HW_Ch05

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Chapter 5

Matter Waves. Home Work Solutions

5.1 Problem 5.10(In the text book)An electron has a de Broglie wavelength equal to the diameter of the hydrogen atom. Whatis the kinetic energy of the electron? How does this energy compare with the ground-stateenergy of the hydrogen atom?

Solution

The radius of the hydrogen atom can be taken as the radius of the first orbit or, Bohr radius

a = 0.5292 1010

m. The de Broglie wave length of the given electron is:

= 2a = 1.0584 1010 mUsingp= h/, wherepis the momentum of the electron, the kinetic energy Kof the electronis:

K = p2

2me

= h2c2

2me

c2

2

= (1.240 106eV m)2

2 5.11 105(eV) (1.0584 1010 m)2= 134.3eV

The energy of the ground state of the hydrogen atom is 13.6eV, so the kinetic energy of thegiven electron is about times the energy of the ground state of the hydrogen atom.

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2 CHAPTER 5. MATTER WAVES. HOME WORK SOLUTIONS

5.2 Problem 5.14(In the text book)

(a) Show that the formula for low-energy electron diffraction (LEED), when electrons areincident perpendicular to a crystal surface, may be written as

sin = nhc

d

2mec2K

wheren is the order of the maximum, dis the atomic spacing, me is the electron mass,Kis the electrons kinetic energy, and is the angle between the incident and diffractedbeams.

(b) Calculate the atomic spacing in a crystal that has consecutive diffraction maxima at= 24.1 and = 54.9 for 100-eV electrons.

Solution

(a) For a maximum in electron diffraction from crystal planes we have:

n = d sin

sin = n

d

where is the de Broglie wave length of the electron, and = h/p, where p is theelectrons momentum. The kinetic energy of the electron K=p2/2me, we th get:

sin = nd

hp

= n

d

h2Kme

=

nhc

d

2Kmec2

Physics 205:Modern Physics I, Chapter 5 Fall 2004 Ahmed H. Hussein

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5.2. PROBLEM 5.14(IN THE TEXT BOOK) 3

(b) The atomic spacing is;

d= nhc

sin 2Kmec2

For = 24.1 the atomic spacingd1 is:

d1 = n 1.240 106(eV m)

sin(24.1)

2 0.511 106(eV) 100(eV)= 3.00 1010 m= 3.00n

A

For = 54.9 the atomic spacingd1 is:

d2 = (n+ 1) 1.240 106(eV m)

sin(54.9)

2 0.511 106(eV) 100(eV)= 3.00(n+ 1) 1010 m= 3.00(n+ 1)

A

for the same spacing we should have:

d1 = d2

3.00n = 1.50(n+ 1)

n = 1

which means the 24.1 corresponds ton= 1 maximum and 54.9 corresponds ton= 2maximum.


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We wish to measure simultaneously the wavelength and position of a photon. Assume thatthe wavelength measurement gives = 6000

A with an accuracy of one part in a million,that is, /= 106. What is the minimum uncertainty in the position of the photon?

Solution

The momentump of the photon is given by:

p=h

we then get:

p

dp

d

= h2

p = h2

Since,

px 1

2

(px)min = 1

2

|x|min = 2p

=

2(h/2)

= h

4(h/)

=

4(/)

= 6000

A

4 106= 4.78 108

A

= 4.78 102 m


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Robert Hofistadter won the 1961 Nobel prize in physics for his pioneering work in scattering20-GeV electrons from nuclei.

(a) What is the factor for a 20-GeV electron, where = (1 v2/c2)12 ? What is themomentum of the electron in kg m/s?

(b) What is the wavelength of a 20-GeV electron and how does it compare with the size ofa nucleus?

Solution

(a) Using the relativistic definition of the total energy we get:

E = mec2

= E

mec2

= 20 103(MeV)0.511(MeV)

= 3.91 104

Since the energy of the electron is very high, then, mec2 pc, and using the relativistic

definition of the total energy, in terms of the momentum, we get:

E2 = (pc)2 +m2ec4

p E

c

= 20 103(MeV) 1.602 1013(J/MeV)

3 108m/s= 1.07 1017 kg m/s

(b) The de Broglie wavelength of the 20 GeV electron is:


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= h

p

= 6.626 1034(J s)

1.0717 kg m/s= 6.19 1017 m

the size of the nucleus is about 1014 m, so the wavelength of the 20 GeV electron isabout 1000 times smaller than the nucleus.


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Figure 5.1:

(a) Find and sketch the spectral content of the rectangular pulse of width 2- shown inFigure (5.1).

(b) Show that a reciprocity relation t holds in this case. Take t= and define similarly.

(c) What range of frequencies is required to compose a pulse of width 2 = 1s? A pulse

of width 2= 1ns?

Solution

(a) Using Fourier integrals:

g() = 1

2

V(t)eit dt

= 1

2

V(t) (cos t i sin t)dt

V(t)sin t is an odd function in t so its integral vanishes, we then get:

g() = 1

2

V(t) cos t dt


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from Figure (5.1), V(t) = V when < t < and V(t) = 0 everywhere else. Also,V(t) is symmetric around t = 0:

g() = 2

2

0

V cos t dt

=

2

V

sin t

0

=

2

V

sin

A plot ofg() is shown in Figure (5.2).

g()

0

2

2

2

V t,

Figure 5.2:

(b) Since the major contribution to the pulse comes from / < t

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= 2f

= 2f

t =

2f =

=

t

= 1

t

= 1

0.5 106

= 2 106

Hz

and for a 1ns pulse:

2f = 1

t

= 1

0.5 109= 2 109 Hz

Notice that to make the pulse sorter, i.e. localized in a smaller time range you need amuch larger frequency range.


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HW_Ch05