CBSE Class 12 Physics Atoms Notes Set 05

1. Thomson's Model of the Atom

By the end of the nineteenth century it was known that atoms are electrically neutral and contain negatively charged electrons. J. J. Thomson (1898) proposed that the positive charge of an atom is spread uniformly over a sphere of atomic size, with the electrons embedded in it like seeds in a watermelon (the plum pudding model).
1. The model explains the overall neutrality of the atom.
2. It predicts that an α-particle passing through the atom should be deflected only very slightly, because the positive charge is thinly spread.
3. It could not explain the large-angle scattering of α-particles observed later, and was therefore given up.

2. Alpha-Particle Scattering Experiment

In 1911, at Rutherford's suggestion, H. Geiger and E. Marsden directed a narrow beam of α-particles (helium nuclei, charge +2e, energy about 5.5 MeV from a radioactive source) at a very thin gold foil (thickness about 2.1 × 10⁻⁷ m). The scattered α-particles were detected by the flashes (scintillations) they produced on a zinc sulphide screen viewed through a microscope. The screen and microscope could be rotated to count particles scattered through different angles θ.

Observations:
1. Most of the α-particles passed straight through the foil without any deflection.
2. Only about 0.14% of the α-particles were scattered by more than 1°.
3. About one α-particle in 8000 was deflected by more than 90°, and a very few bounced back almost along their original path (θ ≈ 180°).

Gold is used because it is very malleable and can be beaten into an extremely thin foil, so each α-particle usually suffers at most one close encounter with a nucleus. Gold nuclei are also heavy, so they hardly recoil.

3. Rutherford's Nuclear Model of the Atom

To explain these results Rutherford proposed the nuclear model:
1. The entire positive charge and almost all the mass of the atom are concentrated in a tiny central core called the nucleus.
2. The electrons move around the nucleus in orbits, much as planets go round the sun. The electrostatic attraction of the nucleus provides the centripetal force.
3. Most of the atom is empty space. The size of the nucleus is about 10⁻¹⁵ m to 10⁻¹⁴ m, while the size of the atom is about 10⁻¹⁰ m, so the atom is roughly 10⁴ to 10⁵ times larger than its nucleus.
4. The atom as a whole is neutral: the total negative charge of the electrons equals the positive charge Ze of the nucleus, where Z is the atomic number.

Most α-particles pass undeflected because they travel through the empty space far from any nucleus. Large-angle scattering occurs only for the few α-particles that come very close to a nucleus, where the repulsive Coulomb force is very strong.

4. Alpha-Particle Trajectory: Impact Parameter and Distance of Closest Approach

The impact parameter b is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus. The path of the α-particle depends on b.
1. Large b: the α-particle stays far from the nucleus and is deflected very little.
2. Small b: it comes close to the nucleus and is scattered through a large angle.
3. b = 0 (head-on collision): the α-particle is brought momentarily to rest and then rebounds, so θ = 180°.
Since only a very small fraction of the α-particles have a small impact parameter, very few are scattered through large angles.

Distance of closest approach (r₀): In a head-on collision the α-particle moves towards the nucleus until its whole kinetic energy K has been converted into electrostatic potential energy. At that point it stops and turns back. For a nucleus of charge Ze and an α-particle of charge 2e,
K = \( \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r_0} \)  ⇒  r₀ = \( \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{K} \)
1. r₀ is inversely proportional to K: a faster α-particle comes closer to the nucleus.
2. r₀ gives an upper limit to the size of the nucleus. For 5.5 MeV α-particles on gold (Z = 79), r₀ ≈ 4 × 10⁻¹⁴ m.

5. Electron Orbits in Rutherford's Model

For a hydrogen atom (Z = 1) with the electron in a circular orbit of radius r, the electrostatic attraction provides the centripetal force:
\( \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \)  ⇒  r = \( \frac{e^2}{4\pi\varepsilon_0 m v^2} \)
Kinetic energy: K = ½mv² = \( \frac{e^2}{8\pi\varepsilon_0 r} \)
Potential energy: U = −\( \frac{e^2}{4\pi\varepsilon_0 r} \)
Total energy: E = K + U = −\( \frac{e^2}{8\pi\varepsilon_0 r} \)
1. The total energy is negative, which means the electron is bound to the nucleus. Energy must be supplied to remove it.
2. K = −E and U = 2E, so the potential energy is twice the total energy and the kinetic energy equals its magnitude.
3. The experimentally measured energy needed to separate a hydrogen atom into a proton and an electron is 13.6 eV, which gives an orbit radius of about 5.3 × 10⁻¹¹ m.

6. Limitations of Rutherford's Model

1. Stability of the atom: An electron moving in a circle is accelerated. According to classical electromagnetic theory an accelerated charge radiates energy as electromagnetic waves. The electron would therefore lose energy continuously, spiral inwards and fall into the nucleus in a tiny fraction of a second. Real atoms are stable, so the model cannot be fully correct.
2. Nature of the spectrum: As the electron spirals in, its frequency of revolution would change continuously, so the atom would emit a continuous spectrum. Atoms actually emit light only at certain discrete wavelengths (a line spectrum).

7. Atomic Spectra (Qualitative)

1. When an element in the form of a low-pressure gas or vapour is excited (for example by an electric discharge), it emits light only at certain particular wavelengths. Seen through a spectroscope, this appears as a set of bright lines on a dark background, called an emission line spectrum.
2. When white light passes through a gas, the gas absorbs light of the same particular wavelengths, giving dark lines on a bright continuous background, called an absorption spectrum.
3. Each element has its own characteristic line spectrum, which acts like a fingerprint and is used to identify elements, for example in stars.
4. Hydrogen, the simplest atom, has the simplest line spectrum. Explaining it was the main success of Bohr's model.

8. Bohr's Postulates

Niels Bohr (1913) combined Rutherford's nuclear model with the quantum idea of Planck and Einstein. His three postulates are:
1. Stationary orbits: An electron in an atom can revolve only in certain stable orbits without emitting radiant energy, contrary to classical theory. Each such state has a definite total energy.
2. Quantisation of angular momentum: The allowed orbits are those for which the angular momentum of the electron is an integral multiple of h/2π.
L = mvr = \( \frac{nh}{2\pi} \), where n = 1, 2, 3, … is the principal quantum number.
3. Frequency condition: An electron can make a transition from one stationary orbit to another of lower energy. A photon is then emitted whose energy equals the difference between the initial and final energies:
hν = E_i − E_f

9. Radius, Speed and Energy in the nth Orbit

Combining the force equation of Section 5 with the quantisation condition mvr = nh/2π gives the following results for hydrogen (the derivation of the radius is not required; the expressions are used directly).

(i) Radius of the nth orbit:
r_n = \( \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \) = n² a₀, where a₀ = 5.3 × 10⁻¹¹ m = 0.53 Å is the Bohr radius (radius of the innermost orbit). So r_n ∝ n².
(ii) Speed of the electron in the nth orbit:
v_n = \( \frac{e^2}{2\varepsilon_0 n h} \) = \( \frac{2.18 \times 10^6}{n} \) m s⁻¹. So v_n ∝ 1/n; the electron moves fastest in the innermost orbit.
(iii) Total energy in the nth orbit:
E_n = −\( \frac{m e^4}{8 \varepsilon_0^2 n^2 h^2} \) = −\( \frac{13.6}{n^2} \) eV. So E_n ∝ −1/n².
(iv) In every orbit, K_n = −E_n = \( \frac{13.6}{n^2} \) eV and U_n = 2E_n = −\( \frac{27.2}{n^2} \) eV.

Note: 1 eV = 1.6 × 10⁻¹⁹ J. As n increases, the radius increases, the speed and kinetic energy decrease, and the total and potential energies increase (become less negative).

10. Energy Levels of the Hydrogen Atom

The allowed energies E_n = −13.6/n² eV are shown as horizontal lines in an energy-level diagram.
1. Ground state (n = 1): the lowest energy state, E₁ = −13.6 eV, with the electron in the smallest orbit.
2. Excited states (n = 2, 3, 4, …): E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, E₅ = −0.54 eV, …
3. As n → ∞, E_n → 0. An electron with zero (or positive) energy is free; the atom is then ionised.
4. Ionisation energy: the minimum energy needed to remove the electron from the ground state, 13.6 eV for hydrogen. The ionisation potential is 13.6 V.
5. Excitation energy: the energy needed to raise the electron from the ground state to an excited state, for example 10.2 eV to reach n = 2 and 12.09 eV to reach n = 3.
6. The levels crowd together as n increases, because the gap between successive levels decreases.
7. At room temperature almost all hydrogen atoms are in the ground state. An excited atom stays excited for only about 10⁻⁸ s before returning to a lower state.

11. Line Spectra of the Hydrogen Atom (Qualitative)

1. Emission: When an electron jumps from a higher level E_i to a lower level E_f, a photon of energy hν = E_i − E_f is emitted. Since only certain energy levels exist, only certain photon energies, and hence certain wavelengths, can be emitted. This is why the spectrum consists of lines.
2. Wavelength of the emitted photon: λ = \( \frac{hc}{E_i - E_f} \). A convenient form is λ (in nm) = \( \frac{1240}{\Delta E \text{ (in eV)}} \).
3. A larger energy gap gives a photon of higher frequency and shorter wavelength. Jumps ending on n = 1 involve large gaps (ultraviolet), while some jumps ending on n = 2 give visible light; for example, the jump from n = 3 to n = 2 releases 1.89 eV, giving red light of wavelength about 656 nm.
4. Absorption: An atom in a lower state can absorb a photon only if the photon energy exactly equals the difference between that state and a higher allowed state. This explains the dark lines of the absorption spectrum, which appear at the same wavelengths as the corresponding emission lines.
5. If electrons in many atoms are excited to the level n, the total number of different spectral lines that can be emitted as they return to the ground state is \( \frac{n(n-1)}{2} \).

12. de Broglie's Explanation of Bohr's Second Postulate

Bohr's quantisation of angular momentum was a bold assumption. In 1923 Louis de Broglie suggested that an electron has a wave nature, with wavelength λ = h/mv. An electron moving round a circular orbit behaves like a wave travelling round a closed loop. A steady (stationary) state can exist only if the wave joins up with itself, forming a standing wave, so the circumference must contain a whole number of wavelengths:
2πr_n = nλ = \( \frac{nh}{mv_n} \)  ⇒  mv_nr_n = \( \frac{nh}{2\pi} \)
This is exactly Bohr's second postulate. Thus the stationary orbits are those in which the electron wave forms a standing wave, and the number of wavelengths in the nth orbit is n.

13. Limitations of Bohr's Model

1. It applies only to hydrogen and hydrogen-like (single-electron) ions such as He⁺ and Li²⁺. It cannot be extended to atoms with two or more electrons, because it does not account for the forces between electrons.
2. It explains the wavelengths of the hydrogen lines but cannot explain their relative intensities.
3. It mixes classical ideas (definite circular orbits) with quantum conditions in an arbitrary way; a complete explanation needs quantum mechanics.

14. Quick Formula Recap