Atoms

By 1897 J. J. Thomson had shown that every atom contains negatively charged electrons that are far lighter than the atom itself. Since atoms are electrically neutral, they must also contain an equal amount of positive charge, and this positive part must carry almost all the mass. The question the chapter answers is how the positive charge and the electrons are arranged inside the atom.

Thomson's first guess, the plum-pudding model (1898), spreads the positive charge uniformly over the whole atom and embeds the electrons in it like seeds in a watermelon. It explains neutrality but, as you will see, cannot survive the alpha-scattering experiment.

The other clue was atomic spectra. A gas excited in a discharge tube does not emit a continuous rainbow. It emits only a set of sharp, bright lines at definite wavelengths, and each element has its own set of lines, a fingerprint. Any correct atomic model must explain why the emitted radiation comes only at those particular wavelengths. Rutherford's nuclear model, built from the scattering experiment, explains the structure but not the spectrum and not the stability. Bohr's model, built on top of Rutherford's, explains all three for hydrogen. That progression, Thomson to Rutherford to Bohr, is the spine of the chapter, and the examiner expects you to know what each model got right and what forced the next one.

In 1911, at Rutherford's suggestion, Geiger and Marsden fired alpha-particles from a radioactive source (²¹⁴₈₃Bi) at a thin gold foil about 2.1 × 10⁻⁷ m thick. The beam was collimated by a lead-brick slit, and the scattered alpha-particles were counted by the flashes they produced on a zinc sulphide screen viewed through a microscope. The alpha-particles had kinetic energy 5.5 MeV. Gold was chosen because it is very malleable, so an extremely thin foil could be made, and because the gold nucleus is heavy, so it does not recoil and the scattering is clean.

Observations and what each one means, which is exactly how the marking scheme wants it:

• Most alpha-particles passed straight through with little or no deflection. So most of the atom is empty space.
• Only about 0.14% were scattered by more than 1°, and about 1 in 8000 was deflected by more than 90°, some almost straight back. A 5.5 MeV alpha-particle can only be turned back by a very strong repulsive force, so the entire positive charge and almost all the mass of the atom must be concentrated in a tiny region, the nucleus. The plum-pudding model, with its spread-out charge, cannot produce such a force.

Rutherford's nuclear model: the atom has a small, dense, positively charged nucleus of size about 10⁻¹⁵ m to 10⁻¹⁴ m, while the atom is about 10⁻¹⁰ m across, so the nucleus is roughly 10⁴ times smaller. Electrons revolve around the nucleus in orbits, held by the Coulomb attraction, the way planets are held by gravity. The electrons and the nucleus together make the atom neutral.

The alpha-particle experiences only the Coulomb repulsion of the gold nucleus, F = (1/4πε₀) · (2e)(Ze)/r², with Z = 79 for gold. Since the nucleus is about fifty times heavier than the alpha-particle it stays at rest, and the electrons are far too light to deflect the alpha-particle appreciably. Rutherford calculated the number of alpha-particles scattered at each angle from this force law, and Geiger and Marsden's counts matched, which confirmed the model.

The path of an alpha-particle is fixed by its impact parameter b, the perpendicular distance between the initial velocity vector of the alpha-particle and the centre of the nucleus. A particle with large b passes far from the nucleus, feels a weak force and is barely deflected. As b decreases, the deflection grows. For a head-on approach, b = 0, the particle comes to rest momentarily and is scattered back through 180°. Because the nucleus is so small, only a tiny fraction of the alpha-particles have small enough b to suffer large-angle scattering, which is why large deflections are rare. The relation between b and the scattering angle θ, which NCERT states but does not derive, is b = (Ze² cot(θ/2))/(4πε₀ · ½mv²).

Distance of closest approach (derive this, it is a regular 2- or 3-marker). For the head-on case the alpha-particle of kinetic energy K stops at a distance d from the nucleus where all its kinetic energy has become electrostatic potential energy:

K = (1/4πε₀) · (2e)(Ze)/d, so d = 2Ze²/(4πε₀K).

For a 7.7 MeV alpha-particle on gold (NCERT Example 12.1): K = 7.7 × 1.6 × 10⁻¹³ J = 1.23 × 10⁻¹² J, and d = (9 × 10⁹)(2 × 79)(1.6 × 10⁻¹⁹)²/(1.23 × 10⁻¹²) = 3.64 × 10⁻²⁶/1.23 × 10⁻¹² ≈ 3.0 × 10⁻¹⁴ m = 30 fm. This is an upper limit on the size of the gold nucleus; the actual radius is about 6 fm, so the alpha-particle never touches the nucleus and the pure Coulomb calculation is valid.

Two things the examiner looks for: d ∝ 1/K, so doubling the kinetic energy halves the distance of closest approach; and d ∝ Z, so a heavier nucleus turns the particle back sooner.

In Rutherford's picture the electron moves in a circular orbit of radius r around the nucleus, with the Coulomb attraction supplying the centripetal force. For hydrogen (Z = 1):

mv²/r = (1/4πε₀) · e²/r², so r = e²/(4πε₀ m v²).

From the same equation the kinetic energy is K = ½mv² = e²/(8πε₀r). The potential energy of the electron at distance r from the proton is U = −e²/(4πε₀r), negative because the force is attractive. So the total energy is

E = K + U = e²/(8πε₀r) − e²/(4πε₀r) = −e²/(8πε₀r).

Three relations you should be able to quote at once: K = −E, U = 2E, and U = −2K. The total energy is negative, which means the electron is bound to the nucleus; energy equal to |E| must be supplied to free it. If E were positive the electron would escape and there would be no atom.

Now the fatal problem, which is the standard answer to "state the limitations of Rutherford's model":

• An electron in a circular orbit is accelerated, and classical electromagnetic theory says an accelerated charge radiates. The electron would lose energy continuously, its orbit would shrink, and it would spiral into the nucleus in about 10⁻⁸ s. The atom would collapse. Real atoms are stable.
• As the electron spiralled in, its frequency of revolution would change continuously, so the radiation emitted would have a continuous spread of frequencies. But atoms emit line spectra, with only certain discrete frequencies.

Rutherford's model gives the structure but has no answer to either objection. That is where Bohr comes in.

When a gas or vapour is excited at low pressure by passing an electric current, it emits radiation. Passed through a prism or grating, this radiation shows a set of bright lines on a dark background, an emission line spectrum. The pattern of wavelengths is characteristic of the element, so a spectrum identifies the element the way a fingerprint identifies a person. Conversely, if white light is passed through the same cool gas, dark lines appear in the continuous spectrum at exactly the wavelengths the gas would emit; this is the absorption spectrum.

Hydrogen, the simplest atom, has the simplest spectrum. In 1885 Balmer found that the visible lines of hydrogen, H_α at 656.3 nm (red), H_β at 486.1 nm (blue-green), H_γ at 434.1 nm (violet) and H_δ at 410.2 nm (violet), fit a single empirical formula:

1/λ = R (1/2² − 1/n²), n = 3, 4, 5, ...

with R, the Rydberg constant, equal to 1.097 × 10⁷ m⁻¹. Putting n = 3 gives 656.3 nm, n = 4 gives 486.1 nm and so on. As n increases, the lines crowd closer together and become fainter, converging to the series limit at n → ∞, where 1/λ = R/4 and λ = 364.6 nm.

Other series were found later in the ultraviolet and infrared and are named after their discoverers. Rydberg generalised Balmer's formula to 1/λ = R (1/n_f² − 1/n_i²) with n_i > n_f. These were empirical fits with no physical reason behind them. Explaining where this formula comes from, and why R has the value it has, is the test that Bohr's model passes and Rutherford's fails.

In 1913 Niels Bohr kept Rutherford's nucleus and orbiting electron but added three postulates that classical physics does not permit. Write them exactly like this in the exam.

1. Stationary orbits. An electron in an atom can revolve only in certain stable orbits without emitting radiation, even though it is accelerating. Each such orbit has a definite total energy; these are the stationary states of the atom.
2. Quantisation of angular momentum. The electron revolves only in those orbits for which its angular momentum is an integral multiple of h/2π: L = mvr = nh/2π, n = 1, 2, 3, ... This picks out the allowed orbits from the continuum that Rutherford's model permits.
3. Frequency condition. An electron may jump from a higher-energy orbit to a lower one; when it does, a photon is emitted whose energy equals the energy difference of the two states: hν = E_i − E_f. (Absorption of a photon of the same energy causes the reverse jump.)

Derivation of r_n and v_n (the 3-marker). From the Coulomb force as centripetal force, mv²/r = e²/(4πε₀r²) ... (1). From the second postulate, mvr = nh/2π ... (2). Divide (1) by (2)² after squaring (2): from (2), v = nh/(2πmr); substitute in (1): m n²h²/(4π²m²r³) = e²/(4πε₀r²), giving

r_n = (n²/m) (h/2π)² (4πε₀/e²) = n² ε₀h²/(π m e²).

Then from (2), v_n = nh/(2πmr_n) = (1/n) (e²/4πε₀) (1/(h/2π)) = e²/(2ε₀nh).

So r_n ∝ n² and v_n ∝ 1/n. For n = 1, r₁ = 5.29 × 10⁻¹¹ m, the Bohr radius a₀, which matches the known size of the hydrogen atom; and v₁ = 2.18 × 10⁶ m s⁻¹, about c/137. The number v₁/c = e²/(2ε₀hc) ≈ 1/137 is the fine-structure constant α; its smallness justifies the non-relativistic treatment. The orbital period T_n = 2πr_n/v_n ∝ n³, with T₁ = 1.53 × 10⁻¹⁶ s.

Derivation of E_n (the 5-marker, continued from the radius). Kinetic energy K_n = ½mv_n² = e²/(8πε₀r_n) and potential energy U_n = −e²/(4πε₀r_n), so the total energy is E_n = −e²/(8πε₀r_n). Substituting r_n = n²ε₀h²/(πme²):

E_n = −me⁴/(8 n² ε₀² h²).

Putting in the constants, E_n = −2.18 × 10⁻¹⁸/n² J = −13.6/n² eV. The energy is negative in every orbit, so the electron is bound. E_n ∝ −1/n²: it rises towards zero as n increases, and the levels get closer and closer together. At n → ∞ the energy is zero and the electron is free.

The lowest state n = 1 is the ground state, with E₁ = −13.6 eV. Higher states are excited states: E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, E₅ = −0.54 eV. Memorise these five; almost every transition numerical uses them. The minimum energy needed to free the electron from the ground state is the ionisation energy, 13.6 eV, which agrees with experiment and was one of Bohr's great successes. The energy needed to take the atom from the ground state to the first excited state, 13.6 − 3.4 = 10.2 eV, is the first excitation energy. The ionisation potential is 13.6 V and the first excitation potential is 10.2 V.

The energy-level diagram (Fig. 12.4 in NCERT) is drawn as horizontal lines, n = 1 at the bottom labelled −13.6 eV, n = 2 above it at −3.4 eV, and so on, with n = ∞ at 0 eV at the top. The spacing shrinks rapidly upward. At room temperature almost all hydrogen atoms sit in the ground state; they must be excited by a collision or a photon before they can emit.

In every state K_n = −E_n = 13.6/n² eV and U_n = 2E_n = −27.2/n² eV. In the ground state, K = 13.6 eV and U = −27.2 eV, which is NCERT Exercise 12.4 and a routine 2-mark question.

Bohr's third postulate turns the energy levels into a spectrum. For a jump from a higher level n_i to a lower level n_f, the photon energy is hν = E_i − E_f = (me⁴/(8ε₀²h²)) (1/n_f² − 1/n_i²). With ν = c/λ,

1/λ = (me⁴/(8ε₀²h³c)) (1/n_f² − 1/n_i²) = R (1/n_f² − 1/n_i²).

This is exactly Rydberg's empirical formula, and the theory now predicts the constant: R = me⁴/(8ε₀²h³c), which NCERT evaluates as 1.03 × 10⁷ m⁻¹, close to the measured 1.097 × 10⁷ m⁻¹. Explaining Balmer's formula from first principles is the strongest evidence for Bohr's model. Use R = 1.097 × 10⁷ m⁻¹ in numericals, or the shortcut E (in eV) = 1240/λ (in nm).

Each series is a set of transitions ending on the same lower level:

• Lyman series: n_f = 1, n_i = 2, 3, 4, ...; ultraviolet. First line 121.6 nm, series limit 91.2 nm.
• Balmer series: n_f = 2, n_i = 3, 4, ...; visible (and near UV). First line H_α 656.3 nm, limit 364.6 nm.
• Paschen series: n_f = 3; infrared. First line 1875 nm.
• Brackett series: n_f = 4; infrared.
• Pfund series: n_f = 5; infrared.

The first line of a series (smallest n_i) has the longest wavelength; the series limit (n_i → ∞) has the shortest wavelength, λ_min = n_f²/R. The number of possible lines when atoms are excited to level n and then de-excite by every route is n(n − 1)/2: six lines from n = 4, three from n = 3.

Absorption works the same way in reverse: a ground-state atom absorbs a photon only if its energy equals E_n − E₁ for some n. A photon of 12.75 eV lifts it to n = 4; a photon of 12.5 eV, which matches no level, is not absorbed at all.

Limitations of Bohr's model (learn two): it applies only to hydrogen and hydrogen-like one-electron ions, not to atoms with more than one electron, because it ignores electron-electron interactions; and it gives the frequencies of the lines but says nothing about their relative intensities, nor about fine structure, nor about the splitting of lines in magnetic fields.

Bohr could not say why angular momentum should come only in multiples of h/2π. In 1923 de Broglie supplied the reason: the electron is a wave as well as a particle, with wavelength λ = h/mv, and an orbit is allowed only if the electron wave fits into it as a standing wave.

Think of a string fixed at both ends. It vibrates only in modes where a whole number of half-wavelengths fit the length. For a wave running round a closed loop the condition is that a whole number of wavelengths fit the circumference; otherwise the wave interferes destructively with itself after each round and dies out. So for the nth orbit,

2πr_n = nλ, n = 1, 2, 3, ...

Substituting λ = h/mv_n: 2πr_n = nh/(mv_n), which rearranges to mv_nr_n = nh/2π. That is Bohr's second postulate exactly. Quantisation of angular momentum is nothing more than the statement that the electron's matter wave must be a standing wave around the orbit.

The diagram the examiner wants: a circle for the orbit with a wavy line drawn around it that closes on itself after exactly n wavelengths (draw n = 4 or so, the standing-wave pattern with crests and troughs alternating around the circle). Label 2πr = nλ.

Two more points that fetch marks. First, the electron speed in the ground state, 2.18 × 10⁶ m s⁻¹, gives λ = h/mv = 6.63 × 10⁻³⁴/(9.1 × 10⁻³¹ × 2.18 × 10⁶) = 3.3 × 10⁻¹⁰ m, which is exactly the circumference 2π × 5.29 × 10⁻¹¹ m of the first orbit, as it should be for n = 1. Second, the de Broglie picture removes the arbitrariness but not the other limitations of the Bohr model; a full treatment needs quantum mechanics, where the electron has no definite orbit at all, only a probability distribution.