Atoms: Complete Concepts and Practice

Big picture

What does an atom look like inside? Experiments showed that atoms are electrically neutral and contain electrons, but the arrangement of positive charge was unknown. Rutherford's alpha-scattering experiment revealed a tiny, dense, positively charged nucleus. Bohr then explained why electrons don't spiral into the nucleus and why hydrogen emits light only at specific wavelengths.

This chapter is formula-rich and numerical: the energy of the nth orbit, wavelengths of spectral lines, and the distance of closest approach are standard exam questions.

Learning goals

• Describe Thomson's model and why it failed.

• Explain the alpha-particle scattering experiment, its observations and conclusions.

• Calculate the distance of closest approach and explain impact parameter.

• State Bohr's postulates and derive the radius and energy of the nth orbit.

• Draw the energy level diagram of hydrogen and explain its spectral series.

• Use the Rydberg formula to find wavelengths.

• Explain Bohr's quantisation using de Broglie waves, and state the model's limitations.

Core ideas

1. Thomson's model

J.J. Thomson (1898) proposed that an atom is a sphere of uniformly distributed positive charge with electrons embedded in it, like seeds in a watermelon ("plum pudding model"). It explained neutrality but couldn't explain the results of alpha scattering or line spectra.

2. Alpha-particle scattering experiment (Geiger–Marsden, under Rutherford)

A beam of alpha particles (5.5 MeV) was directed at a very thin gold foil (about 2.1 × 10⁻⁷ m thick), and the scattered particles were detected on a zinc sulphide screen at different angles.

Observations:

• Most alpha particles passed straight through or were deflected by very small angles.

• A few were deflected by large angles.

• About 1 in 8,000 was deflected by more than 90°, and some bounced almost straight back (about 180°).

Conclusions (Rutherford's nuclear model):

• Most of the atom is empty space.

• All the positive charge and almost all the mass are concentrated in a tiny central nucleus (size about 10⁻¹⁵ to 10⁻¹⁴ m, compared with the atom's 10⁻¹⁰ m).

• Electrons revolve around the nucleus, held by electrostatic attraction, just as planets orbit the sun.

The number of particles scattered at angle θ varies as 1/sin⁴(θ/2), which matched the experiment and confirmed a point-like nucleus.

3. Distance of closest approach and impact parameter

Distance of closest approach (r₀). For a head-on collision, the alpha particle stops momentarily when all its kinetic energy has turned into electric potential energy:

K = (1/4πε₀) × (2e)(Ze) ÷ r₀, so r₀ = (1/4πε₀) × 2Ze² ÷ K

Example. A 5.5 MeV alpha particle approaches a gold nucleus (Z = 79).

r0 = (9 × 10^9) × 2 × 79 × (1.6 × 10^-19)² / (5.5 × 10^6 × 1.6 × 10^-19)
   = (9 × 10^9) × 158 × 1.6 × 10^-19 / (5.5 × 10^6)
   ≈ 4.1 × 10^-14 m

This gives an upper limit on the size of the nucleus.

Impact parameter (b). The perpendicular distance of the initial velocity of the alpha particle from the centre of the nucleus. Small b → large scattering angle; b = 0 is a head-on collision (scattered back at 180°). Large b → small deflection.

Limitation of Rutherford's model. According to classical electromagnetism, an accelerating (revolving) electron should radiate energy continuously, spiral into the nucleus, and emit a continuous spectrum. In reality, atoms are stable and emit line spectra.

4. Bohr's postulates (for the hydrogen atom)

1. Electrons revolve in certain stable orbits (stationary states) without radiating energy.

2. Only those orbits are allowed in which the angular momentum is an integral multiple of h/2π: mvr = nh/2π, where n = 1, 2, 3 … (the principal quantum number).

3. An electron emits or absorbs radiation only when it jumps from one orbit to another. The energy of the photon equals the difference in energies: hν = Ei − Ef.

5. Radius, velocity and energy of the nth orbit

Combining the centripetal force condition (mv²/r = ke²/r²) with mvr = nh/2π gives:

Points to note:

• Radius ∝ n²: r₂ = 4r₁, r₃ = 9r₁.

• Energy is negative, meaning the electron is bound. Kinetic energy = −E, and potential energy = 2E.

• Ground state (n = 1): E₁ = −13.6 eV. Excited states: E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, and E∞ = 0.

• Ionisation energy of hydrogen in the ground state = 13.6 eV (energy needed to take the electron from n = 1 to infinity). From n = 2, it is 3.4 eV.

6. Hydrogen spectrum and the Rydberg formula

When an electron falls from a higher orbit n₂ to a lower orbit n₁, it emits a photon of wavelength λ given by:

1/λ = R (1/n₁² − 1/n₂²), where R = 1.097 × 10⁷ m⁻¹ (Rydberg constant)

Worked example 1: first Balmer line (Hα), n = 3 → 2.

Energy of photon = 13.6 (1/2² - 1/3²) = 13.6 (1/4 - 1/9) = 1.89 eV
λ = 1240 / 1.89 nm ≈ 656 nm (red)

(The shortcut λ in nm = 1240 ÷ E in eV comes from hc = 1240 eV nm.)

Worked example 2: first Lyman line, n = 2 → 1.

E = 13.6 (1 - 1/4) = 10.2 eV
λ = 1240 / 10.2 ≈ 122 nm (ultraviolet)

Series limit: the shortest wavelength of a series, when n₂ → ∞. For Balmer: 1/λ = R/4, λ ≈ 365 nm.

Number of spectral lines when an electron falls from level n to the ground state (all possible jumps): n(n − 1)/2. From n = 4, there are 6 lines.

7. De Broglie's explanation of Bohr's postulate

An electron behaves as a wave of wavelength λ = h/mv. A stable orbit is one in which a whole number of wavelengths fits around the circumference (a standing wave):

2πr = nλ = nh/mv, which gives mvr = nh/2π

This is exactly Bohr's quantisation condition, which de Broglie's wave idea explains naturally.

8. Limitations of the Bohr model

• It applies only to hydrogen-like atoms (one electron: H, He⁺, Li²⁺). It fails for atoms with more electrons.

• It can't explain the relative intensities of spectral lines.

• It can't explain the fine structure of lines or the splitting of lines in electric and magnetic fields.

• It mixes classical ideas (orbits) with quantum ideas without a full theory; it was later replaced by quantum mechanics.

A reliable problem-solving routine

1. For energy questions, use En = −13.6/n² eV; for photons, ΔE = 13.6 (1/n₁² − 1/n₂²) eV.

2. For wavelength, use λ (nm) = 1240 ÷ E (eV), or the Rydberg formula.

3. Identify the series from the lower level n₁.

4. For radius, use rn = 0.53 n² Å.

5. For the closest approach, equate kinetic energy to electric potential energy.

Connections that matter

• Dual nature of radiation and matter gives de Broglie's relation used here.

• Nuclei studies the structure of the nucleus that Rutherford discovered.

• Electrostatics supplies the Coulomb force and potential energy.

Common mistakes

• Taking energy as positive. Bound-state energies are negative.

• Writing radius ∝ n. It is ∝ n².

• Assigning the Balmer series to UV. Balmer is visible; Lyman is UV.

• Forgetting the factor 2 for the alpha particle's charge in r₀.

• Saying a large impact parameter gives large scattering. It is the opposite.

Revision checklist

• I can describe the alpha-scattering experiment and Rutherford's conclusions.

• I can calculate the distance of closest approach.

• I can state Bohr's postulates and use the formulas for radius and energy.

• I can name the spectral series and calculate wavelengths.

• I can explain quantisation using de Broglie waves and state the model's limitations.