1. Electromagnetic Waves
Light has long been known to have a wave-like character. Frequency, ν, and wavelength, λ, are related by ν = c ⁄ λ, where c is the speed of light (approximately 3 × 108m s−1).
Example 1: The frequency of BBC Radio 4 on FM is approximately 93MHz. What is its wavelength?
Solution: The frequency is 93 × 106 = 9.3 × 107Hz. The wavelength is therefore:
Exercise 1: Various types of electromagnetic radiation are described below. If the frequency is given, calculate the wavelength and vice versa.
Solution: First, convert the wavelength into S.I. units:
600nm = 600 × 10−9 = 6 × 10−7m
Then:
Solution:
This very small scale, 1Å, is comparable to atomic spacing in crystals, and this fact is the basis of X-ray crystallography.
Solution: Note that the wavelength was given in μm! We need to first convert it into metres by dividing by a factor of 106, i.e. λ = 1.5 × 10−6m.
Solution:
Click on questions to reveal solutions
Quiz 1: Which of the following might emit electromagnetic radiation with frequency ν = 1016Hz?
Solution: This frequency corresponds to λ ≈ 3 × 10−8m. It is thus ultra-violet radiation, which is emitted by the Sun.
Quiz 2: If the gap between two planes in a particular crystal is 0.75nm, what frequency of X-rays would have a wavelength of half this size?
Solution: Convert the gap into metres: 0.75nm = 7.5 × 10−10m
So half this is ½ × 7.5 × 10−10m and the equivalent frequency is:
2 The Photoelectric Effect
As well as its wave nature, light has a particle like character which is revealed in the photoelectric effect. Einstein's equation for the photoelectric effect is:
- where E is the kinetic energy of electrons emitted from a surface irradiated by light of frequency ν, h (≈ 6.6 × 10−34Js) is Planck's constant and W is a (material specific) constant called the work function.
Example 2: If a metal with W = 3.3 × 10−19J is irradiated by light of frequency ν = 1015Hz, find the energy of the emitted photoelectrons.
Solution: From Einstein's equation we have:
Note that since one electron-volt of energy is 1eV = 1.6 × 10−19J, we could re-express this as E ≈ 2eV.
Quiz 3: If the energy of the photo-electrons emitted from a metal is twice the work function, by what factor must the frequency of the incident radiation be increased to double the energy of the photo-electrons?
Solution: We are told that E = 2W. Therefore 2W = hν - W, so:
If we want to double the energy of the photo-electrons, then the new energy must be E' = 4W. This implies:
So the necessary ratio of the frequencies is given by:
3. The de Broglie Wavelength
As well as light having a particle nature, quantum theory says that matter has a wave-like nature. This is expressed for a particle with momentum p by:
- where λ is the de Broglie wavelength and h is Planck's constant.
Quiz 4: To understand why we do not see the wave nature of normal matter around us, estimate the wavelength of a 100g pebble thrown through the air with speed, v = 2ms−1. (Recall that momentum and velocity are related by p = mv.)
Solution: First re-express the mass in S.I. units:
Therefore:
And so:
- which is much smaller than we could hope to measure.
At atomic and subatomic scales we can see wave-like properties of matter, e.g. electron diffraction.
Example 3: If in a demonstration of electron diffraction, the electrons' de Broglie wavelength was about 5 × 10−2Å, find their kinetic energy.
Solution:
| From: λ | = | hp | with λ = 5 × 10−10m |
|---|---|---|---|
| p | = | hλ = 6.6 × 10−346 × 10−12 ≈ 1.3 × 10−22kg m s−1 | |
| ≈ 1.3 × 10−22kg m s−1 | |||
The kinetic energy is E = ½ mv2 = p2 ⁄ (2me), and so, using the electron mass me = 9.1 × 10−31kg, we get:
| E | = | (1.3 × 10−22)22 × 9.1 ×10−31 |
|---|---|---|
| = | 1.718.2 × 10−13 ≈ 9 × 10−15 J |
Recalling that 1eV = 1.6 × 10−19J, then E ≈ 6 × 104eV.
Quiz 5: What is the wavelength of a 1 keV electron?
Solution: From E = p2 ⁄ (2me), we have: p = √2meE and so:
- where we used 1000eV = 1.6 × 10−16J. This corresponds to λ ≈ 0.4Å.
Quiz 6: The ratio of the proton and electron masses is given by:
If an electron and a proton are to have the same de Broglie wavelength, how must their energies be related?
Solution: If they have the same wavelengths, then they must have the same momentum, p. So their kinetic energies are given by:
Thus, their ratio is:
4 The Balmer Series
Quiz 7: Balmer's original formula for the visible lines of the Hydrogen spectrum expressed their wavelengths in terms of a constant K:
- for n = 3, 4 ... This can also be expressed in terms of the wave number ν̄ = 1⁄λ. Which of the formulae below is correct?
Quiz 8: What are the shortest and longest wavelengths for lines in the Balmer series?
Solution: We have the expression:
The shortest wavelength is found for the largest value of n, i.e., n → 1. For very large n the fraction tends to one and we get λmax = K.
To obtain the longest wavelength we insert the smallest possible value of n, which, in the Balmer series, is n = 3. The fraction becomes 9 ⁄ 5 and we have λmin = 9 K ⁄ 5.
Note that the longest wavelength corresponds to a photon with enough energy to excite an electron into the next state, while a photon with the shortest wavelength in the series can eject an electron from the atom, a process called photo-ionisation.
5. Rotational and Vibrational Spectra
One of the characteristic predictions of quantum mechanics is that many energies are only allowed to have specific discrete values. For example the rotational energy levels of linear molecules are roughly:
- where J is an (integer) quantum number and I is the moment of inertia of the molecule.
Quiz 9: What is the difference in the energy between two such adjacent energy levels, EJ+1 − EJ?
Solution: If: EJ = h28π2I J(J+1) , then: EJ+1 = h28π2I (J+1)(J+2) , so we have:
| EJ+1 − EJ | = | h28π2I (J+1)(J+2) − h28π2I J(J+1) |
| = | h28π2I (J+1)[(J+2) − J] | |
| = | h28π2I (J+1) 2 | |
| = | h24π2I (J+1) |
- where we have factored out the common term h24π2I (J+1)
Quiz 10: A more accurate model of rotating molecules builds in stretching effects via:
- where K is a (small) constant. What is the difference between two such energy levels now?
Solution: The first term in all the answers is just the answer of the previous quiz. What we need to calculate is the difference between the correction terms, −KJ2(J + 1)2. In the next level (J → J + 1) this term becomes: −K(J + 1)2(J + 2)2, so the difference is:
This can now be simplified using the techniques from the Factorisation module:
| = | −K(J + 1)2[(J + 2)2 − J2] |
| = | −K(J + 1)2[J2 + 4J + 4 − J2] |
| = | −K(J + 1)2[4J + 4] |
| = | −4K(J + 1)3 |
where, to expand the quadratic term, we used the FOIL technique (see the Brackets module).
Quiz 11: The energy levels of vibrational modes in the simple harmonic oscillator are given by En = (n+1⁄2)hν. What is the difference between two consecutive levels?
Solution: The consecutive energy levels are:
and:
so their difference is:
6. The Uncertainty Principle
Heisenberg's uncertainty principle tells us that the product of the uncertainty in the position, Δx, and the uncertainty in the momentum, Δp must satisfy:
- where ℏ (pronounced 'hbar') is used as shorthand for the combination h/2π as this appears so frequently in Quantum Theory.
Example 3: If an electron is bound in an atom of diameter roughly 1Å what is the minimum uncertainty in its velocity?
Solution: The electron's positional uncertainty Δx ≈ 10−10m. From the uncertainty principle we have:
| Δp | ≥ | ℏ2Δx |
| ≥ | 5.3 × 10−25kg m s−1 |
Momentum and velocity are related by p = mv, the electron mass is 9.1 × 10−31kg, and so the uncertainty in its velocity must be greater than Δv = Δp ⁄ m ≈ 6 × 105 m s−1.
Quiz 12: If the uncertainty in the position of an object is half a micron, which of the following is closest to the minimum uncertainty in its momentum?
Solution: From the uncertainty principle:
and since ℏ ≈ 10−34J s and Δx is given as 5 × 10−7m the answer follows directly.
This is incredibly tiny: for comparison, the diameter of an atomic nucleus is about 10−14m.
7. Wave Functions and Probabilities
In quantum mechanics the probability density is given by the square modulus of the wave function, ψ:
- where the * denotes complex conjugation (see the Complex Numbers module).
Quiz 13: In a tunnelling process the wave function may have the form:
What is the probability density in this region?
Solution: Here the wave function is real so we just need to square it:
| |ψ(x)|2 | = | A exp(−kx) × A exp(−kx) |
| = | A2 exp(−kx − kx) | |
| = | A2 exp(−2kx) |
- where we have used the rule aman = a(m + n) (see the module on Powers).
Quiz 14: If the wave function of a particle moving with a specific momentum in one dimension is ψ(x, t) = A exp(−i(kx − ωt)) what is its probability density?
Solution: Here the wave function is complex. We replace i → −i to form its conjugate:
so, multiplying ψ and ψ∗ gives:
| |ψ(x)|2 | = | A exp(i(kx − ωt)) × A exp(−i(kx − ωt)) |
| = | A2 exp(0) | |
| = | A2 |
- where we again used the rule aman = a(m+n).
Note that the probability density is independent of x and t, i.e., we have no information about where the particle is. This is a consequence of the uncertainty principle, since we know its momentum exactly and cannot know both quantities at once!
8. Quiz on Mathematics & Quantum Theory
Choose the solutions from the options given