刁勇强 34-35-A17
CHAPTER 2 Theory of Solids
This value of energy may be given in the more common unit of electron-volts(see Appendix D).
The electron-volt(ev) is the unit of energy equal to 1.6×10-19J.We can then write
15
4
19
2.8110
1.75101.610E eV
--?=
=?? ? Comment
The reciprocal relation between photon energy and wavelength is demonstrated:a large energy corresponds to a short wavelength.
Exercise Problem
EX2.1 Determine the energy(in eV)of a photon having wavelengths of(a) 0
10,000λ=A and (b)0
10λ=A .3
.()1.24,()1.2410n s a e V b e V ??
A ???
2.1.2 wave- particle Duality principle
We have seen in the previous paragraphs that light waves,in the photoelectric
effect,behave as if they are particles.In 1924,de Broglie postulated the existence of matter waves.He suggested that since waves exhibit particle-like behavior, then particles should be expected to show wave-like properties. The hypothesis of de Broglie is the existence of a wave-particle duality principle.The momentum of a photon is given by
h
p λ=
(2.2)
Where λis the wavelength of the light wave. Then,de Broglie hypothesized that the wavelength of a particle can be expressed as
h p
λ=
(2.3)
Where p is the momentum of the particle and λ is known as the de Broglie wavelength of the matter wave.
EXAMPLE 2.2 OBJECTIVE
Determine the de Broglie wavelength of a particle.
Consider an electron traveling at a velocity of 107cm/s=105m/s. ? Solution
The momentum is given by P = mv =(9.11×10-31)(105)=9.11×10-26Kg-m/s Then the de Briglie wavelength is
349
26
6.62510
7.72109.1110
h m
p λ---?=
=
=??
2.1 Principles of Quantum Mechanics or
72.7
λ=A
Comment
This calculation shows the order of magnitude of the de Broglie wavelength for a “typical”electron.
Exercise Problem
EX2.2(a) Find the momentum and energy of a particle with mass 5×10-31Kg and a de
Broglie wavelength of
180A.(b)An electron has a kinetic energy of 20 meV.Derermine
the de Broglie wavelength.
2
6
3
2
6
.(
)
3.6
8
1
/
,
4.6
5
1
;(
)
7.6
4
1
/
,
8
6.8n s
a
p
k g
m
s
E
e
V
b
p
k g
m
s
λ
-
-
-?
?
A
=
?
-
=
?
=
?
-
=?
??
?
A
To gain some appreciation of the frequencies and wavelengths involved in the wave-particle duality principle,Figure2.2 shows the electromagnetic frequency spectrum. We
see that a wavelength of
72.7A obtained in Example 2.2 is in the ultraviolet
range.Typically,we will be considering wavelengths in the ultraviolet and visible range.These wavelengths are very short compared with the usual radio spectrum range.
In some cases,electromagnetic waves behave as if are particles(photons)and,in some cases,particles behave as if they are waves,This wave-particle duality principle of quantum mechanics applies primarily to small particles such as electrons,but it has also been shown to apply to protons and neutrons.For very large particles,
Figure 2.2| The electromagnetic frequency spectrum.
第二章 固态原理
能量的这种价值也许会被呈现在更为普遍的电子伏特部件里（见附录D ）。 这个电子伏特部件的能量单位等于16×10-19J 。我们可以写
15
4
19
2.8110
1.75101.610E eV
--?=
=??
? 注释：
光子能跟波长的倒数关系证明：一个大能源对应一个短的波长。
练习题
例2.1：确定在电子伏特中的光子能有（a ）0
10,000λ=A 和（b ）0
10λ=A 的波长。 3
.()1.24,()1.2410
n s a e V b e V ??A ???
2.1.2 粒子波的对偶原理
我们在前一段落所看到的光波，在光电效应里表现得像是颗粒。在1924年，德布洛伊假定了物质波的存在。他暗示了，自从这种波展现出了类似微粒的行为，微粒就被指望有着波状的性能。德布洛伊的假设是波粒子对偶原理的存在。给出的光子动量是：
h
p λ=
（2.2） λ是光波的波长。而德布洛伊假设微粒的波长可以表示为：
h
p λ=
（2.3） P 是指微粒动量，而λ被称为是德布洛伊的物质波波长。
例2.2 目标
确定一个微粒的德布洛伊波长。
考虑一个电子以速度107cm/s=105m/s 的运行。
? 解决方案 给出的动量为
P = mv =(9.11×10-31)(105)=9.11×10-26Kg-m/s
而德布洛伊波长为
34
9
26
6.62510
7.72109.1110
h m
p λ---?=
=
=??
2.1 量子力学原理或
72.7
λ=A
注释
这个计算表明了一个“典型”电子的德布洛伊波长的数量级。
练习题
例2.2（a）找出质量为5×10-31Kg和德布洛伊波长为
180A的微粒的动量和能量。
（b）一个电子拥有20meV的动能，确定它的德布洛伊波长。
2
6
3
2
6
.(
)
3.6
8
1
/
,
4.6
5
1
;(
)
7.6
4
1
/
,
8
6.8n s
a
p
k g
m
s
E
e
V
b
p
k g
m
s
λ
-
-
-?
?
A
=
?
-
=
?
=
?
-
=?
??
?
A
在粒子波的对偶原理中获得一些有关频率和波长的鉴别，图2.2显示的是电磁频
谱。在图2.2中我们看到的波长
72.7A是在紫外线范围内获得的。典型地，我么将
会考虑到在紫外线范围和可视范围内的波长。这些波长比起在通常的无线频谱范围内的较为短。
有时候，电磁波表现得就像是微粒（光子），还有的时候，微粒表现得又像是电磁波。这种量子力学的粒子波原理主要适用于像电子那样的小微粒，但是它也曾被显示适用于光子和中子。对于非常大的颗粒，
图2.2 | 电磁频谱。