Atoms

INTRODUCTION TO ATOMIC STRUCTURE

**Thomson's Plum Pudding Model (1898):** J.J. Thomson proposed that an atom consists of a uniformly distributed positive charge throughout its volume with negatively charged electrons embedded within it (like seeds in a watermelon). This model was later disproven by experimental evidence.

**Atomic Spectra and Historical Context:**

Rarefied gases emit **discrete wavelengths** of light forming **emission line spectra** (series of bright lines), not continuous spectrum

Each element has a **characteristic spectrum** serving as a fingerprint for identification

In 1885, Johann Balmer obtained an empirical formula for hydrogen wavelengths, suggesting intimate relationship between atomic structure and emitted radiation

Classical electromagnetic theory predicted atoms should emit continuous spectrum, contradicting observations — this inconsistency motivated new atomic models

**Size of Atom vs Nucleus:**

Atom radius: ~10⁻¹⁰ m (Angstrom)

Nucleus radius: ~10⁻¹⁵ m (fermi or fm)

Ratio: Nucleus is 10,000 to 100,000 times smaller than atom

**Atom is mostly empty space**

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RUTHERFORD'S NUCLEAR MODEL AND ALPHA-PARTICLE SCATTERING EXPERIMENT

**Historical Background:** Ernst Rutherford suggested the classic alpha-particle scattering experiment to investigate atomic structure. Hans Geiger and Ernst Marsden conducted this groundbreaking experiment around 1911.

Experimental Setup and Procedure

**Components:**

**Radioactive source:** ²¹⁴₈₃Bi emitting 5.5 MeV α-particles

**Target:** Thin gold foil of thickness 2.1 × 10⁻⁷ m

**Detection:** Rotatable zinc sulphide (ZnS) screen with microscope

Apparatus enclosed in **vacuum chamber** to prevent collision with air molecules

**Method:**

α-particles emitted from radioactive source collimated into narrow beam using lead bricks

Beam directed perpendicular onto thin gold foil

Scattered α-particles strike ZnS screen producing bright light flashes (scintillations)

Flashes observed through microscope; number of particles counted at different scattering angles

Experimental Observations

**Key Results:**

Most α-particles (majority) pass straight through foil undeviated

Only ~0.14% of incident α-particles scatter by more than 1°

**About 1 in 8000 particles deflect by more than 90°** (some rebound back)

Scattering follows specific angular distribution matching theoretical predictions

**Figure 12.3 Analysis:** Experimental data points agree with theoretical curve based on assumption that target atom has small, dense, positively charged nucleus.

Rutherford's Interpretation and Nuclear Model

**Rutherford's Key Deduction:** To deflect an α-particle through large angles (especially >90°), it must experience enormous repulsive force. This is only possible if:

**Entire positive charge and most of atomic mass concentrated in tiny nucleus at center**

Electrons at relatively large distance from nucleus in surrounding space

**The Nuclear Model Postulates:**

1. **Nucleus:** Small, dense, positively charged, contains nearly all atomic mass; radius ~10⁻¹⁵ m

2. **Electrons:** Orbit nucleus at distance ~10⁻¹⁰ m (similar to planets around sun)

3. **Mostly empty space:** Most of atom's volume is empty; explains why most α-particles pass through unaffected

4. **Light electrons don't affect α-particles:** Electrons too light to appreciably scatter heavy α-particles

**Why Most Particles Pass Through:** Since nucleus occupies tiny fraction of atomic volume and foil is thin, probability of α-particle hitting nucleus is very small.

Force on Alpha-Particle in Nuclear Field

When α-particle (charge = 2e) approaches nucleus (charge = Ze), Coulomb repulsive force acts:

**F = (1/4πε₀) × (2e)(Ze)/r² = (1/4πε₀) × 2Ze²/r²** ... (12.1)

Where:

**F** = magnitude of repulsive force (N)

**1/4πε₀** = 8.99 × 10⁹ N⋅m²/C²

**Z** = atomic number of target nucleus

**e** = elementary charge = 1.6 × 10⁻¹⁹ C

**r** = distance between α-particle center and nucleus center (m)

Force is directed along line joining the two charged objects and continuously changes as α-particle approaches and recedes.

Impact Parameter and Scattering Angle

**Impact Parameter (b):** Perpendicular distance between initial velocity vector of α-particle and nucleus center (Fig. 12.4)

**Key Relationships:**

**Small impact parameter (b ≈ 0):** α-particle comes very close to nucleus, experiences large repulsive force, large scattering angle (θ ≈ π, head-on collision, particle rebounds back)

**Large impact parameter (b large):** α-particle far from nucleus, weak force, small scattering angle (θ ≈ 0, nearly undeviated)

**Particle beam has distribution of impact parameters** → particles scatter at various angles with different probabilities

**Physical Significance:** Small fraction of particles backscattered indicates nucleus is tiny — most particles miss it.

Determination of Nucleus Size

**Rutherford Scattering provides upper limit on nuclear radius.** When α-particle approaches nucleus head-on (b = 0), it decelerates due to Coulomb repulsion, momentarily stops at distance d from nucleus, then rebounds.

Using **conservation of energy:**

Initial energy = Final energy

KE_initial + PE_initial = KE_final + PE_final

½m_α v₀² + 0 = 0 + (1/4πε₀)(2e)(Ze)/d

**d = (1/4πε₀) × 2Ze²/K_α** ... (distance of closest approach)

Where K_α = kinetic energy of α-particle.

**Worked Example:**

For 7.7 MeV α-particle on gold (Z = 79):

K = 7.7 MeV = 1.2 × 10⁻¹² J

d = (9.0 × 10⁹) × (2)(79)(1.6 × 10⁻¹⁹)²/(1.2 × 10⁻¹²)

d = 3.0 × 10⁻¹⁴ m = 30 fm

**Nuclear radius < distance of closest approach** (α-particle reverses motion before touching nucleus).

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CLASSICAL ATOMIC MODEL: LIMITATIONS

Electron Orbits in Rutherford Model

**Dynamical Stability Condition:** Electrostatic force of attraction provides centripetal force for circular orbit:

**F_e = F_c**

**(1/4πε₀) × e²/r² = m_e v²/r** ... (12.2)

Where:

**F_e** = electrostatic attraction force

**F_c** = centripetal force

**m_e** = electron mass = 9.1 × 10⁻³¹ kg

**v** = orbital velocity (m/s)

**r** = orbital radius (m)

**From Eq. (12.2):**

**r = (1/4πε₀) × e²/(m_e v²)** ... (12.3)

Energy of Electron in Hydrogen Atom

**Kinetic Energy (K):**

K = ½m_e v² = (1/8πε₀) × e²/r

**Potential Energy (U):**

U = -(1/4πε₀) × e²/r

**Total Energy (E):**

E = K + U = (1/8πε₀) × e²/r - (1/4πε₀) × e²/r

**E = -(1/8πε₀) × e²/r** ... (12.4)

**Negative E indicates electron is bound to nucleus** (requires energy to remove electron).

Fundamental Problems with Classical Model

**Problem 1: Radiation and Energy Loss**

According to classical electromagnetic theory, any accelerating charged particle emits electromagnetic radiation

Orbiting electron continuously accelerates (centripetal acceleration = v²/r)

**Electron should continuously radiate energy, lose energy, and spiral into nucleus**

Prediction: Atom should collapse in ~10⁻¹¹ seconds

**Problem 2: Continuous vs Discrete Spectrum**

As electron spirals inward, orbital frequency continuously changes

Emitted radiation frequency equals orbital frequency (classical prediction)

**Should produce continuous spectrum**, not observed discrete lines

Actual observation: **Discrete line spectrum**

**Problem 3: Stability**

No mechanism in classical theory to prevent electron from falling into nucleus

Atoms observed to be stable

**Conclusion:** Classical physics cannot explain atomic structure. New quantum ideas needed.

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ATOMIC SPECTRA

Types of Spectra

**Emission Line Spectrum:**

Produced when atomic gas/vapor excited by electric current (glow tubes, neon signs, mercury vapor lamps)

Appears as **bright discrete lines on dark background**

Each element has **characteristic set of wavelengths**

Example: Hydrogen spectrum (Fig. 12.5) shows series of lines in visible and ultraviolet regions

**Absorption Spectrum:**

White light passes through cool gas and analyzed

**Dark lines appear exactly at wavelengths emitted in emission spectrum**

Indicates gas absorbs photons of specific energies

"Fingerprint" of element: different elements have different spectral patterns

Historical Significance of Hydrogen Spectrum

Hydrogen simplest atom (one electron)

Shows clear pattern of discrete lines

Balmer (1885): Empirical formula for line wavelengths

Suggests **quantization of atomic structure** — only certain discrete electron orbits allowed

Spectrum depends on internal structure of atom

**Key Insight:** Discrete spectrum indicates **discrete energy levels in atom**, contrary to classical prediction of continuous spectrum.

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BOHR MODEL OF HYDROGEN ATOM

Historical Context and Motivation

Niels Bohr (1885-1962) developed new model incorporating **quantum concepts** to explain:

1. **Stability of atoms** — why electrons don't spiral into nucleus

2. **Discrete line spectra** — why only certain wavelengths emitted

3. Limitations of classical theory

**Bohr's Key Innovation:** Apply **quantum condition** (discrete energy states) to classical orbit equations.

Bohr's Postulates

**Postulate 1: Discrete Orbits**

Electron moves in **discrete circular orbits** around nucleus

Only certain orbits are allowed; characterized by **principal quantum number n = 1, 2, 3, ...**

Not all classical orbits possible; only those satisfying quantum condition

**Postulate 2: No Radiation in Orbit**

Contrary to classical theory: **Electron in allowed orbit does NOT radiate energy**

Electron remains in orbit indefinitely

Each orbit has definite fixed energy

**Postulate 3: Quantum Condition (Angular Momentum Quantization)**

**Angular momentum of electron must be integral multiple of ℏ** (reduced Planck's constant)

**L = n(ℏ) = n(h/2π)** where n = 1, 2, 3, ...

**m_e vr = n(h/2π)** ... Angular momentum quantization

Where:

**h** = Planck's constant = 6.626 × 10⁻³⁴ J⋅s

**ℏ** = h/2π = 1.055 × 10⁻³⁴ J⋅s

**m_e** = electron mass

**v** = electron orbital velocity

**r** = orbital radius

**n** = principal quantum number (1, 2, 3, ...)

**Postulate 4: Radiation and Energy Transitions**

**When electron transitions between orbits, it emits or absorbs photon**

**Photon energy = difference in orbital energies**

**hν = E₁ - E₂** where E₁ > E₂ (emission); hν = E₂ - E₁ (absorption)

Photon frequency ν = (E_final - E_initial)/h

**No radiation occurs within single orbit**

Derivation of Bohr's Formulas

**Step 1: Force Balance**

Centripetal force = Electrostatic attraction:

**m_e v²/r = (1/4πε₀) × e²/r²** ... (From Rutherford)

This gives:

**m_e v r = (1/4πε₀) × e²/v** ... (i)

**Step 2: Apply Quantum Condition**

Angular momentum quantization:

**m_e v r = n(h/2π)** ... (ii)

**Step 3: Combine (i) and (ii)**

From (i) and (ii):

**(1/4πε₀) × e²/v = n(h/2π)**

**v = (1/4πε₀) × 2πe²/(nh)** ... (iii)

**Step 4: Find Orbital Radius**

From force balance: **v² = (1/4πε₀) × (e²/m_e r)**

Substituting (iii):

**r = (1/4πε₀) × (n²h²)/(4π²m_e e²)**

**r_n = n² × (h²/4π²ε₀ m_e e²) = n² × a₀** ... (12.5)

Where **a₀ = (1/4πε₀) × h²/(πm_e e²) = 0.529 × 10⁻¹⁰ m** is **Bohr radius** (radius of first orbit, n=1)

**r_n = n² a₀** ... (Bohr's radius formula)

For hydrogen:

n = 1: r₁ = a₀ = 0.53 Å (ground state)

n = 2: r₂ = 4a₀ = 2.12 Å

n = 3: r₃ = 9a₀ = 4.77 Å

**Step 5: Find Orbital Velocity**

From Eq. (iii):

**v_n = (1/4πε₀) × 2πe²/(nh) = (1/4πε₀) × e²/(nℏ)**

Defining: **v₁ = (1/4πε₀) × e²/ℏ = e²/(4πε₀ × h/2π)**

**v_n = v₁/n** ... (12.6)

For hydrogen:

n = 1: v₁ = 2.2 × 10⁶ m/s ≈ c/137 (c = speed of light)

n = 2: v₂ = v₁/2 = 1.1 × 10⁶ m/s

Velocity **decreases** as n increases

**Step 6: Find Total Energy**

Kinetic energy:

**K_n = ½m_e v_n² = ½m_e × (v₁/n)² = (1/2n²) × m_e v₁²**

Potential energy:

**U_n = -(1/4πε₀) × e²/r_n = -(1/4πε₀) × e²/(n²a₀)**

Total energy:

**E_n = K_n + U_n = (1/2n²) × m_e v₁² - (1/n²) × (m_e v₁²)**

**E_n = -(1/2n²) × m_e v₁² = -13.6/n² eV** ... (12.7)

Where **E₁ = -13.6 eV** is **Rydberg energy** (ionization energy of hydrogen from ground state).

**Alternative form:**

**E_n = -[m_e e⁴/(32π²ε₀² ℏ²n²)] = -13.6/n² eV**

For hydrogen:

n = 1: E₁ = -13.6 eV (ground state, most stable)

n = 2: E₂ = -3.4 eV

n = 3: E₃ = -1.51 eV

n = ∞: E_∞ = 0 eV (ionized, electron free)

**Key Properties:**

**Energy negative** indicates bound state

**Energy becomes less negative with increasing n** (higher orbits have more energy, closer to ionization)

**Energy spacing decreases** as n increases: ΔE ∝ 1/n²

All energies quantized (discrete values only)

Photon Emission and Absorption

**When electron transitions from orbit n_f to orbit n_i (n_f > n_i):**

**ΔE = E_{n_f} - E_{n_i} = -13.6(1/n_f² - 1/n_i²) eV** ... Energy difference

**For emission (electron falls from higher to lower orbit, n_f > n_i):**

**hν = E_{n_i} - E_{n_f} = 13.6(1/n_i² - 1/n_f²) eV**

**Frequency: ν = R_H c(1/n_i² - 1/n_f²)** ... (12.8)

Where **R_H = 1.097 × 10⁷ m⁻¹** is **Rydberg constant for hydrogen**

**Wavelength:**

**1/λ = R_H(1/n_i² - 1/n_f²)** ... (12.9)

**Example Spectral Series:**

**Lyman series** (n_i = 1, n_f = 2,3,4,...): UV lines (λ ≈ 91-121 nm)

**Balmer series** (n_i = 2, n_f = 3,4,5,...): Visible lines (λ ≈ 656-365 nm) — H_α, H_β, etc.

**Paschen series** (n_i = 3, n_f = 4,5,6,...): IR lines

**Brackett series** (n_i = 4, n_f = 5,6,7,...): IR lines

Bohr Model Success and Limitations

**Successes:**

1. **Explains discrete line spectrum** of hydrogen — correct wavelengths

2. **Predicts stability** — electrons in quantized orbits don't radiate

3. **Explains ionization energy** — requires 13.6 eV to remove electron from ground state

4. **Quantitatively accurate for hydrogen** and hydrogen-like ions (He⁺, Li²⁺, etc.)

5. **Introduces quantum concepts** to atomic physics

**Limitations (why Bohr model incomplete):**

1. **Fails for multi-electron atoms** — cannot account for chemical properties or predict spectra of atoms beyond hydrogen

2. **Doesn't explain line widths and intensities** properly

3. **Doesn't account for fine structure** — some spectral lines split into closely spaced lines

4. **No physical justification for quantization** — postulates added artificially without derivation

5. **Doesn't explain chemical bonding**

6. **Incorrect angular momentum** for ground state (should be 0 in quantum mechanics, but Bohr gives ℏ)

7. **Orbits not physically real** — electron position/momentum cannot be simultaneously defined

8. **Resolved by Quantum Mechanics** — Schrödinger equation provides complete quantum theory

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WORKED NUMERICAL EXAMPLES

**Example 12.1: Bohr Radius and Scale Comparison**

*Problem:* Rutherford's nuclear model has nucleus radius ~10⁻¹⁵ m and electron orbit radius ~10⁻¹⁰ m (ratio 10⁵). If our solar system had same proportions with sun radius 7 × 10⁸ m, what would be Earth's orbital radius? (Actual Earth orbit: 1.5 × 10¹¹ m)

*Solution:*

If Earth's orbit is 10⁵ times sun's radius:

Orbit radius = 10⁵ × 7 × 10⁸ m = 7 × 10¹³ m

This is 100× greater than actual orbit

**Earth would be much farther from sun**

**Atom contains vastly more empty space than solar system**

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**Example 12.2: Distance of Closest Approach**

*Problem:* What is the distance of closest approach for a 7.7 MeV α-particle scattered by gold nucleus (Z = 79)?

*Solution:*

At closest approach, all kinetic energy converts to potential energy (conservation of energy):

K_initial = U_final

½m_α v₀² = (1/4πε₀) × (2e)(Ze)/d

Where we use: for α-particle charge = 2e; for gold nucleus charge = Ze

Solving for d:

**d = (1/4πε₀) × 2Ze²/K**

Substituting values:

K = 7.7 MeV = 7.7 × 1.6 × 10⁻¹³ J = 1.23 × 10⁻¹² J

Z = 79

1/4πε₀ = 8.99 × 10⁹ N⋅m²/C²

e = 1.6 × 10⁻¹⁹ C

d = (8.99 × 10⁹) × (2)(79)(1.6 × 10⁻¹⁹)²/(1.23 × 10⁻¹²)

d = (8.99 × 10⁹) × (158) × (2.56 × 10⁻³⁸)/(1.23 × 10⁻¹²)

d = 3.0 × 10⁻¹⁴ m = **30 fm**

**Nuclear radius < 30 fm** (α-particle stops before touching nucleus)

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**Example 12.3: Hydrogen Atom Parameters**

*Problem:* Ionization energy of hydrogen is 13.6 eV. Find orbital radius and velocity in ground state.

*Solution:*

From Eq. (12.4), total energy:

E₁ = -(1/8πε₀) × e²/r₁ = -13.6 eV = -13.6 × 1.6 × 10⁻¹⁹ J = -2.176 × 10⁻¹⁸ J

Solving for r₁:

r₁ = -(1/8πε₀) × e²/E₁

r₁ = -(8.99 × 10⁹) × (1.6 × 10⁻¹⁹)²/[2 × (-2.176 × 10⁻¹⁸)]

r₁ = -[8.99 × 10⁹ × 2.56 × 10⁻³⁸]/[-4.352 × 10⁻¹⁸]

r₁ = **5.3 × 10⁻¹¹ m = 0.53 Å** (Bohr radius a₀)

For velocity, from Eq. (12.3): m_e v²r = (1/4πε₀)e²

v = √[(1/4πε₀)e²/(m_e r₁)]

v = √[(8.99 × 10⁹ × (1.6 × 10⁻¹⁹)²)/(9.1 × 10⁻³¹ × 5.3 × 10⁻¹¹)]

v = **2.2 × 10⁶ m/s**

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**Example 12.4: Hydrogen Spectrum (Balmer Series)**

*Problem:* Calculate wavelength of Hα line (n=3 to n=2 transition in Balmer series).

*Solution:*

Using Eq. (12.9):

1/λ = R_H(1/n_i² - 1/n_f²)

For Hα: n_i = 2, n_f = 3

1/λ = 1.097 × 10⁷ (1/4 - 1/9)

1/λ = 1.097 × 10⁷ (9 - 4)/(36)

1/λ = 1.097 × 10⁷ × 5/36

1/λ = 1.524 × 10⁶ m⁻¹

λ = 1/(1.524 × 10⁶) = **6.56 × 10⁻⁷ m = 656 nm** (Red light, visible region)

Energy of photon:

E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(6.56 × 10⁻⁷) = 3.03 × 10⁻¹⁹ J = **1.89 eV**

Or: ΔE = 13.6(1/4 - 1/9) = 13.6 × 5/36 = **1.89 eV** ✓

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**Example 12.5: Ionization Energy**

*Problem:* How much energy required to ionize hydrogen atom from ground state?

*Solution:*

Ionization means removing electron from n=1 to n=∞:

E_ionization = E_∞ - E₁ = 0 - (-13.6 eV) = **13.6 eV**

In joules: 13.6 × 1.6 × 10⁻¹⁹ = **2.18 × 10⁻¹⁸ J**

This matches experimentally observed ionization energy.

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KEY FORMULAS SUMMARY

| Formula | Name | Application |

|---------|------|-------------|

| F = (1/4πε₀) × 2Ze²/r² | Coulomb force on α-particle | Scattering calculation |

| r_n = n²a₀ where a₀ = 0.53 Å | Bohr radius formula | Orbital radius (n-th level) |

| v_n = v₁/n where v₁ = 2.2 × 10⁶ m/s | Bohr velocity formula | Orbital speed (n-th level) |

| E_n = -13.6/n² eV | Bohr energy formula | Total energy (n-th level) |

| hν = E_{n_i} - E_{n_f} | Photon emission | Wavelength/frequency of lines |

| 1/λ = R_H(1/n_i² - 1/n_f²) | Rydberg formula | Hydrogen spectral lines |

| R_H = 1.097 × 10⁷ m⁻¹ | Rydberg constant | Spectral line calculations |

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EXAM-IMPORTANT POINTS

1. **Rutherford's key deduction:** Most α-particles pass through because nucleus tiny; backscattering proves concentrated charge

2. **Classical model failure:** Accelerating electron should radiate continuously and spiral into nucleus — contradicts observation

3. **Bohr's revolutionary idea:** Quantization of angular momentum (L = nℏ) explains discrete orbits and line spectrum

4. **Energy quantization:** Only discrete energies allowed; transitions between them emit/absorb photons of specific frequencies

5. **Negative energy significance:** E < 0 means bound state; E = 0 means ionized

6. **For any hydrogen transition:** Always use ΔE = 13.6(1/n_i² - 1/n_f²) eV

7. **Bohr radius a₀ = 0.53 Å:** Ground state (n=1) orbital radius

8. **Ionization energy = 13.6 eV** for hydrogen ground state

9. **Bohr model accurate only for hydrogen and hydrogen-like ions** (He⁺, Li²⁺, etc.) with single electron

10. **Modern quantum mechanics** (Schrödinger equation) provides complete theory; Bohr model is stepping stone