# 1. Electromagnetic Waves

Light has long been known to have a wave-like character. Frequency, ν, and wavelength, λ, are related by ν = cλ, 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:

λ = c/ν = 3 × 108/9.3 × 107 = 3 × 10/9.3 ≈ 3.2m

Exercise 1: Various types of electromagnetic radiation are described below. If the frequency is given, calculate the wavelength and vice versa.

(a)Visible light with λ = 600nm

Solution: First, convert the wavelength into S.I. units:

600nm = 600 × 10−9 = 6 × 10−7m

Then:

ν = c/λ = 3 × 108/6 × 10−7 = 3 × 1015/6 = 3 × 1014Hz
(b)X-rays with ν = 3 × 1018Hz

Solution:

λ = c/ν = 3 × 108/3 × 1018 = 10−18m

This very small scale, 1Å, is comparable to atomic spacing in crystals, and this fact is the basis of X-ray crystallography.

(c)Infra-red radiation with λ = 1.5μm

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.

μ = c/λ = 3 × 108/1.5 × 10−6 = 2 × 1014Hz
(d)Gamma rays with ν = 1021Hz

Solution:

λ = c/ν = 3 × 108/1021 = 3 × 10−13m

Click on questions to reveal solutions

Quiz 1: Which of the following might emit electromagnetic radiation with frequency ν = 1016Hz?

(a)The SunCorrect - well done!
(b)An X-ray laserIncorrect - please try again!
(c)A gamma-ray sourceIncorrect - please try again!

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?

(b)3 × 1015HzIncorrect - please try again!
(c)1.5 × 1016HzIncorrect - please try again!
(d)8 × 1017HzCorrect - well done!

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:

ν = c/λ = 3 × 108/½ × 7.5−10 = 8 × 1017 Hz

# 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:

E = W

- 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:

E = 6.6 × 10−34 × 1015 − 3.3 × 10−19 = (6.6 − 3.3) × 10−19 = 3.3 × 10−19J

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?

(a)2 ⁄ 3Incorrect - please try again!
(b)3 ⁄ 2Incorrect - please try again!
(c)5 ⁄ 3Correct - well done!
(d)3 ⁄ 5Incorrect - please try again!

Solution: We are told that E = 2W. Therefore 2W = - W, so:

= 3W

If we want to double the energy of the photo-electrons, then the new energy must be E' = 4W. This implies:

hν' = 4W + W = 5W

So the necessary ratio of the frequencies is given by:

hν'/ = ν'/ν = 5W/3W = 5 ⁄ 3

# 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:

λ = h/p

- 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.)

(a)1 ⁄ 100mIncorrect - please try again!
(b)3 × 10−33mCorrect - well done!
(c)5 × 10−55μmIncorrect - please try again!
(d)4 × 10−44mIncorrect - please try again!

Solution: First re-express the mass in S.I. units:

m = 100 ⁄ 1000 = 0.1kg

Therefore:

p = mv = 0.1 × 2 = 0.2kg m s−1

And so:

λ = h/p = 6.6 × 10−34/0.2 ≈ 3 × 10−33m

- 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 × 10−2, find their kinetic energy.

Solution:

From: λ p = h/p with λ = 5 × 10−10m = h/λ = 6.6 × 10−34/6 × 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)2/2 × 9.1 ×10−31 = 1.7/18.2 × 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?

(a)0.4Correct - well done!

Solution: From E = p2 ⁄ (2me), we have: p = 2meE and so:

λ = h/p = 6.6 × 10−34/2 × 9.1 × 10−31 × 1.6 × 10−16

- where we used 1000eV = 1.6 × 10−16J. This corresponds to λ ≈ 0.4.

Quiz 6: The ratio of the proton and electron masses is given by:

mp/me = 1836.15

If an electron and a proton are to have the same de Broglie wavelength, how must their energies be related?

(a) Ep/Ee = 1 Incorrect - please try again!
(b) Ep/Ee = 1836.15 Incorrect - please try again!
(c) Ep/Ee = 1836.15 Incorrect - please try again!
(d) Ep/Ee = 1/1836.15 Correct - well done!

Solution: If they have the same wavelengths, then they must have the same momentum, p. So their kinetic energies are given by:

Ep = p2/2mp and Ee = p2/2me

Thus, their ratio is:

Ep/Ee = me/mp = 1/1836.15

# 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:

λ = K n2/n2 - 4

- for n = 3, 4 ... This can also be expressed in terms of the wave number ν̄ = 1λ. Which of the formulae below is correct?

(a)ν̄ = − 1/4K Incorrect - please try again!
(b)ν̄ = K (1 + 4/n2) Incorrect - please try again!
(c)ν̄ = 1/K ( 1 − 4/n2 ) Correct - well done!
(d)ν̄ = 1/K 4n2/K Incorrect - please try again!

Quiz 8: What are the shortest and longest wavelengths for lines in the Balmer series?

(a)K and 9/5 KCorrect - well done!
(b)0 and 9/5 KIncorrect - please try again!
(c)4/5 K and 4/5 KIncorrect - please try again!
(d)9/13 K and KIncorrect - please try again!

Solution: We have the expression:

λ = K n2/n2 - 4

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:

EJ = h2/8π2I J(J + 1)

- 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+1EJ?

(a) h2/4π2I (J + 1)(J + 2) Incorrect - please try again!
(b) h2/8π2I Incorrect - please try again!
(c) h2/4π2I (J + 1) Correct - well done!
(d) h2/2I(J + 2)/J Incorrect - please try again!

Solution: If: EJ = h2/8π2I J(J+1) , then: EJ+1 = h2/8π2I (J+1)(J+2) , so we have:

 EJ+1 − EJ = h2/8π2I (J+1)(J+2) − h2/8π2I J(J+1) = h2/8π2I (J+1)[(J+2) − J] = h2/8π2I (J+1) 2 = h2/4π2I (J+1)

- where we have factored out the common term h2/4π2I (J+1)

Quiz 10: A more accurate model of rotating molecules builds in stretching effects via:

EJ = h2/8π2I J(J + 1) − KJ2(J + 1)2

- where K is a (small) constant. What is the difference between two such energy levels now?

(a) h2/4π2I (J + 1) − K(J + 1)2 Incorrect - please try again!
(b) h2/4π2I (J + 1) − 4K(J + 1)3 Correct - well done!
(c) h2/4π2I (J + 1) − 4K(J + 2)3 Incorrect - please try again!
(d) h2/4π2I (J + 1) − 2K Incorrect - please try again!

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 (JJ + 1) this term becomes: K(J + 1)2(J + 2)2, so the difference is:

K(J + 1)2(J + 2)2 − (−KJ2(J + 1)2) = −K(J + 1)2(J + 2)2 + KJ2(J + 1)2

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+12). What is the difference between two consecutive levels?

(a) 0 Incorrect - please try again!
(b) (n-12) Incorrect - please try again!
(c) Correct - well done!
(d) (2n+1) Incorrect - please try again!

Solution: The consecutive energy levels are:

En+1 = ({n+1} - 12) = (n + 12)

and:

En = (n - 12)

so their difference is:

En+1 - En = (n + 12) - (n - 12) =

# 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:

ΔxΔp/2

- where (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 1 what is the minimum uncertainty in its velocity?

Solution: The electron's positional uncertainty Δx ≈ 10−10m. From the uncertainty principle we have:

 Δp ≥ ℏ/2Δx ≥ 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 = Δpm ≈ 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?

(a)3.314 × 10−39 kg m s−1Correct - well done!
(b)10−28 kg m s−1Incorrect - please try again!
(c)10−33 kg m s−1Incorrect - please try again!
(d)10−40 kg m s−1Incorrect - please try again!

Solution: From the uncertainty principle:

Δp/x

and since ≈ 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:

|ψ(x)|2  = ψ(x)ψ(x)

- where the * denotes complex conjugation (see the Complex Numbers module).

Quiz 13: In a tunnelling process the wave function may have the form:

ψ(x) = A exp(-kx)

What is the probability density in this region?

(a)A2exp(−2kx)Correct - well done!

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(kxωt)) what is its probability density?

(a)A2 exp(−(ωt-kx)2)Incorrect - please try again!
(b)A2 exp(2(ωt-kx))Incorrect - please try again!
(c)A2Correct - well done!
(d)A2 exp(ωt+kx)Incorrect - please try again!

Solution: Here the wave function is complex. We replace i → −i to form its conjugate:

ψ(x, t)  = A exp(+i(kxωt))

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

1. Estimate the wavelength of a (visible) photon with ν = 6 × 1010Hz.
(a)5
(b)2 × 102
(c)5 × 10−3
(d)6nm
2. Estimate the de Broglie wavelength of 100eV electrons.
(a)4km
(b)0.25 × 10−6m
(c)10−10m
(d)1.6 × 10−3
3. What is the minimum uncertainty in the momentum of a proton inside a nucleus of radius 1 × 10−15m?
(a)5 × 10−20kg m s−1
(b)3.3 × 10−16kg m s−1
(c)J s
(d)0
4. If: El = 2/2I l(l+1) what is E3E2?
(a)18ℏ4I2
(b)3 ⁄ 2
(c)2
(d)6I