Chapter 27: Atoms

Plum Pudding Model

Model Description

• Proposed by J.J. Thomson in 1898.
• Atom consists of a sphere of positive charge with electrons embedded like seeds in a watermelon.
• Positive charge is uniformly distributed throughout the atom's volume.
• Electrons are negatively charged and scattered inside the positive sphere.
• Later experiments disproved this model due to different charge distributions.

Solved Examples

• Ex 1: Identify the charge distribution in the plum pudding model.
  Answer: Positive charge is spread uniformly; electrons embedded inside.

Practice Set

• Q1: Who proposed the plum pudding model?
  A) Ernest Rutherford
  B) J.J. Thomson
  C) Niels Bohr
  D) Hans Geiger
• Q2: In the plum pudding model, electrons are:
  A) Orbiting nucleus
  B) Embedded in positive charge
  C) Outside the atom
  D) Absent
• Q3: Which observation disproved the plum pudding model?
  A) Electrons are positively charged
  B) Positive charge is concentrated in nucleus
  C) Atom is indivisible
  D) Electrons do not exist

Answer Key

• Q1: B - J.J. Thomson proposed the plum pudding model.
• Q2: B - Electrons are embedded in a sphere of positive charge.
• Q3: B - Experiments showed positive charge is concentrated in nucleus, not spread out.

Memory Aids

• "Plum pudding" = electrons like plums in positive pudding.
• Positive charge: uniform sphere; electrons: embedded.

Rutherford Gold Foil Experiment

Experiment Setup & Observations

• Alpha particles (5.5 MeV) from radioactive source directed at thin gold foil (~2.1 × 10⁻⁷ m thick).
• Alpha particles collimated by lead bricks into narrow beam.
• Scattered alpha particles detected by rotatable zinc sulphide screen and microscope.
• Most alpha particles passed through foil undeflected, indicating atom is mostly empty space.
• Some alpha particles deflected at various angles; very few (1 in 8000) deflected by 180°.
• Deflections indicated a small, dense, positively charged nucleus at atom's center.

Solved Examples

• Ex 1: What does the large angle deflection of alpha particles indicate?
  Answer: Presence of a small, dense, positively charged nucleus.
• Ex 2: Why do most alpha particles pass through the foil undeflected?
  Answer: Most of the atom's volume is empty space.

Practice Set

• Q1: What is the main conclusion from the gold foil experiment?
  A) Atom is solid
  B) Atom has a nucleus
  C) Electrons are embedded in positive charge
  D) Alpha particles are neutral
• Q2: What fraction of alpha particles are deflected by 180°?
  A) 1 in 1000
  B) 1 in 8000
  C) 1 in 10000
  D) 1 in 5000
• Q3: The thickness of gold foil used in the experiment is approximately:
  A) 2.1 × 10⁻⁷ m
  B) 2.1 × 10⁻⁵ m
  C) 2.1 × 10⁻⁹ m
  D) 2.1 × 10⁻⁶ m

Answer Key

• Q1: B - The experiment proved the existence of a nucleus.
• Q2: B - About 1 in 8000 alpha particles deflect by 180°.
• Q3: A - Gold foil thickness was 2.1 × 10⁻⁷ m.

Memory Aids

• Gold foil experiment = discovery of nucleus.
• Most alpha particles pass through = atom mostly empty.
• Few deflect back = small dense nucleus.

Rutherford Scattering Formula

Key Relations

• Impact parameter (b): perpendicular distance of alpha particle's initial velocity from nucleus center.
• Relation between impact parameter and scattering angle \(\theta\):
  \[ b = \frac{Ze^2 \cot(\theta/2)}{4 \pi \epsilon_0 E_k} \]
• \(Z\): atomic number, \(e\): electron charge, \(\epsilon_0\): permittivity of free space, \(E_k\): kinetic energy of alpha particle.
• Special cases:
  • If \(b=0\), then \(\theta=180^\circ\) (head-on collision, particle rebounds).
  • If \(b=\infty\), then \(\theta=0^\circ\) (no deflection).
• Distance of closest approach \(r_0\):
  \[ r_0 = \frac{2Ze^2}{4 \pi \epsilon_0 E_k} \]
• Number of scattered particles \(N \propto \frac{1}{\sin^4(\theta/2)}\).

Solved Examples

• Ex 1: Calculate impact parameter for \(\theta=60^\circ\), given \(Z, e, \epsilon_0, E_k\).
• Ex 2: Find distance of closest approach for alpha particle with kinetic energy \(E_k\).

Practice Set

• Q1: For \(b=0\), scattering angle \(\theta\) is:
  A) 0°
  B) 90°
  C) 180°
  D) 45°
• Q2: Number of scattered particles \(N\) varies with angle as:
  A) \(\sin^2 \theta\)
  B) \(\cos^2 \theta\)
  C) \(1/\sin^4(\theta/2)\)
  D) \(\tan \theta\)
• Q3: Distance of closest approach \(r_0\) is:
  A) \(\frac{Ze^2}{4 \pi \epsilon_0 E_k}\)
  B) \(\frac{2Ze^2}{4 \pi \epsilon_0 E_k}\)
  C) \(\frac{Ze^2}{2 \pi \epsilon_0 E_k}\)
  D) \(\frac{4Ze^2}{4 \pi \epsilon_0 E_k}\)

Answer Key

• Q1: C - Head-on collision causes 180° deflection.
• Q2: C - Number of scattered particles \(N \propto 1/\sin^4(\theta/2)\).
• Q3: B - Distance of closest approach \(r_0 = \frac{2Ze^2}{4 \pi \epsilon_0 E_k}\).

Memory Aids

• Impact parameter \(b\) and angle \(\theta\) linked by cotangent.
• Distance of closest approach \(r_0\) doubles numerator compared to \(b\) formula.
• Scattering intensity \(\propto 1/\sin^4(\theta/2)\).

Rutherford Atomic Model

Postulates

• Positive charge and most mass concentrated in a tiny nucleus (~10⁻¹⁴ m diameter).
• Electrons revolve around nucleus in empty space (~10⁻¹⁰ m atom diameter).
• Electrostatic force between nucleus and electrons provides centripetal force for orbit.
• Atom is electrically neutral: total positive charge = total negative charge.

Drawbacks

• Accelerated electrons should radiate energy and spiral into nucleus, collapsing atom (not observed).
• Predicted continuous energy spectrum, but observed atomic spectra are discrete.

Solved Examples

• Ex 1: Explain why Rutherford model predicts atom collapse.
  Answer: Accelerated electrons radiate energy, lose energy, spiral inward.

Practice Set

• Q1: In Rutherford model, where is positive charge located?
  A) Spread uniformly
  B) In nucleus
  C) In electron cloud
  D) Absent
• Q2: Why does Rutherford model fail?
  A) Electrons do not revolve
  B) Electrons radiate energy and atom collapses
  C) Nucleus is large
  D) Atom is not neutral
• Q3: Atomic spectra predicted by Rutherford model are:
  A) Discrete lines
  B) Continuous spectrum
  C) No spectrum
  D) Random lines

Answer Key

• Q1: B - Positive charge is concentrated in nucleus.
• Q2: B - Accelerated electrons radiate energy causing collapse.
• Q3: B - Model predicts continuous spectrum, but actual is discrete.

Memory Aids

• Rutherford model = nucleus center + revolving electrons.
• Drawback: electron radiation causes collapse.
• Atomic spectra are discrete, not continuous.

Electron Orbits and Energy

Key Formulas

• Electrostatic force provides centripetal force:
  \[ \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r} \]
• Orbit radius and velocity relation:
  \[ r = \frac{e^2}{4 \pi \epsilon_0 m v^2} \]
• Kinetic energy (K):
  \[ K = \frac{1}{2} m v^2 = \frac{e^2}{8 \pi \epsilon_0 r} \]
• Potential energy (U):
  \[ U = - \frac{e^2}{4 \pi \epsilon_0 r} \]
• Total energy (E):
  \[ E = K + U = - \frac{e^2}{8 \pi \epsilon_0 r} \]
• Negative total energy implies electron is bound to nucleus.

Solved Examples

• Ex 1: Calculate total energy for electron in orbit radius \(r\).
  Use \(E = - \frac{e^2}{8 \pi \epsilon_0 r}\).
• Ex 2: Show that kinetic energy is half magnitude of potential energy but positive.

Practice Set

• Q1: The force keeping electron in orbit is:
  A) Gravitational
  B) Electrostatic attraction
  C) Nuclear force
  D) Magnetic
• Q2: Total energy of electron in orbit is:
  A) Positive
  B) Zero
  C) Negative
  D) Infinite
• Q3: If orbit radius decreases, total energy:
  A) Increases (less negative)
  B) Decreases (more negative)
  C) Remains same
  D) Becomes zero

Answer Key

• Q1: B - Electrostatic force provides centripetal force.
• Q2: C - Total energy is negative, indicating bound state.
• Q3: B - Smaller radius means more negative total energy.

Memory Aids

• Total energy \(E = -\) (half potential energy magnitude).
• Electron bound if total energy negative.
• Electrostatic force = centripetal force.

Atomic Spectra

Emission Line Spectrum

• Excited atomic gas emits light at specific wavelengths.
• Emission line spectrum: bright lines on dark background.
• Hydrogen emission spectrum shows distinct lines grouped in series.

Explanation of Hydrogen Spectrum

• Electrons transition between energy levels, emitting photons.
• Series: Lyman (UV), Balmer (visible), Paschen (IR).
• Lyman: 91-122 nm; Balmer: 365-656 nm; Paschen: 820-1875 nm.
• Each line corresponds to electron dropping to a lower energy level.

Solved Examples

Ex 1

• Calculate wavelength for Balmer series with n=3.
• Use formula: 1/\(\lambda\) = R (1/2² - 1/3²), R = 1.097 × 10⁷ m⁻¹.
• Calculate \(\lambda\) = 656.3 nm (red line).

Practice Set

• Q1: Which series of hydrogen spectrum lies in visible region?
  • A) Lyman
  • B) Balmer
  • C) Paschen
  • D) Brackett
• Q2: The wavelength of Hα line in Balmer series is approximately?
  • A) 434.1 nm
  • B) 486.1 nm
  • C) 656.3 nm
  • D) 820 nm
• Q3: For Balmer series, if n=4, calculate the wavelength using R = 1.097 × 10⁷ m⁻¹.
  • A) 486.1 nm
  • B) 434.1 nm
  • C) 410.2 nm
  • D) 397.0 nm

Answer Key

• Q1: B) Balmer - Visible region lines belong to Balmer series.
• Q2: C) 656.3 nm - Hα line wavelength in Balmer series.
• Q3: C) 410.2 nm - Calculated using formula 1/\(\lambda\) = R (1/4 - 1/16).

Memory Aids

• Balmer series formula: 1/\(\lambda\) = R (1/2² - 1/n²), n=3,4,...
• Lyman series ends at n=1 (UV), Balmer at n=2 (visible).
• Paschen, Brackett, Pfund series lie in infrared region.

Spectral Series

Definition and Patterns

• Hydrogen spectral lines form series with decreasing spacing.
• Each series corresponds to electron transitions to a fixed lower energy level.
• Balmer series observed in visible region; others in UV and IR.
• Empirical formula for Balmer series: 1/\(\lambda\) = R (1/2² - 1/n²), n=3,4,...

Other Spectral Series Formulas

• General formula: 1/\(\lambda\) = R (1/n_f² - 1/n_i²)
• Lyman: n_f=1, n_i=2,3,... (UV)
• Balmer: n_f=2, n_i=3,4,... (Visible)
• Paschen: n_f=3, n_i=4,5,... (IR)
• Brackett: n_f=4, n_i=5,6,... (IR)
• Pfund: n_f=5, n_i=6,7,... (IR)

Solved Examples

Ex 1

• Calculate wavelength for Lyman series with n=2.
• Use formula: 1/\(\lambda\) = R (1/1² - 1/2²).
• Calculate \(\lambda\) ≈ 121.6 nm (UV line).

Practice Set

• Q1: Which series corresponds to electron transitions ending at n=1?
  • A) Balmer
  • B) Lyman
  • C) Paschen
  • D) Brackett
• Q2: The Paschen series lies in which region of the electromagnetic spectrum?
  • A) Ultraviolet
  • B) Visible
  • C) Infrared
  • D) X-ray
• Q3: Using R = 1.097 × 10⁷ m⁻¹, calculate wavelength for Brackett series with n=6.
  • A) 4050 nm
  • B) 2625 nm
  • C) 1458 nm
  • D) 405 nm

Answer Key

• Q1: B) Lyman - Transitions ending at n=1.
• Q2: C) Infrared - Paschen series is in IR region.
• Q3: B) 2625 nm - Calculated using 1/\(\lambda\) = R (1/4² - 1/6²).

Memory Aids

• Remember series by final energy level n_f: Lyman=1, Balmer=2, Paschen=3, etc.
• Wavelength decreases as n_i increases within a series.
• Rydberg constant R = 1.097 × 10⁷ m⁻¹ used in all formulas.

Bohr Model of the Hydrogen Atom

Bohr's Postulates

• Electrons revolve in stable orbits without emitting energy.
• Electron angular momentum is quantized: \( mvr = n\frac{h}{2\pi} \), where \( n=1,2,3,... \).
• Electrons jump between orbits by absorbing or emitting photons.
• Energy of emitted/absorbed photon: \( h\nu = E_i - E_f \).
• Orbit radius: \( r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z} = 0.53 \frac{n^2}{Z} \text{Å} \).
• Velocity and angular momentum depend on quantum number \( n \).

Solved Examples

Ex 1

• Find radius of 3rd orbit in hydrogen atom (Z=1).
• Use \( r_n = 0.53 n^2 / Z \) Å.
• \( r_3 = 0.53 \times 3^2 / 1 = 4.77 \text{Å} \).

Practice Set

• Q1: What is the angular momentum of electron in 2nd orbit?
  A) \( \frac{h}{2\pi} \) B) \( 2\frac{h}{2\pi} \) C) \( 3\frac{h}{2\pi} \) D) \( \frac{h}{4\pi} \)
• Q2: Energy of photon emitted when electron falls from 3rd to 1st orbit in hydrogen is?
  A) 12.1 eV B) 10.2 eV C) 13.6 eV D) 1.51 eV
• Q3: Radius of 4th orbit in hydrogen atom is?
  A) 0.53 Å B) 8.5 Å C) 9.0 Å D) 0.85 Å

Answer Key

• Q1: B. Angular momentum \( = n\frac{h}{2\pi} = 2\frac{h}{2\pi} \) for \( n=2 \).
• Q2: A. \( E = 13.6(1 - \frac{1}{3^2}) = 12.1 \text{ eV} \).
• Q3: B. \( r_4 = 0.53 \times 4^2 = 8.5 \text{ Å} \).

Memory Aids

• Angular momentum quantized as multiples of \( \frac{h}{2\pi} \).
• Orbit radius \( \propto n^2 \).
• Energy levels: \( E_n = -13.6 / n^2 \) eV.

Energy Transitions and Photon Emission

Key Points

• Electron transitions between orbits emit or absorb photons.
• Photon energy: \( h\nu = E_i - E_f \).
• Energy levels are discrete; transitions correspond to specific photon frequencies.
• Electron moving from higher to lower orbit emits photon.
• Electron moving from lower to higher orbit absorbs photon.

Solved Examples

Ex 1

• Calculate photon energy when electron falls from \( n=3 \) to \( n=2 \) in hydrogen.
• Use \( E_n = -13.6 / n^2 \) eV.
• \( E_3 = -1.51 \) eV, \( E_2 = -3.4 \) eV.
• Energy emitted \( = E_2 - E_3 = 1.89 \) eV.

Practice Set

• Q1: Photon energy emitted when electron falls from 2nd to 1st orbit?
  A) 10.2 eV B) 13.6 eV C) 3.4 eV D) 1.51 eV
• Q2: Frequency of photon emitted for 1.89 eV energy? (Use \( h=6.63\times10^{-34} \) Js)
  A) \( 4.57\times10^{14} \) Hz B) \( 3.2\times10^{14} \) Hz C) \( 1.89\times10^{15} \) Hz D) \( 2.5\times10^{14} \) Hz
• Q3: Electron absorbs photon and jumps from 1st to 3rd orbit. Energy absorbed is?
  A) 12.1 eV B) 10.2 eV C) 13.6 eV D) 1.89 eV

Answer Key

• Q1: A. \( 13.6(1 - \frac{1}{2^2}) = 10.2 \) eV.
• Q2: A. \( \nu = E/h = (1.89 \times 1.6\times10^{-19}) / 6.63\times10^{-34} = 4.57\times10^{14} \) Hz.
• Q3: A. Energy absorbed equals energy difference \( 12.1 \) eV.

Memory Aids

• Energy difference \( = h\nu \) for photon emission/absorption.
• Higher to lower orbit: photon emitted.
• Lower to higher orbit: photon absorbed.

Energy Levels and Quantization

Key Concepts

• Electron energy levels in hydrogen: \( E_n = -13.6 / n^2 \) eV.
• Ground state (\( n=1 \)) energy: \( -13.6 \) eV.
• Energy levels get closer as \( n \) increases.
• Electron can only have discrete energy values (quantization).
• Ionization energy: 13.6 eV to remove electron from ground state.
• Radius of orbit increases as \( r_n \propto n^2 \).

Solved Examples

Ex 1

• Calculate energy of electron in 4th orbit.
• Use \( E_4 = -13.6 / 4^2 = -0.85 \) eV.

Practice Set

• Q1: Energy of electron in 5th orbit?
  A) -0.54 eV B) -1.36 eV C) -3.4 eV D) -13.6 eV
• Q2: Ionization energy from 2nd orbit?
  A) 3.4 eV B) 13.6 eV C) 1.51 eV D) 0.85 eV
• Q3: Radius of 1st orbit (Bohr radius)?
  A) 0.53 Å B) 1.06 Å C) 2.12 Å D) 0.26 Å

Answer Key

• Q1: A. \( E_5 = -13.6 / 25 = -0.54 \) eV.
• Q2: A. Energy difference \( = 13.6/2^2 = 3.4 \) eV.
• Q3: A. Bohr radius \( a_0 = 0.53 \) Å.

Memory Aids

• Energy levels: \( E_n = -13.6 / n^2 \) eV.
• Radius: \( r_n = 0.53 n^2 \) Å.
• Energy quantization means discrete allowed energies only.

The Line Spectra of the Hydrogen Atom

Energy Transitions and Rydberg Formula

• Electron transitions: from higher energy level \( n_i \) to lower \( n_f \) (\( n_f < n_i \)) emit photon.
• Energy difference: \( h\nu_{if} = E_{n_i} - E_{n_f} = \frac{me^4}{8\varepsilon_0^2 h^2} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Frequency formula: \( \nu_{if} = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Discrete frequencies arise due to integer quantum numbers \( n_i, n_f \).
• Spectral series: Lyman (\( n_f=1 \)), Balmer (\( n_f=2 \)), Paschen (\( n_f=3 \)), Brackett (\( n_f=4 \)), Pfund (\( n_f=5 \)).

Solved Examples

Ex 1

• Find frequency of photon emitted when electron falls from \( n_i=3 \) to \( n_f=2 \) in hydrogen.
• Use \( \nu_{if} = \frac{me^4}{8\varepsilon_0^2 h^3} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \).
• Calculate \( \nu_{if} \) by substituting values for constants and quantum numbers.

Practice Set

• Q1: Which series corresponds to electron transitions ending at \( n=2 \)?
  A) Lyman
  B) Balmer
  C) Paschen
  D) Brackett
• Q2: The frequency of emitted photon when electron falls from \( n=4 \) to \( n=1 \) is proportional to:
  A) \( \frac{1}{1^2} - \frac{1}{4^2} \)
  B) \( \frac{1}{4^2} - \frac{1}{1^2} \)
  C) \( \frac{1}{2^2} - \frac{1}{3^2} \)
  D) \( \frac{1}{3^2} - \frac{1}{2^2} \)
• Q3: The Rydberg formula for hydrogen spectrum is:
  A) \( h\nu = E_{n_f} - E_{n_i} \)
  B) \( h\nu = E_{n_i} - E_{n_f} \)
  C) \( h\nu = E_{n_i} + E_{n_f} \)
  D) \( h\nu = E_{n_f} + E_{n_i} \)

Answer Key

• Q1: B) Balmer - Transitions ending at \( n=2 \) form Balmer series.
• Q2: A) \( \frac{1}{1^2} - \frac{1}{4^2} \) - Frequency proportional to difference in inverse squares of \( n_f \) and \( n_i \).
• Q3: B) \( h\nu = E_{n_i} - E_{n_f} \) - Energy difference equals photon energy emitted.

Memory Aids

• Rydberg formula key: \( h\nu = \text{constant} \times (1/n_f^2 - 1/n_i^2) \).
• Series names by \( n_f \): Lyman=1, Balmer=2, Paschen=3, Brackett=4, Pfund=5.
• Electron falls → photon emitted with discrete frequency.

Continuous and Characteristic X-rays

X-ray Production and Spectrum

• In Coolidge tube, high-energy electrons strike metal target producing X-rays.
• Cutoff wavelength: \( \lambda_{min} = \frac{hc}{eV} \), where \( V \) is accelerating voltage.
• Minimum wavelength depends only on \( V \), not target material.
• Continuous spectrum arises from electron deceleration (bremsstrahlung).
• Characteristic X-rays (\( K_\alpha, K_\beta \)) arise from electron transitions between atomic shells.

Electron Transitions and X-ray Emission

• Incident electron knocks out inner shell (K-shell, \( n=1 \)) electron, creating vacancy.
• Electron from higher shell (e.g., L-shell, \( n=2 \)) fills vacancy, emitting characteristic X-ray photon.
• Wavelengths of X-rays given by \( \lambda = \frac{hc}{E_i - E_f} \), where \( E_i, E_f \) are shell energies.
• Examples: \( K_\alpha: \lambda = \frac{hc}{E_K - E_L} \), \( K_\beta: \lambda = \frac{hc}{E_K - E_M} \), \( L_\alpha: \lambda = \frac{hc}{E_L - E_M} \).

Solved Examples

Ex 1

• Calculate \( \lambda_{min} \) for X-rays if accelerating voltage \( V = 20,000 \) V.
• Use \( \lambda_{min} (nm) = \frac{1242}{V (V)} = \frac{1242}{20000} = 0.0621 \) nm.

Practice Set

• Q1: The cutoff wavelength \( \lambda_{min} \) depends on:
  A) Target material
  B) Accelerating voltage
  C) Both A and B
  D) Neither A nor B
• Q2: Characteristic X-rays are produced when:
  A) Electrons decelerate
  B) Electrons transition between atomic shells
  C) Electrons ionize atoms
  D) Electrons accelerate
• Q3: Wavelength of \( K_\alpha \) X-ray is given by:
  A) \( \frac{hc}{E_L - E_M} \)
  B) \( \frac{hc}{E_K - E_L} \)
  C) \( \frac{hc}{E_K - E_M} \)
  D) \( \frac{hc}{E_M - E_N} \)

Answer Key

• Q1: B) Accelerating voltage - \( \lambda_{min} \) depends only on \( V \).
• Q2: B) Electrons transition between atomic shells - characteristic X-rays arise from these transitions.
• Q3: B) \( \frac{hc}{E_K - E_L} \) - defines \( K_\alpha \) wavelength.

Memory Aids

• \( \lambda_{min} = \frac{hc}{eV} \) - cutoff wavelength formula.
• Characteristic X-rays named by shells involved: \( K_\alpha \) (L→K), \( K_\beta \) (M→K).
• Continuous spectrum from electron braking; characteristic peaks from shell transitions.

De-Broglie’s Explanation of Bohr’s Second Postulate

Key Concepts

• Bohr’s second postulate: Electron’s angular momentum is quantised: \( L_n = n \frac{h}{2\pi} \), where \( n=1,2,3... \)
• De-Broglie hypothesis: Electrons behave as waves with wavelength \( \lambda = \frac{h}{p} = \frac{h}{mv} \)
• Standing waves form when circumference of orbit equals an integral multiple of wavelength: \( n\lambda = 2\pi r_n \)
• From above, angular momentum quantisation follows: \( m v_n r_n = n \frac{h}{2\pi} \)
• Only orbits with standing electron waves are stable, explaining quantised energy levels

Solved Examples

• Ex 1: Find the angular momentum of electron in 1st orbit.
  • Use \( L = n \frac{h}{2\pi} \), for \( n=1 \)
  • \( L = \frac{6.626 \times 10^{-34}}{2\pi} = 1.055 \times 10^{-34} \text{ Js} \)

Practice Set

• Q1: According to de-Broglie, the wavelength of electron is inversely proportional to
  • A) velocity
  • B) momentum
  • C) mass
  • D) energy
• Q2: For an electron in 3rd orbit, the circumference is
  • A) \( 3 \lambda \)
  • B) \( \lambda /3 \)
  • C) \( 2\pi \lambda \)
  • D) \( \lambda \)
• Q3: The quantisation of angular momentum is due to
  • A) Particle nature of electron
  • B) Wave nature of electron
  • C) Nuclear charge
  • D) Electron spin

Answer Key

• Q1: B - Wavelength \( \lambda = \frac{h}{p} \), inversely proportional to momentum.
• Q2: A - Circumference \( = 2\pi r_n = n \lambda \), for \( n=3 \), circumference is \( 3 \lambda \).
• Q3: B - Quantisation arises from electron’s wave nature forming standing waves.

Memory Aids

• Angular momentum quantised as \( n h / 2\pi \)
• Electron wavelength \( \lambda = h / mv \)
• Standing wave condition: orbit circumference = integral multiples of \( \lambda \)

Limitations of Bohr’s Theory

Key Points

• Valid only for single-electron atoms (H, He⁺, Li²⁺)
• Cannot explain fine structure of spectral lines (closely spaced lines)
• Fails to explain molecular bonding and stability
• Cannot fully explain Zeeman and Stark effects (splitting under magnetic/electric fields)
• Does not explain electron distribution in different orbits
• Ignores electron-electron interactions in multi-electron atoms
• Cannot explain varying intensities of spectral lines in hydrogen emission

Solved Examples

• Ex 1: Identify if Bohr’s theory explains helium atom spectrum.
  • Helium has two electrons; Bohr’s theory applies only to single-electron atoms.
  • Hence, Bohr’s theory fails for helium spectrum.

Practice Set

• Q1: Bohr’s theory fails to explain spectra of
  • A) Hydrogen
  • B) He⁺
  • C) Lithium atom
  • D) Hydrogen-like ions
• Q2: Fine structure of spectral lines is due to
  • A) Electron spin
  • B) Elliptical orbits
  • C) Nuclear spin
  • D) Electron charge
• Q3: Which effect is not fully explained by Bohr’s theory?
  • A) Zeeman effect
  • B) Photoelectric effect
  • C) Compton effect
  • D) Blackbody radiation

Answer Key

• Q1: C - Lithium has multiple electrons; Bohr’s theory fails.
• Q2: B - Fine structure explained by elliptical orbits (Sommerfeld).
• Q3: A - Zeeman effect splitting not fully explained by Bohr’s model.

Memory Aids

• Bohr’s theory = single-electron atoms only
• Fine structure = elliptical orbits (Sommerfeld)
• Zeeman & Stark effects need quantum mechanics

Moseley’s Law

Key Concepts

• Frequency of X-ray spectrum: \( \sqrt{\nu} = a(Z - b) \), where \( a,b \) are constants
• Effective nuclear charge: \( Z - b \), with \( b \approx 1 \)
• Energy released during transition \( n=2 \to n=1 \):
  • \( \Delta E = h \nu = R h c (Z - b)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \)
  • \( \sqrt{\nu} = \sqrt{\frac{3 R c}{4}} (Z - b) \)

Solved Examples

• Ex 1: Calculate \( \sqrt{\nu} \) for element with \( Z=30 \), \( b=1 \), \( a=constant \).
  • \( \sqrt{\nu} = a (30 - 1) = 29a \)

Practice Set

• Q1: In Moseley’s law, \( b \) is approximately
  • A) 0
  • B) 1
  • C) 2
  • D) 3
• Q2: The frequency of X-rays increases with
  • A) Increasing atomic number
  • B) Decreasing atomic number
  • C) Increasing wavelength
  • D) Decreasing energy
• Q3: Energy released in \( n=2 \to n=1 \) transition is proportional to
  • A) \( (Z - b) \)
  • B) \( (Z - b)^2 \)
  • C) \( \sqrt{Z - b} \)
  • D) \( 1/(Z - b) \)

Answer Key

• Q1: B - Constant \( b \approx 1 \) accounts for screening effect.
• Q2: A - Frequency increases with atomic number \( Z \).
• Q3: B - Energy proportional to square of effective nuclear charge \( (Z - b)^2 \).

Memory Aids

• Moseley: \( \sqrt{\nu} \propto (Z - 1) \)
• Effective nuclear charge = \( Z - b \)
• X-ray frequency increases with atomic number

Bragg’s Law

Key Concepts

• Condition for X-ray diffraction from crystal planes:
  • \( 2 d \sin \theta = n \lambda \), where \( n=1,2,3... \)
  • \( d \): distance between crystal planes
  • \( \theta \): glancing angle of incidence/reflection
  • \( \lambda \): wavelength of X-rays
• Constructive interference occurs when path difference equals integral multiples of wavelength
• Used to determine crystal structure and interplanar spacing

Solved Examples

• Ex 1: Find \( d \) if first order diffraction \( (n=1) \) occurs at \( \theta=30^\circ \) with \( \lambda=0.154 \text{ nm} \).
  • Use \( 2 d \sin 30^\circ = 1 \times 0.154 \)
  • \( d = \frac{0.154}{2 \times 0.5} = 0.154 \text{ nm} \)

Practice Set

• Q1: Bragg’s law relates \( d \), \( \theta \), and
  • A) frequency
  • B) wavelength
  • C) amplitude
  • D) velocity
• Q2: For \( n=2 \), the path difference is
  • A) \( \lambda \)
  • B) \( 2 \lambda \)
  • C) \( \lambda/2 \)
  • D) \( 4 \lambda \)
• Q3: If \( \theta \) increases, to satisfy Bragg’s law, \( d \) must
  • A) Increase
  • B) Decrease
  • C) Remain same
  • D) Become zero

Answer Key

• Q1: B - Bragg’s law involves wavelength \( \lambda \).
• Q2: B - Path difference \( = n \lambda = 2 \lambda \) for \( n=2 \).
• Q3: B - \( d \sin \theta = \frac{n \lambda}{2} \), if \( \theta \) increases, \( d \) decreases.

Memory Aids

• Bragg’s law: \( 2 d \sin \theta = n \lambda \)
• Diffraction occurs at specific angles \( \theta \)
• Used to find crystal spacing \( d \)