Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: quantum physics, photons, line spectra and duality (203-SN3-RE)

Quantum physics at the CEGEP level asks for no new intuition: it asks you to know which relation applies to which object, and in which unit. Marks are lost on a power of ten, a forgotten square, an inverse not taken, rarely on the idea.

This sheet gathers the traps of 203-SN3-RE tests (Waves and Modern Physics, formerly 203-NYC) on the photon, the Bohr model, line spectra and wave-particle duality, the decision tree that picks the relation, a model write-up and a worked example of electron diffraction.

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The thread of the chapter

At the scale of the atom, energy is exchanged only in packets, and it is a wave condition that fixes those packets: a bound electron is a standing wave, only certain wavelengths fit, so only certain energies exist. The traps are about units and powers, never about the idea.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

The photon and the energy levels

• Photon: $E=hf=\dfrac{hc}{\lambda}$ and $p=\dfrac{h}{\lambda}$ . Shortcut: $E(\mathrm{eV})=\dfrac{1240}{\lambda(\mathrm{nm})}$ ; $1$ eV $=1.60\times 10^{-19}$ J.
• Hydrogen (Bohr): $E_{n}=-\dfrac{13.6}{n^{2}}$ eV, $r_{n}=n^{2}a_{0}$ with $a_{0}=0.0529$ nm. One-electron ion of charge $Z$ : $E_{n}=-\dfrac{13.6\,Z^{2}}{n^{2}}$ .
• A transition emits or absorbs a photon whose energy is EXACTLY the gap: $h f=|E_{upper}-E_{lower}|$ . Above the ionization energy, any energy will do.
• Rydberg formula: $\dfrac{1}{\lambda}=R_{H}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)$ , $R_{H}=1.097\times 10^{7}$ m⁻¹. Lyman $n_{1}=1$ , Balmer $n_{1}=2$ , Paschen $n_{1}=3$ .

A drop emits, a climb absorbs, and each arrow has the exact length of a gap between two levels: 10.2 eV for 2 to 1, 1.89 eV for 3 to 2.

Energies are negative because zero is the free electron, at rest at infinity: a bound electron must be given energy to escape.

Duality and quantization by waves

• Every particle has a de Broglie wavelength $\lambda=\dfrac{h}{p}$ ; electron accelerated through $V$ : $\lambda=\dfrac{h}{\sqrt{2m_{e}eV}}$ .
• Electron in a box of width $L$ : standing wave, $L=n\dfrac{\lambda}{2}$ , hence $E_{n}=\dfrac{n^{2}h^{2}}{8mL^{2}}$ with $n\ge 1$ .
• X-rays: cutoff $\lambda_{min}=\dfrac{hc}{eV}$ ; Moseley's law $E_{K\alpha}\approx 10.2\,(Z-1)^{2}$ eV.
• The intensity of light sets the NUMBER of photons; their energy depends only on the frequency.

The walls impose nodes: only

\lambda=2L/n

fit (blue), hence

E_{n}\propto n^{2}

and never

n=0

. A wave that does not vanish at the wall (red) does not exist.

To quantize is to force a wave to fit in a closed space: only certain wavelengths work, so only certain energies exist.

The X-ray spectrum: the cutoff and the lines

• The continuous background, bremsstrahlung, stops abruptly at $\lambda_{min}=\dfrac{hc}{eV}$ : an electron gives at most ALL of its energy $eV$ to a single photon.
• $\lambda_{min}$ depends only on the voltage: $0.0413$ nm at $30$ kV, $0.0827$ nm at $15$ kV.
• The K $\alpha$ and K $\beta$ lines depend only on the metal of the anode: for copper, K $\alpha$ is near $0.155$ nm (Moseley), as soon as the voltage is enough to ionize the K shell.

Doubling the voltage halves

\lambda_{min}

, from

0.083

0.041

nm, without moving the copper lines: the cutoff tells the voltage, the lines tell the metal.

The blackbody: Planck against Rayleigh-Jeans

• Planck's law: $B_{P}=\dfrac{2hc^{2}}{\lambda^{5}}\cdot\dfrac{1}{e^{x}-1}$ with $x=\dfrac{hc}{\lambda kT}$ . The classical Rayleigh-Jeans law, $B_{RJ}=\dfrac{2ckT}{\lambda^{4}}$ , diverges as $\lambda\to 0$ : the ultraviolet catastrophe.
• Their ratio is $\dfrac{B_{RJ}}{B_{P}}=\dfrac{e^{x}-1}{x}$ : $28.6$ at $500$ nm for $5800$ K, $1.14$ at $10$ µm. The two laws meet only at long wavelengths, where $x\ll 1$ .
• Planck assumes that the energy of a mode of frequency $f$ is exchanged only in quanta $hf$ : when $hf\gg kT$ , these quanta are no longer excited and the spectrum shuts off.

5800

K, the classical law (red) shoots off to infinity on the ultraviolet side; Planck (blue) peaks near

500

nm and then falls back. The ratio of the two is

28.6

500

nm.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Leaving the wavelength in nanometres in hc over lambda

the whole question, and a photon a billion times too weak

What not to write

“ $E=\dfrac{6.63\times 10^{-34}\times 3.00\times 10^{8}}{550}=3.6\times 10^{-28}$ J.”

What to write

“ $\lambda=550\times 10^{-9}$ m: $E=3.62\times 10^{-19}$ J, that is $2.25$ eV.”

Why: With $h$ in J·s and $c$ in m/s, $\lambda$ must be in metres. The shortcut $\dfrac{1240}{\lambda(\mathrm{nm})}$ in eV avoids the conversion, but only if you keep the two units it assumes.

2. Believing that more intense light gives more energetic photons

2 to 3 marks on the reasoning question

What not to write

“Doubling the intensity of the lamp doubles the kinetic energy of the ejected electrons.”

What to write

“Doubling the intensity doubles the number of photons, hence the number of electrons and the current; the maximum kinetic energy depends only on the frequency.”

Doubling the intensity doubles the NUMBER of photons (green); each keeps its

1.9

eV, fixed by the frequency. The wrong answer (red) doubles the energy of each photon.

Why: A photon acts alone: its energy does not add to that of a neighbour. This is precisely the observation that wave theory could not explain and that the photon explains.

3. Forgetting to invert the Rydberg formula

the whole question

What not to write

“ $\lambda=R_{H}\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.52\times 10^{6}$ m.”

What to write

“The formula gives $\dfrac{1}{\lambda}=1.52\times 10^{6}$ m⁻¹, so $\lambda=6.56\times 10^{-7}$ m $=656$ nm.”

The Balmer lines crowd together toward the limit at 365 nm: none is shorter, and H alpha at 656 nm is the longest of the series.

Why: The unit of R, metres to the power minus one, announces the inverse of a length. A wavelength of light of a thousand kilometres should make you reread the line.

4. Making the Bohr radius grow as n

2 marks

What not to write

“Orbit $n=3$ : $r_{3}=3a_{0}=0.159$ nm.”

What to write

“ $r_{n}=n^{2}a_{0}$ : $r_{3}=9\times 0.0529=0.476$ nm.”

On the left, radii in 1, 2, 3: wrong. On the right, radii in 1, 4, 9: the third orbit is nine times larger than the first.

Why: The condition $m_{e}vr=n\hbar$ combined with the electric force gives a radius in $n^{2}$ and a speed in $\dfrac{1}{n}$ . The orbits spread apart faster and faster, while the energy levels crowd together.

5. Using Z instead of Z squared for a one-electron ion

2 marks, and every line of the ion wrong

What not to write

“He⁺: $E_{ion}=13.6\times 2=27.2$ eV.”

What to write

“ $E_{ion}=13.6\,Z^{2}=13.6\times 4=54.4$ eV.”

The electron of He⁺ is bound by

54.4

eV (green), four times more than in hydrogen: double the charge AND an orbit twice as close.

27.2

eV (red) is the wrong answer.

Why: A nucleus with twice the charge attracts twice as strongly AND the electron is twice as close: the two effects multiply.

6. Doubling the voltage to halve the wavelength

2 marks

What not to write

“To go from 0.167 to 0.084 nm, accelerate the electrons through 108 V instead of 54 V.”

What to write

“ $\lambda\propto\dfrac{1}{\sqrt{V}}$ : you need $4\times 54=216$ V.”

\lambda

varies as

1/\sqrt{V}

: doubling the voltage,

108

V (red), gives only

0.118

nm; it must be quadrupled,

216

V (green), to reach

0.084

nm.

Why: Momentum varies as the square root of the kinetic energy, hence as the square root of the voltage. Halving the wavelength takes four times the energy.

7. Forgetting screening in Moseley's law

2 marks, and an element misidentified

What not to write

“Copper, $Z=29$ : $E_{K\alpha}=10.2\times 29^{2}=8.58$ keV.”

What to write

“ $E_{K\alpha}=10.2\,(Z-1)^{2}=10.2\times 28^{2}=8.00$ keV, that is $0.155$ nm.”

Why: The electron that falls into the K shell sees the nucleus through the electron that remains there: the effective charge is Z minus 1. Forgetting that 1 shifts the identification of an element.

Which method to choose

Which relation, depending on what the problem involves

The object described by the problem

If a wavelength or a frequency of light, a photon energy → $E=\dfrac{hc}{\lambda}$ , or $\dfrac{1240}{\lambda(\mathrm{nm})}$ in eV

Example: $\dfrac{1240}{656}=1.89$ eV
If lines of hydrogen, levels n → $E_{n}=-\dfrac{13.6}{n^{2}}$ eV or Rydberg

Example: $3\to 2$ : $13.6\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.89$ eV
If an ion with a single electron (He⁺, Li²⁺) → same model with $Z^{2}$ in the energies, $\dfrac{1}{Z}$ in the radii

Example: Li²⁺: $13.6\times 9=122.4$ eV
If an electron or a neutron that diffracts → $\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}$

Example: $54$ V: $0.167$ nm
If a particle confined between two walls → standing wave: $E_{n}=\dfrac{n^{2}h^{2}}{8mL^{2}}$

Example: $L=1$ nm: $E_{1}=0.377$ eV
If an X-ray tube → $\lambda_{min}=\dfrac{hc}{eV}$ and Moseley $10.2(Z-1)^{2}$ eV

Example: $30$ kV: $0.0413$ nm

Before any calculation, choose the energy unit for the whole problem: electronvolts for atoms and photons, joules as soon as a mass or a momentum appears.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Working out a transition, written up for the marker

When to use it: The problem gives a photon absorbed or emitted by a hydrogen atom and asks for the level reached or the line produced.

1 Convert the wavelength to an energy, in eV: $E=\dfrac{1240}{\lambda(\mathrm{nm})}$ .
2 Write the energy balance: $E_{n}=E_{initial}+E_{photon}$ for an absorption, $E_{photon}=E_{upper}-E_{lower}$ for an emission.
3 Solve $-\dfrac{13.6}{n^{2}}=E_{n}$ and check that $n$ is an integer; if not, the photon is not absorbed.
4 Conclude with a sentence that names the transition and the region of the radiation.

Concluding sentence

“The photon of $97.3$ nm carries $12.74$ eV; $-13.6+12.74=-0.86$ eV corresponds to $n=4$ : the electron goes from level 1 to level 4.”

The trap: Rounding a non-integer n to the nearest: a photon of 11.0 eV would give n of about 2.3 and is simply not absorbed.

Marking: Typically 1 mark for the conversion, 1 for the balance, 1 for the integer n, 1 for the conclusion.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The visible photon between 1.8 and 3.1 eV

Any visible photon energy you calculate must fall in this interval; ultraviolet is above it, infrared below.

$3.6\times 10^{-28}$ J for green light: wavelength left in nanometres.
The integer level

After an absorption, the calculated n must be an integer to within 1 %. Otherwise, either the photon is not absorbed, or the energy is wrong.

$n=3.99$ : level 4. $n=2.29$ : no absorption.
Atomic de Broglie wavelengths

An electron of a few tens to a few hundred electronvolts has a wavelength of the order of a tenth of a nanometre, comparable to the distances between atoms.

$0.1$ nm for $150$ V: consistent. $10^{-7}$ m: a power-of-ten error.
Balmer lines in the visible

A line that ends on level 2 must lie between 365 and 656 nm.

A Balmer line at $1875$ nm is really the Paschen transition $4\to 3$ .

The typical problem, taken apart

An electron beam diffracted by a crystal

Electrons start from rest and are accelerated through $150$ V, then strike a crystal whose rows of atoms are spaced $d=0.215$ nm apart. $h=6.63\times 10^{-34}$ J·s, $m_{e}=9.11\times 10^{-31}$ kg, $e=1.60\times 10^{-19}$ C.

Calculate the wavelength of the electrons and the angle of the first maximum, given by $d\sin\theta=\lambda$ .

The situation: electrons accelerated through

150

V strike a crystal whose rows are spaced

d=0.215

nm apart; the angle

\theta

of the first maximum remains to be found.

Step 1

Kinetic energy: $E=eV=1.60\times 10^{-19}\times 150=2.40\times 10^{-17}$ J.

Why

We switch to joules because what follows involves a mass: electronvolts do not mix with kilograms.

Step 2

Momentum: $p=\sqrt{2m_{e}E}=\sqrt{2\times 9.11\times 10^{-31}\times 2.40\times 10^{-17}}=6.61\times 10^{-24}$ kg·m/s.

Why

We go through $p$ rather than $v$ : it is $p$ that enters the de Broglie relation.

Step 3

$\lambda=\dfrac{h}{p}=\dfrac{6.63\times 10^{-34}}{6.61\times 10^{-24}}=1.00\times 10^{-10}$ m $=0.100$ nm.

Why

The result is of the order of interatomic distances: this is what makes diffraction observable.

Step 4

$\sin\theta=\dfrac{0.100}{0.215}=0.466$ , so $\theta=27.8°$ .

Why

The crystal behaves like a grating: the same relation as in optics, with the wavelength of matter.

Step 5

Check: $v=\dfrac{p}{m_{e}}=7.26\times 10^{6}$ m/s, $2.4$ % of $c$ .

Why

Below 5 to 10 % of the speed of light, the non-relativistic calculation is justified; a marker appreciates seeing it checked.

The conclusion, written out

“Electrons of $150$ eV have a wavelength of $0.100$ nm and give a first maximum at $27.8°$ : matter diffracts like light.”

The classic mistake on this problem: Putting the voltage of 150 V directly in place of the energy under the root, without multiplying it by the charge of the electron.

Learn by heart

• $E=\dfrac{hc}{\lambda}$ , $p=\dfrac{h}{\lambda}$ ; $E(\mathrm{eV})=\dfrac{1240}{\lambda(\mathrm{nm})}$ .
• Hydrogen: $E_{n}=-\dfrac{13.6}{n^{2}}$ eV, $r_{n}=n^{2}a_{0}$ ; ion: $Z^{2}$ in the energy.
• Rydberg gives $\dfrac{1}{\lambda}$ : invert at the end.
• A transition absorbs only the exact gap; above $13.6$ eV, ionization.
• de Broglie: $\lambda=\dfrac{h}{\sqrt{2meV}}$ , $\lambda\propto\dfrac{1}{\sqrt{V}}$ .
• Box: $E_{n}=\dfrac{n^{2}h^{2}}{8mL^{2}}$ , never $n=0$ .
• X-rays: $\lambda_{min}=\dfrac{hc}{eV}$ ; Moseley $10.2(Z-1)^{2}$ eV.
• Intensity: number of photons. Frequency: energy of each.

Frequently asked questions

Why are the energy levels of the hydrogen atom negative?

Because the zero of energy is set for a free electron, at rest and infinitely far from the nucleus. An electron bound to the atom has less energy than that free electron: it must be given energy to tear it away. The energy of the ground state, minus 13.6 electronvolts, means that 13.6 electronvolts are needed to ionize the atom.

How do you know whether a photon is absorbed by a hydrogen atom?

You add the energy of the photon to that of the starting level and check whether the result matches an allowed level exactly, that is, whether the quantum number you obtain is an integer. If it is not, the photon passes through the gas without being absorbed, unless it exceeds the ionization energy, in which case it can tear the electron away.

What is the de Broglie wavelength?

It is the wavelength associated with any moving particle, equal to Planck's constant divided by its momentum. It is measurable for electrons, which diffract off crystals like X-rays, but it is tiny for a macroscopic object, which explains why we never see a ball diffract.

Why does the Rydberg formula give an inverse wavelength?

Because it is proportional to the energy gap between two levels, and the energy of a photon is inversely proportional to its wavelength. The Rydberg constant is therefore expressed in metres to the power minus one, and you must take the inverse of the result to get the wavelength itself.

Practise it

Corrected exercises: Quantum physics: photons, line spectra and duality, 203-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys3-quantum-physics. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: quantum physics, photons, line spectra and duality (203-SN3-RE)

The essentials

The photon and the energy levels

Duality and quantization by waves

The X-ray spectrum: the cutoff and the lines

The blackbody: Planck against Rayleigh-Jeans

The mistakes that cost marks

1. Leaving the wavelength in nanometres in hc over lambda

2. Believing that more intense light gives more energetic photons

3. Forgetting to invert the Rydberg formula

4. Making the Bohr radius grow as n

5. Using Z instead of Z squared for a one-electron ion

6. Doubling the voltage to halve the wavelength

7. Forgetting screening in Moseley's law

Which method to choose

Which relation, depending on what the problem involves

How the answer is expected to be written

Working out a transition, written up for the marker

Check before you hand in

The typical problem, taken apart

An electron beam diffracted by a crystal

Learn by heart

Frequently asked questions

Why are the energy levels of the hydrogen atom negative?

How do you know whether a photon is absorbed by a hydrogen atom?

What is the de Broglie wavelength?

Why does the Rydberg formula give an inverse wavelength?

Corrected exercises: Quantum physics: photons, line spectra and duality, 203-SN3-RE

See also

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