PHYS2265 Modern Physics/PHYS2627 Introductory Quantum Physics

2013/14 Conceptual Questions 3

1. Electron 1 is accelerated from rest through a potential difference of 100 V. Electron 2 isaccelerated from rest through a potential difference of 200 V. Afterward, which electronhas a larger de Broglie wavelength? Explain.

Answer:

Electron 1 has a larger de Broglie wavelength.Note that a particle has speed v =

√2K/m and de Broglie wavelength λ = h/(mv) if

it’s kinetic energy is K.Given: K1 = 100 eV and K2 = 200 eV.∴ v1 < v2 ⇒ λ1 > λ2

2. All other things being equal, which would be more likely to exhibit its wave nature —a proton or an electron — and why? By making something unequal, how could you“compensate” so as to make one as wavelike as the other?

Answer:

An electron is more likely to be wavelike. For equal speeds, its momentum p is smaller bya factor of about 2000 (the ratio of proton to electron mass), so its de Broglie wavelengthλ = h/p is longer. To give the proton a wavelength as long as the electron’s, we wouldhave to slow the proton to a speed about 1/2000 that of the electron.

3. Double-slit interference of neutrons occurs because:

(i) The neutrons passing through the two slits repel each other.

(ii) Neutrons collide with each other behind the slits.

(iii) Neutrons collide with the edges of the slits.

(iv) Each neutron goes through both slits.

(v) The energy of the neutrons is quantized.

(vi) Only certain wavelengths of the neutrons fit through the slits.

Which of these (perhaps more than one) statements are correct? Explain your answer.

Answer:

Only (iv) is correct. With a weak neutron beam so that each time only one neutron issent through the apparatus, we would still observe the interference pattern.

4. Why it is not reasonable to do crystallographic studies with protons?

Answer:

At typical laboratory energies, a proton’s de Broglie wavelength is of order 10−12 m. Itis much smaller than the spacings of the planes of atoms in common crystals (like theslits in a grating) which is in order of 10−10 m. So the wave nature of protons would notbe evident in the diffraction by a crystal. We cannot observe an interference pattern toget any information about the structure of the crystal.

5. Below figure is the plot of angular frequency ω(k) versus wave number k for a particularwave phenomenon. How do the phase and group velocities compare, and does the answerdepend on the value of k under consideration? Explain your answer.

Answer:

For small (average) values of k, the slope of the curve, or tangent line, dω/dk is smalland the slope of a line from the origin, ω/k is comparatively large. That is, the groupvelocity is smaller than the phase velocity. When the (average) value of k is large, thesetwo quantities are equal, and so are the phase and group velocities.

6. The location of a particle is measured and specified as being exactly at x = 0, with zerouncertainty in the x direction. How does this affect the uncertainty ∆vy of its velocitycomponent in the y direction?

(i) It does not affect ∆vy.

(ii) It makes ∆vy infinite.

(iii) It makes ∆vy zero.

Answer:

(i) is the correct answer. Heisenberg’s uncertainty principle relates uncertainty in po-sition and velocity along the same axis. The zero uncertainty in position along the xaxis results in infinite uncertainty in its velocity component in the x direction, but it isunrelated to that in the y direction.

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7. An electron and a proton are both confined in boxes of length a. Measurements aremade on the momentum of both particles. Which statement is true about the precisionof these measurements?

(i) The electron’s momentum can be measured more precisely than the proton’s mo-mentum.

(ii) The proton’s momentum can be measured more precisely than the electron’s mo-mentum.

(iii) The electron’s momentum can be measured only as precisely as the proton’s mo-mentum.

Answer:

(iii) is the correct answer. The electron and proton have the same uncertainty in position.According to Heisenberg’s uncertainty principle, they must have the same uncertaintyin momentum.

8. What difficulties does Heisenberg’s uncertainty principle cause in trying to pick up anelectron with a pair of forceps?

Answer:

When the electron is picked up by the forceps, the position of the electron is “localized”(or fixed), i. e. ∆x = 0. Accroding to Heisenberg’s uncertainty principle, the momentumof the electron must be highly uncertain in such case. In effect, a large ∆p means theelectron is “shaking” furiously against the forceps’ tips that tries to hold the electron“tightly”.

9. A low-intensity beam of light is sent toward a narrow single slit. On the far side,individual flashes are seen sporadically at detectors over a broad area that is orders ofmagnitude wider than the slit width. What aspects of this experiment suggest a wavenature for light, and what aspect suggest a particle nature?

Answer:

The flashes are “experiments” in which the light is behaving as particles. Besides, thefact that they are detected at a very wide angle after passing through the slit is amanifestation of a wave behavior: diffraction.

10. A single electron with de Broglie wavelength λ passes through a slit of width d = 2λ.The electron can then strike a fluorescent screen. What will be observed on the screen?

(i) A diffraction pattern.

(ii) A single flash as if the electron has moved in a straight line through the slit.

(iii) A single flash that occurs randomly at anywhere on the screen.

(iv) A single flash that would most likely occur where a corresponding diffraction patternwould have the highest intensity.

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Answer:

(iv) is the correct answer. Since the electron has de Broglie wavelength comparable withthe width of the slit, we cannot neglect the quantum behavior of the electron. Althoughthe electron is detected as particle at a localized spot at some instant of time, theprobability of arrival at that spot is determined by finding the intensity of the diffractedmatter wave.

11. A monenergetic beam of electrons is incident on two slits whose dimensions and spacingare of the same order as of magnitude as the de Broglie wavelength of the electrons asshown in the following figure. Draw the pattern of electrons recorded on the screen forthe following cases:

(a) The classical prediction for both slits open.

(b) The observed pattern for both slits open.

(c) The observed pattern for only slit A or B open.

Answer:

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PHYS2265 Modern Physics/PHYS2627 Introductory Quantum Physics

2013/14 Worked Example 3

1. Show that the de Broglie wavelength of a particle is approximately the same as that ofa photon with the same energy when the energy of the particle is much greater than itsrest energy.

Solution:

If the particle has momentum p and energy E much much greater than its rest energyE0, i. e. E � E0, then we have

E2 = p2c2 + E20

⇒ p =E

c

√1− E2

0

E2≈ E

c

So the de Broglie wavelength of the particle is given by:

λ =h

p≈ hc

E

A photon of wavelength λγ has energy

Eγ = hf =hc

λγ

Therefore, if E = Eγ,

λγ =hc

E≈ λ

2. (a) A particle is “thermal” if it is in equilibrium with its surroundings — its averagekinetic energy would be K = 3kBT/2 where T is its temperature. Show that thewavelength of a thermal particle is given by

λ =h√

3mkBT

(b) Use the result of part (a), find the wavelength of a room-temperature (at tempera-ture T = 22◦C) electron.

Solution:

(a) The average kinetic energy of a thermal particle of momentum p is

K =3

2kBT =

p2

2m

Putting the de Broglie wavelength λ = h/p into the above equation, we have:

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2kBT =

(h/λ2)

2m

⇒ λ =h√

3mkBT

(b) Using the result of part (a), we can find the wavelength of the electron

λ =6.626×10−34 J · s√

3(9.109×10−31 kg)(1.381×10−23 J/K)(295 K)= 6.28×10−9 m

3. The neutron interference pattern in the below figure was made by shooting neutronswith a speed of v = 200 m/s through two slits spaced d = 0.10 mm apart.

(a) What was the kinetic energy of the neutrons in eV?

(b) What was the de Broglie wavelength of the neutrons?

(c) The pattern was recorded by using a neutron detector to measure the neutrondensity at different positions. By making appropriate measurements directly on thefigure, determine how far the detector was behind the slits.

Solution:

(a) A neutron has mass mN = 1.675×10−27 kg. The kinetic energy of each neutronpassing through the slits is equal to

K =1

2mNv

2 =1

2(1.675×10−27 kg)(200 m/s)2 = 3.35×10−23J

(b) The de Broglie wavelength at this speed is

λ =h

mNv=

(6.626×10−34 J · s)(1.675×10−27 kg)(200 m/s)

= 1.98×10−9m

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(c) As shown in the below figure, the fringe spacing in a double-slit interference ex-periment is ∆y = λL/d where L is the distance from the slits to the detector.

From the given figure, we can see that the spacing between the two peaks for ordersm = ±1 (on either side of the central maximum) is 1.4 times of the length of thereference bar, which gives ∆y = 7×10−5 m. Thus the distance from the slits to thedetector is given by

L =d∆y

λ=

(1.0×10−4 m)(7.0×10−5 m)

(1.98×10−9 m)= 3.54 m

4. A narrow beam of electrons of kinetic energy K = 60 keV passes through a thin silverpolycrystalline foil. The inter-atomic spacing of silver crystals is d = 4.08 A (1 A = 1×10−10 mm). Calculate the radius of the first-order diffraction pattern from the principalBragg planes on a screen placed L = 40 cm behind the foil.

Solution:

Below figure shows how the diffraction occurs in the silver polycrystalline foil.

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Each of the electrons has rest energy E0 = 0.511 MeV and momentum

p =1

c

√E2 − E2

0 =1

c

√(K + E0)2 − E2

0

So its de Broglie wavelength is equal to

λ =h

p=

hc√(K + E0)2 − E2

0

=(1240 eV · nm)√

[(60 + 511)2 − 5112]×103 eV= 0.00487 nm

The angle corresponding to the first-order Bragg reflection is

θ = sin−1

(λ

2d

)= sin−1

[(0.00487 nm)

2(0.408 nm)

]= 0.342◦

Thus the radius of the first-order diffraction pattern (see the figure) is given by

R = L tan 2θ = (40×10−2 m) tan(2× 0.342◦) = 4.77×10−4 m

5. For wavelengths less than about 1 cm, the dispersion relation — the relationship betweenangular frequency ω and wave number k — for waves moving on a water surface is

ω =

√γ

ρk3,

where γ and ρ are the surface tension and density of water. Given that γ = 0.072 N/mand ρ = 103 kg/m3, calculate the phase and group velocities for a wave of wavelengthλ = 5 mm.

Solution:

The water wave has wave number

k =2π

λ=

2π

5×10−3 m= 1.257×103 m−1.

From the given relation, we can find the phase velocity vp and the group velocity vg:

vp =ω

k=

√(γ/ρ)k3

k=

√γ

ρk =

√(0.072 N/m)

(103 kg/m3)(1.257×103 m−1) = 0.301 m/s

vg =dω

dk=

d

dk

(√γ

ρk3)

=3

2

√γ

ρk =

3

2

√(0.072 N/m)

(103 kg/m3)(1.257×103 m−1) = 0.451 m/s

6. A stone tossed into a body of water creates a disturbance at the point of impact thatlasts for ∆t = 4.0 s. Measurements indicates that the wave speed is v = 25 cm/s.

(a) Over what distance on the surface of water does the wave group extend?

(b) An observer counts 12 wave crests in the group. Estimate the precision with whichthe wavelength can be determined.

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Solution:

(a) The disturbance lasts for ∆t = 4.0 s at the point of impact. So the wave groupextends on the water surface for a distance of

∆x = v∆t = (25 cm/s)(4.0 s) = 100 cm.

(b) Since the wave group consists of 12 wave crests, the period of the wave T = ∆t/12.So the wavelength is

λ = vT = v

(∆t

12

)= (25 cm/s)

(4.0 s

12

)=

25

3cm.

Using the classical uncertainty relationship ∆x∆k ≥ 1/2, we can estimate the valueof ∆k:

∆k ≈ 1

2∆x=

1

2(100 cm)= 0.005 cm−1.

Since k = 2π/λ, the uncertainty in λ is given by:

∆k =2π

λ2∆λ

⇒ ∆λ =λ2

2π∆k =

[(25/3) cm]2

2π(0.005 cm−1) = 0.0553 cm

It is the precision with which the wavelength can be determined.

7. Heisenberg’s uncertainty relationship for the position and momentum is

∆x∆px ≥~2.

Use it to show that for a particle moving in a circle,

∆L∆θ ≥ ~2

where ∆L and ∆θ denote the uncertainty in the angular momentum and angular positionof the particle, respectively.

Solution:

Since the particle moves in a circle, let’s apply the uncertainty principle to the directiontangential to the circle. Then we obtain

∆s∆ps ≥~2.

where s is measured along the circumference of the circle and ps is the tangential com-ponent of its momentum.

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If the particle has mass m, its angular momentum is related to its linear momentum by

L = mvR = psR

⇒ ∆ps =∆L

R

Besides, the angular displacement is related to the arc length by θ = s/R which implies

∆s = R∆θ.

Hence,

∆s∆ps =

(∆L

R

)(R∆θ) = ∆L∆θ ≥ ~

2

For a state of fixed angular momentum (e. g. an electron in a Bohr orbit, which will bediscussed in Chapter 7), the uncertainty in the angular momentum is zero (i. e. ∆L = 0).Therefore, the uncertainty in the angular position, ∆θ, is infinite, so that the positionof the particle in the orbit cannot be determined.

8. A π meson and a proton can briefly join together to form a ∆ particle. A measurementof the energy of the π-p system shows a peak of the reaction probability at 1236 MeV,corresponding to the rest energy of the ∆ particle, with an experimental spread of120 MeV. What is the lifetime of the ∆ particle?

Solution:

The given information tells us that the uncertainty in the measured energy is ∆E =120 MeV. Using the Heisenberg uncertainty relationship for the energy and time, wecan find the uncertainty in time ∆t:

∆E∆t ≥ ~2

⇒ ∆t ≈ ~2∆E

=(1.055 × 10−34 J · s)

2(120 MeV)(1.602 × 10−13 J/MeV= 2.74×10−24 s

The lifetime of the ∆ particle is taken as the uncertainty in time ∆t = 2.74×10−24 s.

9. The position of a particle of momentum p is measured by passing it through a slit ofwidth d. Estimate the corresponding uncertainty induced in the particle’s momentumbased on the given information. Compare the result with Heisenberg uncertainty’sprinciple.

Solution:

When monochromatic waves of wavelength λ pass through a slit of width d, a diffractionpattern will be produced on a screen as shown in the following figure. According to thediffraction theory, the angle corresponding to the first point of zero intensity equals

sinα =λ

d.

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Due to its associated de Broglie wave, whose wavelength is λ = h/p, the particle will bediffracted as it passes through the slit and hence will acquire some unknown momentumin the x-direction. Although we don’t know exactly where the particle will strike thescreen, the most probable place for it to hit will be somewhere within the central regionof the diffraction pattern. Therefore we can be reasonably certain that the x-componentof the particle’s momentum has a magnitude between 0 and p sinα, i. e.

∆px ≈ p sinα = p

(h/λ

d

)=h

d

This uncertainty can be made as small as desired by increasing d. However, sinced = ∆x, the uncertainty in the particle’s x-position, we find that

∆x∆px ≈ h ∼ ~2

which is consistent with Heisenberg’s uncertainty principle.

10. A woman on a ladder drops small pellets toward a spot on the floor.

(a) Show that, according to Heisenberg’s uncertainty principle, the miss distance mustbe at least

∆xmin =

(2~m

)1/2(2H

g

)1/4

where H is the initial height of each pellet above the floor and m is the mass ofeach pellet.

(b) If H = 2.0 m and m = 0.50 kg, what will be the uncertainty ∆x?

Solution:

(a) The woman tries to hold a pellet directly above the spot on the floor within somehorizontal region ∆xi. Heisenberg’s uncertainty principle requires her to give a

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pellet some x velocity at least as large as

vx =~

2m∆xi

As shown in the above figure, the pellet would hit the floor at time t given by:

H =1

2gt2 ⇒ t =

√2H

g

So the miss distance is given by

∆x = ∆xi + vxt = ∆xi +~

2m∆xi

√2H

g.

The minimum value of the function ∆x occurs for

d(∆x)

d(∆xi)= 0

⇒ 1− ~2m(∆xi)2

√2H

g= 0

⇒ ∆xi =

(~

2m

)1/2(2H

g

)1/4

Plugging it back into the equation for ∆x, we obtain the minimum miss distance

∆xmin =

(2~m

)1/2(2H

g

)1/4

.

(b) Putting H = 2.0 m and m = 0.50 kg into the equation found in part (a) yields

∆xmin =

[2(1.055 × 10−34 J · s)

(0.50 kg)

]1/2 [2(2.0 m)

(9.807 m/s2)

]1/4= 1.64×10−17 m.

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11. Electromagnetic waves strike a single slit of width a = 1µm (1µm = 1× 10−6 m).

(a) Determine the angular full width (angle from the first minimum on one side ofthe center to the first minimum on the other) in degrees of the central diffractionmaximum if the waves are (i) visible light of wavelength 500 nm and (ii) X-rays ofwavelength 0.05 nm.

(b) Which case in part (a) more clearly demonstrates a wave nature?

Solution:

(a) The first diffraction minimum occurs when

a sin θ = λ

⇒ θ = sin−1

(λ

a

)The angle from the first minimum on one side to the first minimum on the otherside is thus:

∆θ = 2 sin−1

(λ

a

)(i) For visible light of wavelength 500 nm, the angular full width is:

∆θ = 2 sin−1

(λ

a

)= 2 sin−1

(500×10−9 m

1×10−6 m

)= 60◦

(ii) For X-rays of wavelength 0.05 nm, the angular full width is:

∆θ = 2 sin−1

(λ

a

)= 2 sin−1

(0.05×10−9 m

1×10−6 m

)= 0.00573◦

(b) Diffraction, which is a wave phenomenon, is more pronounced for the long wave-length: the visible light. A particle nature (moving in a straight line, not diffracting)is more evident when the wavelength is very small compared to dimensions of theapparatus, i. e. λX-ray � a.

12. In a double slit interference experiment with electrons, you find the most intense fringeto be at x = 7.0 cm. There are slightly weaker fringes at x = 6.0 and 8.0 cm, still weakerfringes at x = 4.0 and 10.0 cm, and two very weak fringes at x = 1.0 and 13.0 cm. Noelectrons are detected at x < 0 cm or x > 14 cm.

(a) Sketch a graph of the probability density |ψ(x)|2 for these electrons.

(b) Sketch a possible graph of the wave function ψ(x).

(c) Are there other possible graphs for ψ(x)? If so, draw one.

Solution:

(a) We assume that |ψ(x)|2 = 0 at some point in between each of the peaks. Thefollowing is the graph of |ψ(x)|2.

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(b) Two factors are important for drawing ψ(x). First, the value at each point is thesquare root of the value on the |ψ(x)|2 graph. Second, and especially important,is that the wave function ψ(x) oscillates between positive and negative values eachtime |ψ(x)|2 reaches zero. Below is a possible graph of ψ(x).

(c) Multiplying ψ(x) by −1 does not change |ψ(x)|2. So another possible graph forψ(x) is the “upside down” version of the graph in part (b), i. e.

13. Electrons are accelerated through a potential difference, producing a monoenergeticbeam. This is directed at a double slit apparatus of d = 0.010 mm slit separation. Abank of electron detectors is 10 m beyond the double slit. With slit 1 alone open, 100electrons per second are detected at all detectors. With slit 2 alone open, 900 electronsper second are detected at all detectors. Now both slits are open.

(a) How many electrons per second will be detected at the central detector?

(b) How many electrons per second will be detected at the detector X where the firstminimum occurs?

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Solution:

(a) With slit 1 or 2 alone open, the probability of an electron reaching the screen is

P1 ∝ |Ψ1|2 or P2 ∝ |Ψ2|2.

It’s given that

P2

P1

=N2

N1

=900

100= 9

⇒ |Ψ2||Ψ1|

=

√P2

P1

= 3

where N1 and N2 are the number of electron reaching the screen with slit 1 or 2alone open.

With both slits open, the probability of an electron reaching the screen is P ∝|Ψ1 + Ψ2|2. Constructive interference occurs when the wave functions are in phase.So the probability for an electron reaching a maximum is equal to

Pmax ∝ (|Ψ1|+ |Ψ2|)2

Hence, the number of electron reaching the central detector (i. e. a maximum),Nmax, is given by

Nmax

N1

=Pmax

P1

=(|Ψ1|+ |Ψ2|)2

|Ψ1|2

⇒ Nmax = N1

[(|Ψ1|+ |Ψ2|)2

|Ψ1|2

]= N1

(1 +|Ψ2||Ψ1|

)2

= (100)(1 + 3)2 = 1600

(b) Destructive interference occurs when the wave functions are exactly out of phase.So the probability for an electron reaching a minimum is equal to:

Pmin ∝ (|Ψ1| − |Ψ2|)2

Hence, the number of electron reaching the detector X (i. e. a minimum), Nmin, isgiven by:

Nmin

N1

=Pmin

P1

=(|Ψ1| − |Ψ2|)2

|Ψ1|2

⇒ Nmin = N1

[(|Ψ1| − |Ψ2|)2

|Ψ1|2

]= N2

(1− |Ψ2||Ψ1|

)2

= (100)(1− 3)2 = 400

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