Quantum Physics Answer Sheet 1

1. Since c = λν we can find the frequency ν by simply rearranging so that ν = c/λ.

So for λ = 1mm then ν = 3× 1011Hz the energy of a photon is E = hc/λ which is 1.24meV

So for λ = 500nm then ν = 6 × 1014Hz the energy of a photon is E = hc/λ which is2.49 eV

So for λ = 200nm then ν = 1.5 × 1015Hz the energy of a photon is E = hc/λ which is6.22 eV

So for λ = 10nm then ν = 3 × 1016Hz the energy of a photon is E = hc/λ which is124.3 eV

So for λ = 0.1nm then ν = 3 × 1018Hz the energy of a photon is E = hc/λ which is1.24 keV

[1eV = 1.60× 10−19J]

2. In lectures we took the definition of action as being the integral over a path ofmomentum×position, for the three cases we can take the velocity and therefore the momentum as be-ing constant over the path so it is simple to evaluate the action in each case (i.e. no trickyintegration involved) as we only need an order of magnitude estimate.

(a) (I don’t know much about cricket - but "hit for six" means the ball will travel about100 m (or more) before hitting the ground) and we can guess the mass of the ball is about1kg and that it travels at a velocity of about 20ms−1. So p = 1 × 20 = 20kgms−1 overa distance 100m: S = p × x = 20 × 100 = 2 × 103kgm2s−1, so very much in the"‘classical"’ regime.

(b) We can compute the electron’s momentum as

p =√

2meE =√

2× 9.11× 10−31 × 1.6× 10−19 × 5× 103 ≈ 3.8× 10−23kgms−1 andso the action S = 3.8 × 10−23 × 0.3 = 1.1 × 10−23kgm2s−1 which is small, but stillclassical.

(c) Again we can compute the electron’s momentum as

p =√

2meE =√

2× 9.11× 10−31 × 0.4× 1.6× 10−19 = 3.58 × 10−25kgms−1 and sothe action will be S = 3.58 × 10−25 × 10−9 = 3.58 × 10−34kgm2s−1 this is now smallenough for us to expect significant quantum effects.

3. Number of atoms per unit volume is:

n =Mass per unit volume

Mass per atom=

2700

27× 1.66× 10−27= 6.02× 1028 m−3 .

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Volume per atom is 1/n = 1.66 × 10−29 m3. If we assume that each atom is a sphere ofradius r, we have

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3πr3 = 1.66× 10−29

and hence r = 1.58× 10−10 m = 1.58 Å. This is an overestimate because it is impossibleto fill space with spheres (there are always gaps between them). A different estimate maybe obtained by assuming that the spheres are stacked in a simple cubic lattice. Each cubeof side 2r then contains one atom of radius r and the volume per atom is 8r3. This gives

8r3 = 1.66× 10−29

and hence r = 1.28 × 10−10 m = 1.28 Å. It is possible to pack spheres much moreefficiently than in a simple cubic lattice (the packing in Al is actually face-centred cubic —the most efficient regular packing of spheres), and so the true answer must lie somewherebetween these two estimates.

In any case X-ray wavelengths of less than ≈ 10−10m will be required for accurate struc-ture determination using diffraction.

4. For a very rough estimate, equate kBT (or 3kBT/2 if you insist) to 13.6 eV. This gives:

T =13.6× 1.6× 10−19

1.38× 10−23≈ 160, 000 K .

5. (a) The phase velocity of a wave is given by vphase = ω/k, reading from the expressionk = 2m−1 and ω = 3rads−1, so v = 3/2 = 1.5ms−1.

(b) The intensity of a wave is just αα∗ where α = 0.5, so intensity is 0.25.

6. To work out the wavelength, note that

ψ(x+

2π

k, t)

= a cos(−k

[x+

2π

k

]− ωt+ φ

)= a cos(−kx− ωt+ φ− 2π) = ψ(x, t) .

Hence, ψ(x, t) changes by one full period as x increases by 2π/k at constant t. In otherwords, the wavelength λ = 2π/k.

Similarly, to work out the time period, note that

ψ(x, t+

2π

ω

)= a cos

(−kx− ω

[t+

2π

ω

]+ φ

)= a cos(−kx− ωt+ φ− 2π) = ψ(x, t) .

Hence, ψ(x, t) changes by one full period as t increases by 2π/ω at constant x. In otherwords, the time period T = 2π/ω. The frequency ν = 1/T is therefore given by ν =ω/2π.

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To work out the velocity of the crests, consider the crest at the point xcrest where theargument of the cosine function is zero:

−kxcrest − ωt+ φ = 0 .

This equation rearranges to

xcrest =φ

k− ω

kt .

Comparing with the equation xcrest = x0 + vt that describes uniform motion at velocity v,we see that the wave crest is at position x0 = φ/k at time t = 0 and is moving at velocity−ω/k.

7. The phase velocity is

v =ω

k=

√g

k=

√gλ

2π≈√

9.8× 10

2π= 3.96 ms−1 .

8. Assuming the spectrum and temporal profile to both be Gaussian we can use the form ofthe bandwidth theorem for the full-width at half maximum (FWHM) spread from Lecturesthat says ∆t∆ν ≥ 0.44 to compute that for a 3.5 fs duration pulse we will need a frequencybandwidth ∆ν ≥ 0.44/3.5× 10−15 = 1.26× 1014Hz. [Note: In the uncertainty relationsderived later in the course it is the standard deviation of the distributions not the FWHMthat are involved, so it is not trivial to go directly from this result to the energy-timeuncertainty relation - but they are closely related]

The cycle time for an 800 nm laser pulse is only 2.7 fs so the pulse is only 1.3 opticalcycles. This laser is used in our experiments to generate even shorter soft X-ray pulses (ofduration ≈ 250 attosecond).

9. We find λ = c/ν = 3× 108/2.5× 1014 = 1.2× 10−6m, so k = 2π/λ = 5.24× 106m−1.

(a) No diffraction (but the width of the focus of the lens will have a limit set by λ/d), (b)No diffraction detectable, (c) A 10µm slit will lead to significant diffraction, (d) A humanhair (see 1st year lab) is typically 50 − 100µm in diameter so there will be detectablediffraction.

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10. (a) The diffraction angle where the first minimum is located is (from Lectures) sin(θmin) =λ/w where w is the slit width. In fact the intensity is zero at the minimum and for smallangles we have sin θ ≈ θ so θ = 6× 10−7/10−5 = 0.06 radians or 60 mrad.

(b) For the two slits we will obtain Young’s or "cos2" fringes, with a spacing inverselyrelated to the slit separation d. The central maximum will be located at θ = 0 and thensuccessive orders of maxima at θ = ±nλ/d. We find that the angular spacing betweenfringes is therefore θ = 6× 10−7/4× 10−5 = 1.5× 10−2 radians or 15 mrad. That means4 orders of fringes fit between the central diffraction maximum and the first zero (at 60mrad), but the 4th order at 60 mrad won’t be seen as the envelope function is zero here, soinfact there will only be 3 fringes either side of the central ringe i.e. 7 fringes (see figurebelow).

(c) The shifting of phase at one slit by π is the same as shifting the path diffrence by anextra ±λ/2. This means the cos2 fringe pattern is shifted (right or left) by half the normalspacing. Although there is now no central fringe there are now infact 4 fringes either sideof the pattern between the centre and the zeros in the intensity envelope due to the singleslit diffraction spread. The pattern is shifted, and there are 8 fringes between diffractionminima.

This is illustrated in the sketch below:

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11. (a) (i) False, X-ray wavelengths are < 1nm and so are shorter than that of sunlight 500nm. [1 mark]

(ii) True, a synchrotron is used to generate X-rays from electrons accelerated in a circularorbit. [1 mark]

(iii) False, in practice even the shortest electron pulse in an electron tube are a poor choiceto generate < 10fs X-ray pulses. We would need an XFEL! [1 mark]

(b) The Bragg condition for the reflected X-rays to give constructive interference betweenplanes separated by a distance d is 2d sin θ = nλ where n is an integer (here we take n =1). So plugging in the numbers we get d = 1×0.07×10−9/2× sin 10o = 2.02×10−10m.[5 marks]

(c) For the second order n = 2 so that

sin θ2 = 2λ/2d = 2× 0.07× 10−9/2× 2.02× 10−10 = 0.3465

, which corresponds to θ = 20.3o. [2 marks]

12. The wave emerging from the segment ∆y at height y is

Aei(k[ζ−(−y sin θ)]−ωt)∆y = Aei(kζ−ωt+ky sin θ)∆y .

The total wave emerging in the ζ direction is the sum of the waves emerging from all thelittle segments:

ψ(ζ, t) =∑

segments

Aei(kζ−ωt+ky sin θ)∆y ,

where, for each segment (or, equivalently, for each term in the sum), y is the position ofthe centre of that segment. In the limit as ∆y → 0, the sum turns into the integral:

ψ(ζ, t) =∫ d/2

−d/2Aei(kζ−ωt+ky sin θ)dy .

Since ei(kζ−ωt+ky sin θ) = ei(kζ−ωt)eiky sin θ, this is equivalent to

ψ(ζ, t) = Aei(kζ−ωt)∫ d/2

−d/2eiky sin θdy ,

as required.

If you are confused about the relationship between the sum and the integral, considerthe diagram below. The integral of f(y) from a to b is the sum of the areas of thesegments of width ∆y.

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∆ y

f(y)

ya b

Since the area of each segment is approximately f(y)∆y, it follows that∫ b

af(y)dy ≈

∑segments

f(y)∆y .

In fact, the integral is normally defined as the limit of this sum as ∆y tends to zero.

The integral ∫ d/2

−d/2eiky sin θdy

may be written as ∫ d/2

−d/2eαydy ,

where α = ik sin θ. This is easy to integrate:∫ d/2

−d/2eαydy =

[eαy

α

]d/2−d/2

=1

α

(eαd/2 − e−αd/2

)

=e

12ikd sin θ − e− 1

2ikd sin θ

ik sin θ=

d sin(kd sin θ

2

)kd sin θ

2

,

where the last step used the identity eiφ − e−iφ = 2i sinφ discussed in the answer toquestion 7.

The wavefunction ψ(ζ, t) is therefore given by:

ψ(ζ, t) = Adsin

(kd sin θ

2

)kd sin θ

2

ei(kζ−ωt) .

The intensity I is the square modulus of the complex amplitude (which is everything infront of the ei(kζ−ωt) factor). Hence

I =|A|2d2 sin2

(kd sin θ

2

)(kd sin θ

2

)2 ,

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as required.

The diffraction pattern looks like this (note that the horizontal axis shows values of kd sin θ2π

instead of values of kd sin θ2

):

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-4 -3 -2 -1 0 1 2 3 4

Inte

nsity / (

|A

|^2 d

^2 )

k d sin(theta) / ( 2 pi )

Single Slit Diffraction Pattern

The first zero occurs where kd sin θ2π

= 1 and hence where

k =2π

λ=

2π

d sin θ.

This implies thatλ = d sin θ .

The difference in the lengths of the paths emerging from the top and bottom of the slit isone wavelength.

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