Atoms: common doubts, answered
The questions students ask most often about Atoms, each with a short answer. For the full chapter, read the Atoms notes.
About the chapter
What is the difference between the Rutherford and Bohr models of the atom?
Rutherford's model placed a tiny, dense, positive nucleus at the centre with electrons orbiting it, but it used only classical physics, so it could not explain why atoms are stable or why they give line spectra. Bohr kept the nucleus and added quantum rules: only orbits with mvr = nh/2π are allowed, an electron in such an orbit does not radiate, and light is emitted or absorbed only in jumps, with hν = Ei − Ef.
Early models of the atom
Read this section in the notes →What was wrong with Thomson's plum pudding model of the atom?
It could not explain the large-angle scattering of alpha-particles. With the positive charge spread thinly through the whole atom, the force on a passing alpha-particle would be weak everywhere, so none should be turned through a large angle. Geiger and Marsden found that a few were turned right back, which needs a concentrated positive charge. The model also could not account for the line spectra of atoms.
Alpha-particle scattering
Read this section in the notes →Why did most alpha-particles pass straight through the gold foil?
Because an atom is mostly empty space. A nucleus is only about 10⁻¹⁵ m to 10⁻¹⁴ m across, while the atom is about 10⁻¹⁰ m, so a particle aimed at random almost never passes close to a nucleus. Away from it the alpha-particle feels hardly any force, and the light electrons barely affect it. Only about 1 in 8000 was deflected by more than 90°.
What did the alpha-particles that bounced back from the gold foil prove?
They proved that all the positive charge and most of the mass of an atom are packed into a tiny central nucleus. Turning a fast, heavy alpha-particle back needs an enormous repulsive force, which is possible only if it can come very close to a large, concentrated charge. The measured numbers scattered at different angles matched this picture, and so Rutherford is credited with discovering the nucleus.
Alpha-particle trajectory
Read this section in the notes →Why must the foil in Rutherford's scattering experiment be very thin?
So that each alpha-particle is scattered by at most one nucleus. The gold foil used was only about 2.1 × 10⁻⁷ m thick. With a single scattering per particle, the analysis can follow one alpha-particle past one nucleus using Coulomb's law and Newton's second law. In a thick foil many small deflections would pile up at random and hide the single large ones that reveal the nucleus.
What is impact parameter, and how does it affect the scattering angle?
The impact parameter b is the perpendicular distance between the alpha-particle's initial line of motion and the centre of the nucleus. A small b means a close pass and a large deflection; b near zero, a head-on approach, sends the particle almost straight back. A large b leaves it nearly undeviated. Since very few particles happen to have a tiny b, very few are scattered through large angles.
Is the distance of closest approach equal to the radius of the nucleus?
No, it only gives an upper limit on the nuclear size. In a head-on approach the alpha-particle stops where all its kinetic energy has turned into electric potential energy, at d = 2Ze²/(4πε₀K). For a 7.7 MeV alpha-particle on gold this is about 30 fm, while the gold nucleus has a radius of only about 6 fm. The particle turns back well before it reaches the nucleus.
Electron orbits
Read this section in the notes →Why is the total energy of the electron in a hydrogen atom negative?
Because the electron is bound to the nucleus. Energy is measured from zero for an electron at rest infinitely far away. The attractive potential energy, −e²/(4πε₀r), is twice the size of the kinetic energy, e²/(8πε₀r), so the total is E = −e²/(8πε₀r). Energy has to be supplied to free the electron. A positive total energy would mean the electron is not held in a closed orbit.
How are the kinetic, potential and total energies of the electron in hydrogen related?
K = −E and U = 2E, where E is the negative total energy. So the kinetic energy equals the size of the total energy, and the potential energy is twice the total. In the ground state E = −13.6 eV, K = +13.6 eV and U = −27.2 eV. Writing U = −E, or giving the total energy a positive sign, are the usual slips.
Atomic spectra
Read this section in the notes →What is the difference between an emission spectrum and an absorption spectrum?
An emission spectrum shows bright lines on a dark background, given out by an excited gas; an absorption spectrum shows dark lines across a continuous spectrum, where a gas has removed certain wavelengths from white light passing through it. The dark lines fall at exactly the wavelengths of that gas's bright lines, because an atom absorbs the same frequencies it can emit.
Why does a hot solid give a continuous spectrum while a hot gas gives lines?
Because in a solid the atoms are packed closely and strongly influence one another, so a continuous range of wavelengths comes out. In a rarefied gas the atoms are far apart, so the light comes from individual atoms, and each can emit only particular wavelengths set by its own energy levels. That produces separate bright lines, which act as a fingerprint for the element.
Why Rutherford's atom fails
Read this section in the notes →Why couldn't Rutherford's model explain the stability of atoms?
Because an orbiting electron is always accelerating, and classical electromagnetism says an accelerating charge radiates energy. Losing energy steadily, the electron should spiral inward and fall into the nucleus, yet real atoms are stable. The model also predicted the wrong kind of spectrum: the revolution frequency would change continuously as the electron spiralled in, giving a continuous spectrum instead of sharp lines.
Bohr's postulates
Read this section in the notes →What are Bohr's three postulates?
First, an electron can move in certain stable orbits without radiating energy. Second, these are only the orbits in which its angular momentum is a whole multiple of h/2π, so mvr = nh/2π. Third, the atom emits or absorbs light only when the electron jumps between two such orbits, and the photon carries away or brings in the energy difference, hν = Ei − Ef.
Is the frequency of light emitted by a hydrogen atom equal to the electron's frequency of revolution?
No. In the Bohr model the photon's frequency is set by the energy difference between two levels, ν = (Ei − Ef)/h, not by how fast the electron circles. This was a sharp break from classical physics, in which a revolving charge radiates at its revolution frequency. The two values agree only for jumps between neighbouring orbits with very large n.
Energy levels
Read this section in the notes →How do the radius, speed and energy of a Bohr orbit depend on n?
For hydrogen, the radius grows as n², the speed falls as 1/n and the energy goes as −1/n². So rn = n²a₀, with a₀ = 5.3 × 10⁻¹¹ m, and En = −13.6/n² eV. The second orbit is four times as large as the first and its electron moves at half the speed. These are the n-dependences only; for a one-electron ion the nuclear charge changes the values too. Do not make the radius or the speed proportional to n.
Why do the energy levels of hydrogen get closer together as n increases?
Because the energy goes as −1/n², and the differences between successive values of 1/n² shrink rapidly. Going from n = 1 to n = 2 takes 10.2 eV, but from n = 2 to n = 3 only 1.89 eV, and higher gaps are smaller still. Near the top the levels crowd towards E = 0, above which the electron is free and can have any energy.
What is the difference between ionisation energy and excitation energy of hydrogen?
Ionisation energy is the least energy that frees the electron completely, 13.6 eV from the ground state. Excitation energy only lifts the electron to a higher level without freeing it: 10.2 eV to reach n = 2 and 12.09 eV to reach n = 3. An atom that is already excited needs less energy to be ionised, because its electron is less tightly bound.
Line spectra of hydrogen
Read this section in the notes →Why does hydrogen, with only one electron, give so many spectral lines?
Because a sample contains enormous numbers of atoms, excited to many different levels, and their electrons fall back by many different routes. Each atom gives one photon per jump, but different atoms make different jumps at the same time, from various higher levels to various lower ones. Each possible jump has its own frequency, hν = Ei − Ef, so the sample shows many lines together.
How do I calculate the wavelength of light emitted in a hydrogen transition?
Find the photon energy from hν = 13.6 eV (1/nf² − 1/ni²), then use λ = hc/E with hc about 1240 eV nm. A drop from n = 3 to n = 2 gives 1.89 eV, about 656 nm, which is red light. A drop from n = 2 to n = 1 gives 10.2 eV, about 122 nm, in the ultraviolet. Bigger energy jumps always give shorter wavelengths.
De Broglie's explanation
Read this section in the notes →How did de Broglie explain Bohr's quantisation of angular momentum?
He treated the orbiting electron as a wave and required a whole number of wavelengths to fit round the orbit, like a standing wave on a string: 2πrn = nλ. Putting in λ = h/mv gives mvr = nh/2π, exactly Bohr's condition. Only orbits that hold a complete standing wave survive; on any other orbit the wave would interfere with itself and die out.
Limits of the Bohr model
Read this section in the notes →Why doesn't the Bohr model work for helium and other many-electron atoms?
Because it has no way of including the forces between electrons. In a many-electron atom each electron is repelled by the other electrons with forces comparable to its attraction to the nucleus, so the simple orbit picture breaks down. The model ignores this, so it works only for one-electron systems such as H, He⁺ and Li²⁺. It also cannot predict why some spectral lines are brighter than others.
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