Physics notes, chapter 7

Quantum Physics

Textbook Physics Form 5 (DLP), pages 222–250

What this chapter covers

Explain a black body and why classical physics could not explain black-body radiation (7.1)
Describe quantum of energy, photon energy E = hf and the line spectrum (7.1)
Explain wave-particle duality and use de Broglie's wavelength λ = h/p (7.1)
Solve problems on photon energy and the number of photons per second from the power (7.1)
Describe the photoelectric effect and its four characteristics (7.2)
State Einstein's photoelectric equation, work function and threshold frequency (7.3)
Solve problems with hf = W + Kmax and interpret the Kmax–f graph (7.3)
Describe photocells and applications of the photoelectric effect (7.3)

7.1Quantum theory of light

A black body is an idealised body that absorbs all the electromagnetic radiation that falls on it. It also emits thermal radiation that depends only on its temperature, not on the nature of its surface.

A large cavity with a small hole acts as a black body: light entering the hole is reflected again and again inside and is absorbed, so almost none escapes. The ear cavity is a natural example.

Three curves of radiation intensity against wavelength for 4 000 K, 5 270 K and 7 000 K; the peak moves to shorter wavelength as temperature rises — As the temperature rises the peak moves to a *shorter* wavelength and the intensity at every wavelength rises. Each curve falls on the short-wavelength side.

Why classical physics failed

Classical theory said that the vibrating electrons in a hot object can have any energy, so the intensity should keep growing without limit at high frequency. Experiments showed that it does not: each curve comes back down on the left. This conflict started quantum theory.

Quantum of energy and photons

Physicist	Contribution
Max Planck (1900)	Radiation is emitted in discrete packets called quanta; the energy of one quantum is proportional to the frequency
Albert Einstein (1905)	Light travels as particles called photons; explained the photoelectric effect
Niels Bohr	Line spectrum of hydrogen from electrons moving between energy levels
Louis de Broglie (1924)	All particles have wave properties

Photon energy: E = hf = hc ÷ λ, where h = 6.63 × 10⁻³⁴ J s (Planck's constant) and c = 3.00 × 10⁸ m s⁻¹. A higher frequency means a shorter wavelength and a more energetic photon.

A continuous spectrum (white light through a prism, 400–750 nm) has no gaps. A line spectrum from excited atoms is a set of coloured lines with wavelengths unique to each element, so it identifies the element.

Worked example: photon energy

Compare photons of 400 nm and 750 nm. E = hc ÷ λ.
400 nm: E = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J.
750 nm: E = 2.65 × 10⁻¹⁹ J. The shorter wavelength has the higher energy.

Photons per second from power

If a source of power P emits n photons every second, each of energy hf = hc ÷ λ, then

P = n h f = n h c ÷ λ so n = Pλ ÷ (hc)

Worked example

A 50 W lamp gives red light of λ = 7.0 × 10⁻⁷ m. n = (50)(7.0 × 10⁻⁷) ÷ [(6.63 × 10⁻³⁴)(3.00 × 10⁸)] = 1.76 × 10²⁰ s⁻¹.

Use the output power that goes into light. If only 10% of a 100 W bulb's power is light, P = 10 W, not 100 W. Also convert nm to m (× 10⁻⁹).

Wave-particle duality

Light shows wave properties (diffraction, interference) and particle properties (photons). De Broglie proposed that moving particles such as electrons also behave as waves with wavelength

λ = h ÷ p = h ÷ mv. The greater the momentum, the shorter the wavelength.

Electron diffraction through a thin graphite layer (1927) confirmed this. Because h is so small, the wavelength of a large object is far too short to detect. An electron beam has a wavelength thousands of times shorter than light, so an electron microscope gives a much higher magnification than an optical microscope.

Worked example: de Broglie wavelength

An electron (m = 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁶ m s⁻¹. λ = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 2.0 × 10⁶) = 3.64 × 10⁻¹⁰ m.

What is the energy of a photon of frequency 5.49 × 10¹⁴ Hz (green light)?

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E = hf = 6.63 × 10⁻³⁴ × 5.49 × 10¹⁴ = 3.64 × 10⁻¹⁹ J.

7.2The photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when light of suitable frequency shines on it. The emitted electrons are photoelectrons.

In a photocell the illuminated metal (cathode) emits photoelectrons, which are attracted to the positive anode. They flow round the circuit and the milliammeter shows a photoelectric current.

Four characteristics found by experiment

Observation	Classical wave theory predicted
Kinetic energy of photoelectrons rises with the frequency of the light	Frequency should not matter
No emission below a threshold frequency f₀, however bright	Any frequency should work if bright enough
Kinetic energy does not depend on intensity; brighter light gives more photoelectrons per second, not faster ones	Brighter light should give faster electrons
Emission is instantaneous, even for dim light	Dim light should need time to build up energy

Intensity changes the number of photoelectrons (the current). Frequency changes their energy. A bright red light with f below f₀ gives no photoelectrons at all.

Why is emission instantaneous?

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One photon gives all its energy to one electron at once; the electron does not have to collect energy slowly.

7.3Einstein's photoelectric theory

Einstein (1905) explained all four characteristics: each photon of energy hf is absorbed completely by one electron. Part of the energy frees the electron and the rest is its kinetic energy.

Einstein's photoelectric equation: hf = W + Kmax, that is hf = W + ½mv²max.
Work function W is the minimum energy needed to release an electron from the metal. Threshold frequency f₀: W = hf₀ = hc ÷ λ₀.

Two parallel straight lines of Kmax against f starting on the frequency axis at f0 of metal A and metal B — Kmax against f is a straight line. The **gradient is h** and is the same for every metal. The **f-intercept is f₀**; the line for the metal with the larger f₀ (B) has the larger work function. The Kmax-intercept (extended back) is −W.

Condition	Result
f < f₀ (hf < W)	No photoelectrons
f = f₀	Electrons just escape, Kmax = 0
f > f₀	Kmax = hf − W

Worked example

Blue light of f = 6.67 × 10¹⁴ Hz shines on caesium of work function 3.43 × 10⁻¹⁹ J.
hf = 6.63 × 10⁻³⁴ × 6.67 × 10¹⁴ = 4.42 × 10⁻¹⁹ J.
Kmax = 4.42 × 10⁻¹⁹ − 3.43 × 10⁻¹⁹ = 9.92 × 10⁻²⁰ J.

Worked example: using electron-volts

Light of 550 nm shines on a metal with W = 2.00 eV. Given hc = 1243 eV nm: E = 1243 ÷ 550 = 2.26 eV, so Kmax = 0.26 eV = 0.26 × 1.60 × 10⁻¹⁹ = 4.16 × 10⁻²⁰ J. Then v = √(2K ÷ m) = 3.02 × 10⁵ m s⁻¹.

1 eV = 1.60 × 10⁻¹⁹ J. Keep the whole calculation in eV (with hc = 1243 eV nm) or in J, and convert only at the end. Convert Kmax to joules before using ½mv².

Photocells and applications

A photocell has a light-sensitive cathode and an anode in a vacuum tube; a higher light intensity gives a larger photoelectric current. A metal with a higher work function has a shorter threshold wavelength, which is the longest wavelength that still releases electrons (caesium 579 nm, lithium 496 nm).

Solar cells, solar power plants and solar panels (such as on the ISS)
Light detectors in automatic doors: a broken beam cuts the photocell current
Image sensors in digital cameras

The threshold frequency of a metal is 5.0 × 10¹⁴ Hz. Find its work function and the threshold wavelength.

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W = hf₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.32 × 10⁻¹⁹ J. λ₀ = c ÷ f₀ = 3.00 × 10⁸ ÷ 5.0 × 10¹⁴ = 6.0 × 10⁻⁷ m (600 nm).