −½.
Lower (n + l) → lower energy. If equal, lower n → lower energy. Example: 4s (4+0=4) fills before 3d (3+2=5).
Pauli Exclusion Principle
No two electrons in an atom can have the same set of four quantum numbers. Maximum electrons in a shell = 2n².
Hund's Rule
Pairing in degenerate orbitals starts only after each orbital is singly occupied (pairing begins with the 4th, 6th, 8th electron in p, d, f).
Extra Stability Exceptions (Cr & Cu)
Half-filled (p³, d⁵, f⁷) and fully-filled (p⁶, d¹⁰, f¹⁴) subshells are extra stable due to symmetry and maximum exchange energy. Hence: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹.
Energy Ordering
Hydrogen: 1s < 2s = 2p < 3s = 3p = 3d (depends only on n — degenerate). Multi-electron atoms: energy depends on n and l — within a shell, s < p < d < f.
🧠 Mnemonics Cheat-Sheet (Quick Revision)
Concept
Mnemonic
Key Point
Cathode rays
S.V.N.D.
Straight, Visible, Negative, Don't depend on gas/electrode
Discoverers
E-Tom, P-Gold, N-Chad
ElectronThomson, ProtonGoldstein, NeutronChadwick
Isotope/Isobar/Isotone
P.A.N.
Pe = same Protons; Ar = same A; Ne = same Neutrons
Orbits fixed, Energy jumps, Frequency rule, Angular momentum quantised
de Broglie
Feather waves, Rock doesn't
λ = h/mv — small mass → detectable wave
Heisenberg
WHERE + HOW FAST? Pick ONE!
Δx·Δp ≥ h/4π — no trajectories, only probability
Quantum numbers
Nice Little Monkeys Sing
n, l, ml, ms = shell, shape, orientation, spin
Subshell letters
Smart People Don't Fail Geometry
s, p, d, f, g = l = 0, 1, 2, 3, 4
Filling rules
A-P-H: Address, Pairing, House-full
Aufbau order, Pauli (2 max), Hund (singles first)
Nodes
A.T.R.
Angular = l, Total = n−1, Radial = n−l−1
Extra stability
SEE; "Cr likes 5, Cu likes 10"
Symmetry + Exchange = Extra → Cr 3d⁵4s¹, Cu 3d¹⁰4s¹
Chapter Tests
T1.1 The neutron was discovered by Chadwick by bombarding beryllium with:
Incorrect: Cathode rays are electron streams used in Thomson's experiments.
Correct Answer: Bombarding a thin beryllium sheet with α-particles emitted electrically neutral particles with mass slightly greater than protons — neutrons (1932).
Incorrect: β-particles are electrons emitted in radioactivity.
Incorrect: X-rays were not used in Chadwick's experiment.
T1.2 The present accepted value of the charge on the electron is:
Incorrect: The electron is negatively charged (this is the proton's charge).
Incorrect: This is the electron's mass in kg, not its charge.
Correct Answer: Millikan's oil drop experiment gave the charge as −1.602176 × 10⁻¹⁹ C.
Incorrect: This is close to the e/mₑ ratio (C kg⁻¹), not the charge.
T1.3 In a neutral atom, the number of electrons is always equal to:
Correct Answer: Electrical neutrality requires electrons = protons = Z.
Incorrect: Neutrons generally differ from electrons (neutrons = A − Z).
Incorrect: The mass number A = protons + neutrons, not the electron count.
Incorrect: 2n² gives the maximum capacity of a shell, not the electron count of an atom.
T1.4 The maximum number of electrons that the M shell (n = 3) can accommodate is:
Incorrect: 8 is the capacity of the L shell (n = 2).
Incorrect: 32 is the capacity of the N shell (n = 4).
Incorrect: 2 is the capacity of the K shell (n = 1).
T1.5 The correct electronic configuration of fluorine (Z = 9) is:
Incorrect: This accounts for only 8 electrons (oxygen).
Incorrect: This places electrons in 3s before 2p is filled — violates the aufbau principle.
Incorrect: 2p can hold at most 6 electrons.
Correct Answer: Filling in order of increasing energy gives 1s² 2s² 2p⁵ — one unpaired electron in 2p.
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T2.1 The frequency of light with wavelength 600 nm is approximately:
Correct Answer: ν = c/λ = (3 × 10⁸ m s⁻¹)/(600 × 10⁻⁹ m) = 5.0 × 10¹⁴ Hz.
Incorrect: This is 100 times too large — check the nm conversion.
Incorrect: This would correspond to λ ≈ 167 nm (ultraviolet).
Incorrect: This is infrared (about 100 times too small).
T2.2 The threshold frequency of a metal is 7.0 × 10¹⁴ s⁻¹. The kinetic energy of the photoelectron emitted when ν = 1.0 × 10¹⁵ s⁻¹ radiation strikes it is:
Incorrect: This equals hν itself, not the difference h(ν − ν₀).
Incorrect: This is 10,000 times too large (UV wavelength range).
T2.4 The energy required to ionise a hydrogen atom whose electron occupies the n = 2 orbit is:
Incorrect: This is the ionisation energy from the ground state (n = 1).
Incorrect: This corresponds to n² = 2, which is not a valid orbit.
Incorrect: This is the ionisation energy from n = 5 (2.18 × 10⁻¹⁸/25).
Correct Answer: E = 2.18 × 10⁻¹⁸/2² = 5.45 × 10⁻¹⁹ J — 4 times smaller than from n = 1.
T2.5 The ground state electronic configuration of copper (Z = 29) is:
Correct Answer: Cu adopts the fully-filled 3d¹⁰ configuration with 4s¹ due to the extra stability of a completely filled d subshell (symmetry + exchange energy).
Incorrect: This is the simple filling prediction, but the actual configuration shifts one 4s electron to 3d.
Incorrect: This corresponds to manganese (Z = 25).
Incorrect: This would total 30 electrons — that is zinc (Zn).
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T3.1 Using the (n + l) rule, arrange 3d, 4p, 3p and 4d orbitals in order of increasing energy:
Incorrect: 3p has (n + l) = 4, lower than 3d (5) — 3p must come first.
Correct Answer: (n + l) values are 3p = 4, 3d = 5, 4p = 5, 4d = 6. For the tie between 3d and 4p, lower n wins — so 3d < 4p.
Incorrect: When (n + l) is equal (3d and 4p both = 5), the orbital with lower n (3d) has lower energy.
Incorrect: This ordering is almost fully reversed.
T3.2 The wavenumber of the spectral line for the hydrogen transition n = 4 → n = 2 is:
Incorrect: This is the n = 3 → 2 (longest Balmer) wavenumber.
Incorrect: This is R_H/4, corresponding to the n = 2 → 1 Lyman limit divided by 4.
Incorrect: This is the Rydberg constant itself (the series limit, n = ∞ → 1).
T3.3 A golf ball of mass 40 g moves at 45 m/s with speed measurable within 2% accuracy. The uncertainty in its position is:
Correct Answer: Δv = 0.9 m/s; Δx = h/(4πmΔv) = (6.626 × 10⁻³⁴)/(4π × 0.04 × 0.9) = 1.46 × 10⁻³³ m — meaninglessly small for macroscopic objects.
Incorrect: This would result from forgetting the 2% accuracy factor in Δv.
Incorrect: This corresponds to using h/(4πmΔv) with an extra factor error.
Incorrect: 0.9 m/s is the uncertainty in velocity, not position.
T3.4 Which of the following pairs is isoelectronic (same number of electrons)?
Incorrect: Na⁺ has 10 electrons while K⁺ has 18.
Incorrect: Mg²⁺ has 10 electrons while Ca²⁺ has 18.
Incorrect: S²⁻ has 18 electrons while Na⁺ has 10.
Correct Answer: Na⁺ (11 − 1 = 10) and Mg²⁺ (12 − 2 = 10) both have 10 electrons. (K⁺, Ca²⁺, S²⁻ and Ar form the other isoelectronic group with 18 electrons.)
T3.5 The number of unpaired electrons in a chromium atom (Z = 24, [Ar]3d⁵4s¹) is:
Incorrect: 4 unpaired electrons occur in iron (3d⁶4s²).
Correct Answer: The five 3d orbitals each have one electron (Hund's rule) plus the single 4s electron — all with parallel spins — giving 6 unpaired electrons.
Incorrect: This counts only the half-filled 3d⁵ set and misses the 4s¹ electron.
Incorrect: A completely filled configuration (like Kr) has 0 unpaired electrons; Cr is highly paramagnetic.