🔬 Discovery of Sub-atomic Particles

Pehle sochte the — atom indivisible hai. Phir experiments hue aur pata chala — atom ke andar bhi cheezein hain!

Yahi se sub-atomic particles ki story shuru hoti hai.

⚡ Discovery of Electron — Cathode Ray Experiment

— J.J. Thomson, 1897

Experiment:

• Ek sealed glass tube liya
• Dono ends par metal plates lagaye — cathode (−) aur anode (+)
• High voltage pass kiya
• Cathode se ek ray nikli — cathode ki taraf se anode ki taraf

Observations:

• Ray straight line mein chali
• Jab electric field apply kiya → ray positive plate ki taraf mudi → matlab rays negatively charged hain
• Kisi bhi material ka cathode use karo — rays same hain

Conclusion:

Cathode rays = negatively charged particles = ELECTRONS Electron har atom mein hota hai — universal particle hai

📊 Charge to Mass Ratio of Electron

— J.J. Thomson

Thomson ne electric aur magnetic fields use karke calculate kiya:

e/m = 1.758820 × 10¹¹ C/kg

Yeh ratio sabhi cathode materials ke liye same nikla — proof ki electron ek universal particle hai.

💧 Charge on the Electron

— R.A. Millikan, Oil Drop Experiment, 1909

Experiment:

• Oil ki tiny droplets spray ki
• X-rays se droplets ko charge kiya
• Electric field se droplets ko balance kiya

Result:

Charge on electron = −1.6 × 10⁻¹⁹ C

Mass of electron:

m = e/(e/m) = 1.6×10⁻¹⁹ / 1.758×10¹¹ = 9.1094 × 10⁻³¹ kg

🔴 Discovery of Protons and Neutrons

Proton:

• Goldstein (1886) ne canal rays discover ki
• Anode mein hole karke rays observe ki — ye positive thi
• Hydrogen gas use karne par — sabse hafli positive ray mili
• Yahi proton hai
• Charge = +1.6 × 10⁻¹⁹ C
• Mass = 1.6726 × 10⁻²⁷ kg

Neutron:

• James Chadwick, 1932
• Beryllium par alpha particles bombard kiye
• Ek neutral particle nikla — na electric field se deflect hua
• Yahi neutron hai
• Charge = 0
• Mass = 1.6749 × 10⁻²⁷ kg

🏛️ Atomic Models

1. 🍮 Thomson Model of Atom

“Plum Pudding Model”

Model:

• Atom ek positive sphere hai
• Usme electrons embedded hain — jaise plum pudding mein raisins

Limitations:

• Yeh explain nahi kar saka ki positive charge kahan concentrated hai
• Rutherford ke experiment ne ise completely disprove kar diya

2. ☢️ Rutherford’s Nuclear Model

— Gold Foil Experiment, 1911

Experiment:

• Thin gold foil par alpha particles (α) shoot kiye
• ZnS screen se detect kiya

Observations:

• Most particles seedha nikal gaye → atom zyaadatar empty space
• Kuch particles thoda mude → kuch positive charge hai beech mein
• Very few particles 180° wapas aaye → ek concentrated positive center hai

Conclusions — Nuclear Model:

1. Atom ke center mein nucleus hota hai — bahut chhota, bahut dense
2. Nucleus mein protons (+ charge) hote hain
3. Electrons nucleus ke chaaron taraf orbit karte hain
4. Atom zyaadatar empty space hai

🔢 Atomic Number and Mass Number

Atomic Number (Z) = Number of protons in nucleus e.g., Carbon Z = 6

Mass Number (A) = Protons + Neutrons A = Z + N

Number of neutrons (N) = A − Z

🔄 Isotopes and Isobars

Isotopes:

• Same atomic number (Z), different mass number (A)
• Same element, different neutrons
• e.g., ¹H, ²H (Deuterium), ³H (Tritium)
• e.g., ¹²C, ¹³C, ¹⁴C

Isobars:

• Different atomic number (Z), same mass number (A)
• Different elements, same total nucleons
• e.g., ⁴⁰Ar (Z=18) and ⁴⁰Ca (Z=20)

❌ Drawbacks of Rutherford’s Model

1. Electron orbit problem: Classical physics kehta hai — charged particle jo accelerate kare, woh energy radiate karta hai. Toh electron spiral karke nucleus mein gir jaata — atom stable nahi rehta. But atoms stable hain!
2. No explanation of atomic spectra: Atoms specific wavelengths ki light emit karte hain — Rutherford model yeh explain nahi kar saka.

🌊 Developments Leading to Bohr’s Model

Wave Nature of Electromagnetic Radiation

— Maxwell

Light ek electromagnetic wave hai jo electric aur magnetic fields ka oscillation hai.

Wave characteristics:

Relationship:

c = ν × λ c = 3 × 10⁸ m/s

Electromagnetic Spectrum (increasing wavelength):

γ rays → X-rays → UV → Visible → IR → Microwaves → Radio waves

Visible light range = 400 nm (violet) to 750 nm (red)

🎯 Particle Nature — Planck’s Quantum Theory

— Max Planck, 1900

Problem: Classical theory couldn’t explain black body radiation — hot objects emit specific colors.

Planck’s Solution:

Energy is not continuous — it comes in discrete packets called QUANTA (singular: quantum)

E = hν

Jahan:

• E = energy of quantum
• h = Planck’s constant = 6.626 × 10⁻³⁴ J·s
• ν = frequency

Matlab — energy sirf certain fixed values mein hoti hai — quantized!

💡 Photoelectric Effect

— Albert Einstein, 1905

Observation:

• Jab light kisi metal surface par girti hai → electrons emit hote hain
• Lekin sirf tab jab frequency ek minimum value se zyada ho — threshold frequency (ν₀)

Classical theory fail kyu hui:

• Intensity badhao — electrons bahut der baad nikalenge (classical prediction)
• But actually — instantly nikalte hain!

Einstein ka explanation:

Light packets (photons) mein aati hai E = hν Ek photon ek electron ko eject karta hai — agar E ≥ threshold energy

Key points:

• Frequency badhao → electrons ki kinetic energy badhti hai
• Intensity badhao → zyada electrons nikalte hain, energy same
• KE = hν − hν₀ = h(ν − ν₀)

🌓 Dual Behaviour of Electromagnetic Radiation

Light ka dual nature hai:

• Wave nature — diffraction, interference mein dikhta hai
• Particle nature — photoelectric effect mein dikhta hai

Dono ek saath observe nahi ho sakte — situation par depend karta hai

📡 Atomic Spectra — Quantized Energy Levels

Emission Spectrum:

• Element ko heat karo ya excite karo → specific wavelengths ki light emit hoti hai
• Yeh bright lines form karta hai dark background par
• Har element ka spectrum unique hota hai — fingerprint of element

Absorption Spectrum:

• White light element ke through pass karo → specific wavelengths absorb ho jaati hain
• Dark lines on bright background

🌈 Line Spectrum of Hydrogen

Hydrogen ka spectrum series of lines dikhata hai:

Rydberg Formula:

1/λ = R_H (1/n₁² − 1/n₂²)

Jahan R_H = Rydberg constant = 1.097 × 10⁷ m⁻¹

🔵 Bohr’s Model for Hydrogen Atom

— Niels Bohr, 1913

Postulates:

1. Electrons fixed circular orbits mein revolve karte hain — stationary states — energy emit nahi karte

2. Sirf woh orbits allowed hain jisme angular momentum:

mvr = nh/2π (n = 1, 2, 3…)

3. Electron ek orbit se doosre mein jump karta hai energy absorb/emit karke:

ΔE = E₂ − E₁ = hν

Energy of nth orbit (Hydrogen):

Eₙ = −2.18 × 10⁻¹⁸ / n² J

Radius of nth orbit:

rₙ = 52.9 × n² pm (Bohr radius for n=1 = 52.9 pm)

📊 Explanation of Hydrogen Spectrum using Bohr’s Model

ΔE = hν = hc/λ 1/λ = R_H (1/n₁² − 1/n₂²)

• Electron lower orbit → higher orbit: energy absorbed
• Electron higher orbit → lower orbit: energy emitted as photon

Yahi explain karta hai hydrogen ke line spectrum ko! ✅

❌ Limitations of Bohr’s Model

1. Multi-electron atoms explain nahi kar saka
2. Zeeman effect (magnetic field mein spectrum splitting) explain nahi kiya
3. Stark effect (electric field mein splitting) explain nahi kiya
4. 3D orbital shapes explain nahi kiye
5. De Broglie aur Heisenberg ki discoveries ke saath contradictory tha

🌀 Towards Quantum Mechanical Model

🌊 Dual Behaviour of Matter

— Louis de Broglie, 1924

Agar light wave bhi hai aur particle bhi — toh matter bhi wave bhi ho sakta hai!

de Broglie equation:

λ = h/mv = h/p

Jahan:

• λ = wavelength of particle
• h = Planck’s constant
• m = mass
• v = velocity
• p = momentum

Electrons ke liye yeh experimentally prove hua — electron diffraction!

Large objects (ball, car) ki wavelength itni chhoti hoti hai ki measure hi nahi hoti.

🌫️ Heisenberg’s Uncertainty Principle

— Werner Heisenberg, 1927

“It is impossible to simultaneously determine both the exact position AND exact momentum of a sub-atomic particle.”

Mathematical form:

Δx × Δp ≥ h/4π Δx × mΔv ≥ h/4π

Jahan:

• Δx = uncertainty in position
• Δp = uncertainty in momentum

Simple samjho:

• Jitna precisely position jaano → momentum aur uncertain ho jaata hai
• Jitna precisely momentum jaano → position aur uncertain ho jaata hai

💡 Significance of Uncertainty Principle

1. Electron ki exact trajectory nahi de sakte — Bohr model galat!
2. Orbits concept invalid — we can only talk about probability of finding electron
3. Yahi quantum mechanics ka base hai

❌ Reasons for Failure of Bohr Model

1. Bohr model mein electron ki exact position aur momentum dono define kiye gaye — Heisenberg principle violate!
2. Electron wave nature consider nahi kiya — de Broglie ignore!
3. Toh Bohr model fundamentally incomplete tha

🔮 Quantum Mechanical Model of Atom

— Erwin Schrödinger, 1926

Schrödinger Equation:

Ĥψ = Eψ

Jahan:

• Ĥ = Hamiltonian operator (total energy operator)
• ψ (psi) = wave function
• E = total energy

ψ² = Probability density — batata hai ki electron kahan milne ki kitni probability hai

🌐 Orbitals and Quantum Numbers

Orbital = 3D region in space where probability of finding electron is maximum (>90%)

Quantum numbers electron ko completely describe karte hain:

1. Principal Quantum Number (n)

• Shell number
• n = 1, 2, 3, 4… (K, L, M, N shells)
• Energy aur size determine karta hai
• Larger n → larger orbital → higher energy

2. Azimuthal Quantum Number (l)

• Subshell aur shape determine karta hai
• l = 0 to (n−1)

3. Magnetic Quantum Number (mₗ)

• Orbital orientation in space
• mₗ = −l to +l (including 0)
• Total values = (2l + 1)

4. Spin Quantum Number (mₛ)

• Electron ki spin
• mₛ = +½ (spin up ↑) or −½ (spin down ↓)

🔷 Shapes of Atomic Orbitals

s orbital:

• Spherical
• 1s < 2s < 3s (size badhta hai)
• 2s aur 3s mein nodes hote hain (nodes = zero probability region)
• Number of radial nodes = n − l − 1

p orbital:

• Dumbbell shape
• 3 orbitals: pₓ, pᵧ, p_z — perpendicular axes par
• 2 lobes separated by a nodal plane

d orbital:

• 5 orbitals
• dₓᵧ, dᵧ_z, d_xz, dₓ²₋ᵧ², d_z²
• Complex cloverleaf shapes

f orbital:

• 7 orbitals — very complex shapes
• Not required in detail for Class 11

⚡ Energies of Orbitals

For hydrogen (one electron):

Energy depends only on n 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f

For multi-electron atoms:

Energy depends on both n and l Lower (n+l) → lower energy Same (n+l) → lower n → lower energy

Order:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d…

📥 Filling of Orbitals

1. 📈 Aufbau Principle

“Electrons fill orbitals in order of increasing energy”

Memory trick — diagonal rule:

1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f

Diagonally upar se neeche fill karo → 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d…

2. 🚫 Pauli Exclusion Principle

“No two electrons in an atom can have all four quantum numbers identical”

Simple matlab:

Ek orbital mein maximum 2 electrons — aur unki spin opposite honi chahiye (↑↓)

3. 🎯 Hund’s Rule of Maximum Multiplicity

“Electrons fill degenerate orbitals (same energy) one by one first — all with same spin — before pairing begins”

e.g., Nitrogen (N) — 2p has 3 electrons:

↑ ↑ ↑ ← correct (one each, same spin) ↑↓ ↑ _ ← WRONG (pairing before all filled)

📋 Electronic Configuration of Atoms

Notation: nˡˣ (n = shell, l = subshell, x = electrons)

⭐ Stability of Completely Filled and Half Filled Subshells

Cr aur Cu special kyun hain?

Expected:

• Cr = [Ar] 3d⁴ 4s²
• Cu = [Ar] 3d⁹ 4s²

Actual:

• Cr = [Ar] 3d⁵ 4s¹ ← half-filled d
• Cu = [Ar] 3d¹⁰ 4s¹ ← completely filled d

Kyun? — Two reasons:

1. Symmetry: Half-filled (d⁵) aur completely filled (d¹⁰) subshells symmetrical hote hain → extra stability

2. Exchange energy: Jab electrons ek jaisi spin ke saath zyada orbitals mein hote hain → exchange interactions zyada hoti hain → energy kam hoti hai → stability zyada

Basically nature prefer karta hai half-filled ya completely filled subshells — isliye ek electron 4s se 3d mein “jump” kar jaata hai!

📝 Summary Table

💡 Exam Tips:

• ⭐ Bohr model energy formula yaad karo
• ⭐ Quantum numbers ke values aur rules
• ⭐ Electronic configuration of Cr aur Cu — tricky!
• ⭐ de Broglie equation numericals
• ⭐ Heisenberg principle — formula aur significance

NCERT Chemistry Chapter 2 — Solved Numericals

Structure of Atom | Q 2.1 to 2.13

✅ Q2.1

(i) Number of electrons that weigh 1 gram

Mass of 1 electron = 9.1094 × 10⁻³¹ kg = 9.1094 × 10⁻²⁸ g

Number of electrons = 1 g / 9.1094 × 10⁻²⁸ g

= 1.098 × 10²⁷ electrons ✅

(ii) Mass and Charge of 1 mole of electrons

Mass:

1 mole = 6.022 × 10²³ electrons Mass = 6.022 × 10²³ × 9.1094 × 10⁻³¹ kg = 5.486 × 10⁻⁴ kg = 5.486 × 10⁻¹ g ≈ 0.5486 g ✅

Charge:

Charge = 6.022 × 10²³ × 1.6 × 10⁻¹⁹ C = 9.65 × 10⁴ C = 96500 C = 1 Faraday ✅

✅ Q2.2

(i) Total electrons in 1 mole of Methane (CH₄)

Electrons in 1 molecule CH₄:

C = 6 electrons 4H = 4 × 1 = 4 electrons Total = 10 electrons per molecule

In 1 mole CH₄:

= 10 × 6.022 × 10²³ = 6.022 × 10²⁴ electrons ✅

(ii) Neutrons in 7 mg of ¹⁴C

¹⁴C: Z = 6 (protons), A = 14, Neutrons = 14 − 6 = 8 neutrons per atom

Step 1 — Moles of ¹⁴C:

Molar mass of ¹⁴C = 14 g/mol Moles = 7 × 10⁻³ / 14 = 5 × 10⁻⁴ mol

Step 2 — Number of atoms:

Atoms = 5 × 10⁻⁴ × 6.022 × 10²³ = 3.011 × 10²⁰ atoms

(a) Total number of neutrons:

= 3.011 × 10²⁰ × 8 = 2.408 × 10²¹ neutrons ✅

(b) Total mass of neutrons:

= 2.408 × 10²¹ × 1.675 × 10⁻²⁷ kg = 4.03 × 10⁻⁶ kg ✅

(iii) Protons in 34 mg of NH₃

NH₃: N = 7 protons, 3H = 3 × 1 = 3 protons

Total protons per molecule = 10 protons

Step 1 — Moles of NH₃:

Molar mass = 14 + 3 = 17 g/mol Moles = 34 × 10⁻³ / 17 = 2 × 10⁻³ mol

Step 2 — Number of molecules:

= 2 × 10⁻³ × 6.022 × 10²³ = 1.2044 × 10²¹ molecules

(a) Total number of protons:

= 1.2044 × 10²¹ × 10 = 1.2044 × 10²² protons ✅

(b) Total mass of protons:

Mass of proton = 1.6726 × 10⁻²⁷ kg = 1.2044 × 10²² × 1.6726 × 10⁻²⁷ = 2.014 × 10⁻⁵ kg ✅

Will answer change with T and P?

No — number of protons depends on number of molecules (moles), which depends only on mass and molar mass — not on temperature or pressure. ✅

✅ Q2.3 Protons and Neutrons in Nuclei

Formula: Protons = Z, Neutrons = A − Z

✅

✅ Q2.4 Complete Symbol of Atom

Format: ᴬ_Z Symbol

(i) Z = 17, A = 35

Z = 17 → Chlorine (Cl) ³⁵₁₇Cl ✅

(ii) Z = 92, A = 233

Z = 92 → Uranium (U) ²³³₉₂U ✅

(iii) Z = 4, A = 9

Z = 4 → Beryllium (Be) ⁹₄Be ✅

✅ Q2.5 Frequency and Wavenumber of Yellow Light

Given:

• λ = 580 nm = 580 × 10⁻⁹ m = 5.80 × 10⁻⁷ m
• c = 3 × 10⁸ m/s

Frequency (ν):

ν = c/λ = (3 × 10⁸) / (5.80 × 10⁻⁷) ν = 5.17 × 10¹⁴ Hz ✅

Wavenumber (ν̄):

ν̄ = 1/λ = 1 / (5.80 × 10⁻⁷) ν̄ = 1.724 × 10⁶ m⁻¹ = 1.724 × 10⁴ cm⁻¹ ✅

✅ Q2.6 Energy of Photons

Formula: E = hν = hc/λ

h = 6.626 × 10⁻³⁴ J·s

(i) ν = 3 × 10¹⁵ Hz

E = hν = 6.626 × 10⁻³⁴ × 3 × 10¹⁵ E = 1.988 × 10⁻¹⁸ J ✅

(ii) λ = 0.50 Å = 0.50 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (5 × 10⁻¹¹) = 19.878 × 10⁻²⁶ / 5 × 10⁻¹¹ E = 3.976 × 10⁻¹⁵ J ✅

✅ Q2.7 Wavelength, Frequency, Wavenumber (Period = 2.0 × 10⁻¹⁰ s)

Step 1 — Frequency:

ν = 1/T = 1 / (2.0 × 10⁻¹⁰) ν = 5.0 × 10⁹ Hz ✅

Step 2 — Wavelength:

λ = c/ν = (3 × 10⁸) / (5.0 × 10⁹) λ = 0.06 m = 6 × 10⁻² m ✅

Step 3 — Wavenumber:

ν̄ = 1/λ = 1 / (6 × 10⁻²) ν̄ = 16.67 m⁻¹ = 16.67 cm⁻¹ ✅

✅ Q2.8 Number of Photons providing 1 J of energy

Given:

• λ = 4000 pm = 4000 × 10⁻¹² m = 4 × 10⁻⁹ m
• Total energy = 1 J

Energy of 1 photon:

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (4 × 10⁻⁹) = 19.878 × 10⁻²⁶ / 4 × 10⁻⁹ = 4.97 × 10⁻¹⁷ J

Number of photons:

n = Total energy / Energy per photon = 1 / 4.97 × 10⁻¹⁷ = 2.012 × 10¹⁶ photons ✅

✅ Q2.9 Photoelectric Effect — λ = 4 × 10⁻⁷ m, Work function = 2.13 eV

Given:

• λ = 4 × 10⁻⁷ m
• Work function W₀ = 2.13 eV
• 1 eV = 1.6020 × 10⁻¹⁹ J

(i) Energy of photon in eV

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (4 × 10⁻⁷) = 4.969 × 10⁻¹⁹ J E in eV = 4.969 × 10⁻¹⁹ / 1.6020 × 10⁻¹⁹ E = 3.102 eV ✅

(ii) Kinetic Energy of emitted electron

KE = E − W₀ = 3.102 − 2.13 KE = 0.972 eV ✅ In Joules = 0.972 × 1.6020 × 10⁻¹⁹ = 1.558 × 10⁻¹⁹ J

(iii) Velocity of photoelectron

KE = ½mv² 1.558 × 10⁻¹⁹ = ½ × 9.1094 × 10⁻³¹ × v² v² = (2 × 1.558 × 10⁻¹⁹) / (9.1094 × 10⁻³¹) v² = 3.421 × 10¹¹ v = 5.848 × 10⁵ m/s ✅

✅ Q2.10 Ionisation Energy of Sodium

Given:

• λ = 242 nm = 242 × 10⁻⁹ m
• This is just sufficient → all energy = ionisation energy

Energy per photon:

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (242 × 10⁻⁹) = 8.208 × 10⁻¹⁹ J

Ionisation energy per mole:

IE = E × Nₐ = 8.208 × 10⁻¹⁹ × 6.022 × 10²³ = 4.942 × 10⁵ J/mol = 494.2 kJ/mol ✅

✅ Q2.11 Rate of emission of quanta — 25W bulb, λ = 0.57 μm

Given:

• Power = 25 W = 25 J/s
• λ = 0.57 μm = 0.57 × 10⁻⁶ m

Energy per photon:

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (0.57 × 10⁻⁶) = 3.487 × 10⁻¹⁹ J

Rate of emission:

n = Power / E per photon = 25 / 3.487 × 10⁻¹⁹ = 7.169 × 10¹⁹ quanta per second ✅

✅ Q2.12 Threshold Frequency and Work Function

Given:

• λ = 6800 Å = 6800 × 10⁻¹⁰ m = 6.8 × 10⁻⁷ m
• Electrons emitted with zero velocity → all energy = work function

Threshold frequency (ν₀):

ν₀ = c/λ = (3 × 10⁸) / (6.8 × 10⁻⁷) ν₀ = 4.41 × 10¹⁴ Hz ✅

Work function (W₀):

W₀ = hν₀ = 6.626 × 10⁻³⁴ × 4.41 × 10¹⁴ W₀ = 2.922 × 10⁻¹⁹ J ✅

✅ Q2.13 Wavelength of light — n = 4 to n = 2 (Hydrogen)

Formula:

1/λ = R_H (1/n₁² − 1/n₂²) R_H = 1.097 × 10⁷ m⁻¹

Given: n₁ = 2 (lower), n₂ = 4 (higher)

1/λ = 1.097 × 10⁷ × (1/2² − 1/4²) = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × (4/16 − 1/16) = 1.097 × 10⁷ × (3/16) = 1.097 × 10⁷ × 0.1875 = 2.057 × 10⁶ m⁻¹

λ = 1 / 2.057 × 10⁶ λ = 4.86 × 10⁻⁷ m = 486 nm ✅

🔵 This is the Hα line of Balmer Series — visible blue-green light!

✅ Q2.14 Energy to ionise H atom from n = 5

Formula:

Eₙ = −2.18 × 10⁻¹⁸ / n² J

Energy at n = 5:

E₅ = −2.18 × 10⁻¹⁸ / 25 = −8.72 × 10⁻²⁰ J

Ionisation energy from n = 5:

IE = 0 − (−8.72 × 10⁻²⁰) = +8.72 × 10⁻²⁰ J ✅

Ionisation energy from n = 1 (ground state):

IE = 0 − (−2.18 × 10⁻¹⁸) = +2.18 × 10⁻¹⁸ J

Comparison:

IE(n=1) / IE(n=5) = 2.18×10⁻¹⁸ / 8.72×10⁻²⁰ = 25 times

🔴 Ionisation from n=5 requires 25 times less energy than from n=1 ✅

✅ Q2.15 Maximum emission lines — n = 6 to ground state

Formula:

Number of spectral lines = n(n−1)/2

Here n = 6 (highest level) Lines = 6(6−1)/2 = 6×5/2 = 15 lines ✅

These transitions are:

6→5, 6→4, 6→3, 6→2, 6→1 5→4, 5→3, 5→2, 5→1 4→3, 4→2, 4→1 3→2, 3→1 2→1 Total = 5+4+3+2+1 = 15 ✅

✅ Q2.16

(i) Energy of 5th orbit

Given: E₁ = −2.18 × 10⁻¹⁸ J

Eₙ = E₁/n²

E₅ = −2.18 × 10⁻¹⁸ / 5² = −2.18 × 10⁻¹⁸ / 25 = −8.72 × 10⁻²⁰ J ✅

(ii) Radius of 5th Bohr orbit

Formula:

rₙ = n² × 52.9 pm (Bohr radius a₀ = 52.9 pm)

r₅ = 5² × 52.9 = 25 × 52.9 = 1322.5 pm = 13.225 Å ✅

✅ Q2.17 Longest wavelength in Balmer Series

Balmer series: n₁ = 2, transitions from higher levels

Longest wavelength = smallest energy gap = n₂ = 3 (closest to n=2)

Formula:

1/λ = R_H (1/n₁² − 1/n₂²) R_H = 1.097 × 10⁷ m⁻¹

1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × (9−4)/36 = 1.097 × 10⁷ × 5/36 = 1.097 × 10⁷ × 0.13889 = 1.5236 × 10⁶ m⁻¹

λ = 1/1.5236 × 10⁶ λ = 6.564 × 10⁻⁷ m = 656.4 nm

Wavenumber:

ν̄ = 1.5236 × 10⁶ m⁻¹ = 1.524 × 10⁴ cm⁻¹ ✅

🔴 This is the red line of hydrogen spectrum (Hα line)!

✅ Q2.18 Energy to shift electron n=1 to n=5, and wavelength of return

Given: E₁ = −2.18 × 10⁻¹¹ ergs = −2.18 × 10⁻¹⁸ J

Energy at n=1:

E₁ = −2.18 × 10⁻¹⁸ J

Energy at n=5:

E₅ = −2.18 × 10⁻¹⁸ / 25 = −8.72 × 10⁻²⁰ J

Energy required (absorption):

ΔE = E₅ − E₁ = (−8.72×10⁻²⁰) − (−2.18×10⁻¹⁸) = −8.72×10⁻²⁰ + 218×10⁻²⁰ = 2.093 × 10⁻¹⁸ J ✅

Wavelength of light emitted when electron returns (n=5 → n=1):

ΔE = hc/λ λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸) / 2.093×10⁻¹⁸ = 19.878×10⁻²⁶ / 2.093×10⁻¹⁸ = 9.498 × 10⁻⁸ m = 94.98 nm ✅

(UV region — Lyman series)

✅ Q2.19 Energy to remove electron from n=2, longest wavelength

Energy at n=2:

E₂ = −2.18 × 10⁻¹⁸ / 4 = −5.45 × 10⁻¹⁹ J

Energy required to remove (ionise from n=2):

IE = 0 − (−5.45×10⁻¹⁹) = 5.45 × 10⁻¹⁹ J ✅

Longest wavelength of light that can cause this:

λ = hc/ΔE = (6.626×10⁻³⁴ × 3×10⁸) / 5.45×10⁻¹⁹ = 19.878×10⁻²⁶ / 5.45×10⁻¹⁹ = 3.647 × 10⁻⁷ m = 3.647 × 10⁻⁵ cm ✅

✅ Q2.20 Wavelength of electron — v = 2.05 × 10⁷ m/s

de Broglie formula:

λ = h/mv

Given:

• h = 6.626 × 10⁻³⁴ J·s
• m = 9.1094 × 10⁻³¹ kg
• v = 2.05 × 10⁷ m/s

λ = 6.626×10⁻³⁴ / (9.1094×10⁻³¹ × 2.05×10⁷) = 6.626×10⁻³⁴ / 18.674×10⁻²⁴ = 6.626×10⁻³⁴ / 1.8674×10⁻²³ = 3.548 × 10⁻¹¹ m = 35.48 pm ✅

✅ Q2.21 Wavelength from KE — m = 9.1 × 10⁻³¹ kg, KE = 3.0 × 10⁻²⁵ J

Step 1 — Find velocity from KE:

KE = ½mv² v² = 2KE/m = (2 × 3.0×10⁻²⁵) / 9.1×10⁻³¹ = 6.0×10⁻²⁵ / 9.1×10⁻³¹ = 6.593 × 10⁵ v = 8.12 × 10² m/s = 812 m/s

Step 2 — de Broglie wavelength:

λ = h/mv = 6.626×10⁻³⁴ / (9.1×10⁻³¹ × 812) = 6.626×10⁻³⁴ / 7.389×10⁻²⁸ = 8.967 × 10⁻⁷ m ≈ 897 nm ✅

✅ Q2.22 Isoelectronic Species

Count electrons in each:

Isoelectronic groups:

🔵 10 electrons: Na⁺, Mg²⁺

🟢 18 electrons: K⁺, Ca²⁺, S²⁻, Ar ✅

✅ Q2.23

(i) Electronic Configurations of Ions

(a) H⁻ (H gains 1 electron, Z=1, electrons=2)

1s² ✅

(b) Na⁺ (Na loses 1 electron, Z=11, electrons=10)

1s² 2s² 2p⁶ ✅

(c) O²⁻ (O gains 2 electrons, Z=8, electrons=10)

1s² 2s² 2p⁶ ✅

(d) F⁻ (F gains 1 electron, Z=9, electrons=10)

1s² 2s² 2p⁶ ✅

Note: H⁻, Na⁺, O²⁻, F⁻ — sab isoelectronic hain (10 electrons)!

(ii) Atomic Numbers from outermost electron

(a) 3s¹

Configuration so far: 1s² 2s² 2p⁶ 3s¹ Total electrons = 2+2+6+1 = 11 Z = 11 → Sodium (Na) ✅

(b) 2p³

Configuration: 1s² 2s² 2p³ Total = 2+2+3 = 7 Z = 7 → Nitrogen (N) ✅

(c) 3p⁵

Configuration: 1s² 2s² 2p⁶ 3s² 3p⁵ Total = 2+2+6+2+5 = 17 Z = 17 → Chlorine (Cl) ✅

(iii) Atoms from configurations

(a) [He] 2s¹

He = 2 electrons, +1 = 3 electrons Z = 3 → Lithium (Li) ✅

(b) [Ne] 3s² 3p³

Ne = 10, +2+3 = 15 electrons Z = 15 → Phosphorus (P) ✅

(c) [Ar] 4s² 3d¹

Ar = 18, +2+1 = 21 electrons Z = 21 → Scandium (Sc) ✅

✅ Q2.24 Lowest n for g orbitals

For g orbital: l = 4

Rule: l ranges from 0 to (n−1)

For l = 4: n−1 ≥ 4 → n ≥ 5

Minimum n = 5 ✅

✅ Q2.25 Quantum numbers for 3d orbital electron

3d orbital:

mₛ can be +½ or −½ (any one)

✅

✅ Q2.26 Element with 29 electrons and 35 neutrons

(i) Number of protons:

Atom is neutral → Protons = Electrons = 29 ✅ Z = 29 → Copper (Cu)

(ii) Electronic configuration:

Expected: [Ar] 3d⁹ 4s² Actual (exception): [Ar] 3d¹⁰ 4s¹ ✅

(Completely filled d subshell gives extra stability!)

✅ Q2.27 Electrons in H₂⁺, H₂, O₂⁺

✅

✅ Q2.28

(i) n = 3 → possible values of l and mₗ

l values: 0, 1, 2 (l = 0 to n−1)

✅

(ii) Quantum numbers for 3d orbital

n = 3, l = 2 mₗ = −2, −1, 0, +1, +2 ✅

(iii) Which orbitals are possible?

Possible: 2s and 2p only ✅

✅ Q2.29 s, p, d, f notation from quantum numbers

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