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Atoms — Notes

6 sections 7 min readBeta

1. The alpha-scattering experiment and Rutherford’s model

  • Thomson’s model: positive charge spread uniformly through the atom, electrons embedded in it; it predicts only tiny deflections of α particles and none thrown back.
  • Geiger–Marsden experiment (1909–1913): a narrow beam of α particles from a radioactive source in a lead block falls on a very thin gold foil in an evacuated chamber; a zinc sulphide screen with a microscope, turned round the foil, counts the scintillations at each scattering angle θ.
  • Observations → conclusions: (i) most α particles pass straight through or are deflected by a degree or two → the atom is mostly empty space; (ii) a few are deflected through large angles → the positive charge is concentrated in a tiny region; (iii) about 1 in several thousand is turned back (θ > 90°) → that region also holds nearly all the mass.
  • Rutherford’s nuclear model: a tiny, massive, positively charged nucleus (radius about 10−1510^{-15}–10−1410^{-14} m) with electrons revolving round it at distances of about 10−1010^{-10} m; the electrons are too light (about 1/7300 of an α particle’s mass) to deflect α particles appreciably.
  • The mathematical theory of α scattering (the impact-parameter formula and Rutherford’s scattering formula) is not in the ISC syllabus.
  • relative count, falling steeply with θ
Most α particles are deflected through small angles; very few, but some, are thrown back.

Key terms

Scintillation

A tiny flash of light produced when an α particle strikes a zinc sulphide screen.

Nucleus

The tiny central part of an atom that contains all its positive charge and nearly all its mass.

2. Distance of closest approach; limits of Rutherford’s model

  • An α particle (charge 2e) of kinetic energy K aimed head-on at a nucleus (charge Ze) stops momentarily at the distance of closest approach d, where all its kinetic energy has become electrostatic potential energy (nucleus treated as fixed): K=14πε0⋅2Ze2d⇒d=14πε0⋅2Ze2KK = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{d}\quad\Rightarrow\quad d = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{K}
  • d is an upper limit to the radius of the nucleus; d ∝ Z/K. Use 1 MeV = 1.6 × 10⁻¹³ J; e24πε0=1.44\dfrac{e^2}{4\pi\varepsilon_0} = 1.44 MeV fm.
  • Example: 5.0 MeV α on gold (Z = 79): d = 4.55 × 10⁻¹⁴ m ≈ 45.5 fm; the atom (≈ 1.4 × 10⁻¹⁰ m) is about 3000 times larger.
  • Limitations of Rutherford’s model: an orbiting electron is accelerated and should radiate (classical electromagnetism), spiralling into the nucleus in about 10−810^{-8} s, so atoms would be unstable; the radiated frequency would change continuously, giving a continuous spectrum instead of the observed line spectrum.
  • electrostatic potential energy U = k(2e)(79e)/r
A head-on α particle stops where U equals its initial kinetic energy: more energy, closer approach.

Formulas

  • Distance of closest approach of an α particle
    d=14πε0⋅2Ze2Kd = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2Ze^2}{K}

Key terms

Distance of closest approach

The least distance from the centre of a nucleus reached by an α particle aimed head-on at it, where its kinetic energy has all become electrostatic potential energy.

3. Bohr’s postulates; radius and speed of the nth orbit

  • Postulates (hydrogen): (1) the electron revolves in certain circular stationary orbits without radiating, the Coulomb force providing the centripetal force; (2) in these orbits the angular momentum is quantised: mvr=nh2πmvr = \dfrac{nh}{2\pi}, n = 1, 2, 3, … (n the principal quantum number); (3) a photon is emitted or absorbed only when the electron jumps between orbits: hν=En2−En1h\nu = E_{n_2} - E_{n_1}.
  • Radius (derivation): mv2r=e24πε0r2\dfrac{mv^2}{r} = \dfrac{e^2}{4\pi\varepsilon_0 r^2} and v=nh2πmrv = \dfrac{nh}{2\pi m r} give rn=ε0n2h2πme2=0.53 n2r_n = \dfrac{\varepsilon_0 n^2h^2}{\pi m e^2} = 0.53\,n^2 Å; r ∝ n², radii 1 : 4 : 9 : 16; r₁ = 0.53 Å is the Bohr radius.
  • Speed (derivation): dividing mv2r=e24πε0mv^2r = \dfrac{e^2}{4\pi\varepsilon_0} by mvr=nh2πmvr = \dfrac{nh}{2\pi}: vn=e22ε0nh=2.2×106nv_n = \dfrac{e^2}{2\varepsilon_0 nh} = \dfrac{2.2 \times 10^6}{n} m s⁻¹; v ∝ 1/n; v₁ ≈ c/137.
  • Frequency of revolution v2πr∝1n3\dfrac{v}{2\pi r} \propto \dfrac{1}{n^3}; linear momentum mv ∝ 1/n; angular momentum ∝ n.

Formulas

  • Bohr’s quantum condition
    mvr=nh2πmvr = \dfrac{nh}{2\pi}
  • Radius of the nth orbit
    rn=ε0n2h2πme2r_n = \dfrac{\varepsilon_0 n^2 h^2}{\pi m e^2}
  • Speed in the nth orbit
    vn=e22ε0nhv_n = \dfrac{e^2}{2\varepsilon_0 n h}

Key terms

Stationary orbit

An allowed Bohr orbit in which the electron does not radiate energy.

Bohr radius

The radius of the first orbit of hydrogen, about 0.53 Å (5.3 × 10⁻¹¹ m).

4. Energy of the electron; meaning of negative energy

  • Kinetic energy: K=12mv2=e28πε0rK = \dfrac{1}{2}mv^2 = \dfrac{e^2}{8\pi\varepsilon_0 r}; potential energy: U=−e24πε0rU = -\dfrac{e^2}{4\pi\varepsilon_0 r}; total: E=K+U=−e28πε0rE = K + U = -\dfrac{e^2}{8\pi\varepsilon_0 r}, so K = −E and U = 2E.
  • With rnr_n: En=−me48ε02n2h2=−13.6n2E_n = -\dfrac{me^4}{8\varepsilon_0^2 n^2 h^2} = -\dfrac{13.6}{n^2} eV; E₁ = −13.6, E₂ = −3.40, E₃ = −1.51, E₄ = −0.85 eV, E∞ = 0.
  • Negative total energy means the electron is bound to the nucleus: zero is the free electron at rest far away, so energy must be supplied to remove it. The more negative E, the more tightly bound.
  • Ionisation energy from level n: 13.6/n² eV (13.6 eV from the ground state; ionisation potential 13.6 V). Excitation energy: E_n − E₁ (10.2 eV to n = 2, 12.09 eV to n = 3).
  • As n increases, E and U increase (towards zero) while K and v decrease.
  • kinetic energy 13.6/n²
  • potential energy −27.2/n²
  • total energy −13.6/n²
K = −E and U = 2E in every orbit; all three tend to zero as n → ∞.

Formulas

  • Total energy of the nth level
    En=−me48ε02n2h2=−13.6n2 eVE_n = -\dfrac{me^4}{8\varepsilon_0^2 n^2 h^2} = -\dfrac{13.6}{n^2}\ \text{eV}
  • Kinetic and potential energy
    K=−E,U=2EK = -E,\quad U = 2E

Key terms

Ionisation energy

The minimum energy needed to remove the electron from an atom in a given state; 13.6 eV for hydrogen in its ground state.

Excitation energy

The energy needed to raise an electron from the ground state to a higher energy level.

5. Energy-level diagram and the five series

  • The energy-level diagram shows the allowed energies as horizontal lines (n = 1 at −13.6 eV, crowding together towards 0 at n = ∞). A downward jump emits a photon hν=En2−En1h\nu = E_{n_2} - E_{n_1}; an upward jump needs exactly that energy to be absorbed.
  • Series (lines ending on the same lower level n₁): Lyman n₁ = 1, ultraviolet (91.2–121.5 nm); Balmer n₁ = 2, visible and near ultraviolet (364.6–656.3 nm; the lines from n₂ = 3–6 are visible); Paschen n₁ = 3, infrared (820–1875 nm); Brackett n₁ = 4, infrared (1459–4051 nm); Pfund n₁ = 5, far infrared (2279–7458 nm).
  • The infrared series overlap; Lyman and Balmer do not.
  • Atoms excited to level n can emit n(n−1)2\dfrac{n(n-1)}{2} different lines. Cool hydrogen (all in n = 1) absorbs only Lyman lines.

Formulas

  • Bohr’s frequency condition
    hν=En2−En1h\nu = E_{n_2} - E_{n_1}
  • Number of possible lines from level n
    N=n(n−1)2N = \dfrac{n(n-1)}{2}

Key terms

Series limit

The shortest wavelength of a spectral series, for the jump from n₂ = ∞ to the series level n₁.

Energy-level diagram

A diagram showing the allowed energies of an atom as horizontal lines, with transitions as vertical arrows.

6. The Rydberg formula; successes and limits of Bohr’s model

  • From hcλ=En2−En1\dfrac{hc}{\lambda} = E_{n_2} - E_{n_1}: 1λ=R(1n12−1n22)\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) with R=me48ε02ch3=1.097×107R = \dfrac{me^4}{8\varepsilon_0^2ch^3} = 1.097 \times 10^7 m⁻¹ (the Rydberg constant). Equivalently λ (nm) = 1240/ΔE (eV).
  • Method: n₁ is the lower level (it fixes the series), n₂ the upper. Series limit λ∞=n12/R\lambda_\infty = n_1^2/R; first line n₂ = n₁ + 1. To identify a line, decide the series from the wavelength range, then solve for n₂.
  • Example: 5 → 2 gives 1/λ = 1.097 × 10⁷ × 0.21, λ = 434 nm (Balmer, violet). A line at 1282 nm is Paschen 5 → 3.
  • A graph of 1/λ against 1/n₂² for one series is a straight line of slope −R.
  • Successes: stability, size and ionisation energy of hydrogen, all its wavelengths. Limits: only one-electron atoms; no line intensities or fine structure; quantum rules assumed, not explained.
  • Balmer series: 1/λ = R/4 − R/n²
A straight line of slope −R: the Balmer wavelengths give the Rydberg constant.

Formulas

  • Rydberg formula
    1λ=R(1n12−1n22)\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)
  • Rydberg constant from Bohr’s theory
    R=me48ε02ch3R = \dfrac{me^4}{8\varepsilon_0^2 c h^3}

Key terms

Rydberg constant

The constant R = 1.097 × 10⁷ m⁻¹ in the formula for the wave numbers of the hydrogen lines.

Exam tips

  • Pair each α-scattering observation with its conclusion; for closest approach use charge 2e and 1 MeV = 1.6 × 10⁻¹³ J.
  • Bohr derivations: start from mv²/r = e²/(4πε₀r²) and mvr = nh/2π; end with r ∝ n², v ∝ 1/n, E ∝ −1/n².
  • Negative energy = bound electron; ionisation energy from level n is 13.6/n² eV.
  • Rydberg formula with n₁ the lower level; give λ in nm and name the series and region.

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