Title: InkPhysics / Quantum phenomena
On every note of this deck, unless a note says otherwise:
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Front: DEFINITION
Define: Ground state.
Back: The lowest energy level of an atom, in which its electrons normally sit.

Front: DEFINITION
Define: Work function.
Back: The minimum energy needed to release an electron from the surface of a metal.

Front: DEFINITION
Define: Threshold frequency.
Back: The minimum frequency of radiation that will release electrons from a surface: hf₀ equals the work function.

Front: DEFINITION
Define: The electronvolt.
Back: The energy gained by an electron accelerated through a potential difference of one volt: 1 eV = 1.60 × 10^−19 J.

Front: CHECK YOURSELF
In a fluorescent tube, describe the energy story that turns the kinetic energy of a free electron into visible light. Name each process.
Show a hint
There are four hand-offs, and the ultraviolet step is the one candidates forget.
Back: The tube's pd accelerates a free electron, giving it kinetic energy. A collision excites a mercury atom, transferring one exact level gap.
The atom de-excites almost at once, emitting the gap as an ultraviolet photon (hf equal to the gap).
The phosphor coating absorbs the UV, and its own electrons de-excite in smaller steps, emitting visible photons. Kinetic energy became level gaps became light, in exact parcels the whole way.

Front: CHECK YOURSELF
An electron (m = 9.11 × 10^−31 kg) moves at 3.3 × 10^6 m s^−1. Find its de Broglie wavelength, and state what happens to the diffraction pattern if the electrons are accelerated to a higher speed.
Show a hint
Momentum first. Then remember what diffraction needs from a wavelength.
Back: Momentum: mv = 9.11 × 10^−31 × 3.3 × 10^6 = 3.0 × 10^−24 kg m s^−1.
Wavelength: λ = hmv = 6.63 × 10^−34 / 3.0 × 10^−24 = 2.2 × 10^−10 m, about an atomic spacing, which is why crystals diffract electron beams so well.
At higher speed the momentum grows, λ shrinks, and the electrons diffract less: the rings contract towards the centre.

Front: DEFINITION
Define: Photon.
Back: A discrete packet of electromagnetic energy: E = hf.

Front: CHECK YOURSELF
In hydrogen, an electron falls from the −3.4 eV level to the −13.6 eV ground state. Find the frequency of the emitted photon. (h = 6.63 × 10^−34 J s.)
Show a hint
Gap first, in eV; then joules; then h.
Back: Gap: E_1 - E_2 = (−3.4) − (−13.6) = 10.2 eV. The signs take care of themselves.
In joules: 10.2 × 1.60 × 10^−19 = 1.63 × 10^−18 J.
Frequency: f = E/h = 1.63 × 10^−18 / 6.63 × 10^−34 = 2.5 × 10^15 Hz, in the ultraviolet, as a drop to the ground state usually is.

Front: CHECK YOURSELF
Sodium has a work function of 2.3 eV. Light of frequency 7.0 × 10^14 Hz falls on it. Find the maximum kinetic energy of the photoelectrons, in eV and in joules. (h = 6.63 × 10^−34 J s.)
Show a hint
Find the photon energy in joules first, then run the budget.
Back: Photon energy: E = hf = 6.63 × 10^−34 × 7.0 × 10^14 = 4.64 × 10^−19 J = 2.9 eV.
Budget: E_k(max) = hf - φ = 2.9 − 2.3 = 0.6 eV.
In joules: 0.6 × 1.60 × 10^−19 = 9.6 × 10^−20 J. A photon either pays the 2.3 eV toll or nothing happens; brightness never enters the calculation.

Front: DEFINITION
Define: Excitation.
Back: An atomic electron gaining exactly the energy needed to jump to a higher energy level.

Front: DEFINITION
Define: Ionisation.
Back: An electron gaining enough energy to leave the atom entirely, forming an ion.

Front: DEFINITION
Define: De Broglie wavelength.
Back: The wavelength associated with a moving particle: λ = h/p, where p is the particle's momentum.