爱因斯坦的光电方程
阿尔伯特·爱因斯坦在 1905 年，即物理学的奇迹年（奇迹年）发表了一个方程来解释这种效应。根据爱因斯坦的说法，光是一种波，它以能量包或能量量子的形式与物质相互作用。光子是辐射的量子，这个方程被称为爱因斯坦的光电方程。
When a substance absorbs electromagnetic radiation, electrically charged particles are emitted from or inside it, resulting in the photoelectric effect. The phenomenon of emission of electrons by certain substance (metal), when it is exposed to radiations of suitable frequencies is called as photoelectric effect and emitted electrons are called photoelectrons.
这种效应通常被描述为当暴露在光线下时从金属板中喷射出电子。辐射能可以是红外线、可见光或紫外线、X射线或伽马射线；该物质可以是固体、液体或气体；释放的粒子可以是离子或电子。
光电效应的特点：
- 对于每种特定的光敏材料，入射辐射都有一个最小截止频率，称为阈值频率 ν o ，低于该频率时不会产生光电子。
- 阈值频率因金属而异。
- 光电流与特定感光材料的入射光强度和入射辐射频率（高于阈值频率）成正比。
- 高于阈值频率ν 0 ，发射光电子的最大动能随入射辐射频率线性增长，但与入射辐射强度无关。
- 光电子的发射发生在瞬间。
- 金属表面的照射和光电发射之间没有时间延迟。
爱因斯坦的光电方程
1905 年，爱因斯坦扩展了普朗克的概念，推导出了他自己的方程式，该方程式正确地描述了光电效应的特征。
他预设了两件事。
- 频率为 ν 的辐射由离散的量子或光子流组成，能量为 hν，其中 h 是普朗克常数。光子以光速穿过空间。
- 当频率为 ν 的辐射入射到光敏表面时，发射器原子中的光子和电子发生碰撞。在这样的碰撞过程中，光子的全部能量毫无延迟地传递给电子。
光子不是物质粒子，而是能量量子。入射光子被电子吸收的能量hv 以两种方式利用。电子利用它的一些能量从原子中挣脱出来。从给定表面释放自由电子所需的最小能量称为该表面材料的光电功函数φ 0 。
The residual energy (h ν- φ0) emerges as electron kinetic energy. If the electron does not lose any of its energy in impact with the surface and exits with the greatest possible kinetic energy.
∴ 1 / 2 m v2max = hν – φ0 …..(1)
where,
- m = mass of the electron
- vmax = maximum velocity of electron
All the photoelectrons emitted from the metal surfaces do not have the same energy.
The photoelectric current becomes zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy, so that the stopping potential (in volts) is numerically equal to the maximum kinetic energy of photo-electron in eV.
1 / 2 m v2max = ev0 …..(2)
where,
- e = magnitude of the charge on an electron.
From (2) we have,
1 / 2 m v2max = hν – φ0
eV0 = hν – φ0
This equation is Einstein`s photoelectric equation.
With the aid of Einstein’s photoelectric equation, we can now describe all of the features of the photoelectric effect:
If the frequency of incident radiation is decreased, the kinetic energy of photoelectrons also decreases, and finally, it becomes zero for a particular frequency (say ν). ν is called the threshold frequency. Thus,
When ν=νo,
then
KEmax = 1/2 m V2max = 0
Therefore from equation (1), we get
0 = hν0 – φ0
hν0 = φ0
Therefore, Einstein`s equation can be written as,
1/2 m v2max = h(ν – ν0) …..(3)
From the above equation, we can say three points,
- ν > ν0, the photoelectrons are emitted with some velocity,
- ν < ν0, no photoelectrons are emitted, and
- ν = ν0, photoelectrons are just emitted with zero kinetic energy.
A more intense beam, according to quantum theory, includes a higher number of photons. As a result, the number of photon-electron collisions increases, and more photoelectrons are released. This explains why the photoelectric current increases with incoming photon intensity. Because the photoelectric work function (φ0) is constant for every given emitter, equation (3) demonstrates that the KEmax of photoelectrons grows with the frequency of incoming radiation but does not rely on intensity.
由于电子-光子碰撞而发射光电子。一旦辐射撞击光敏表面，就会发生这种碰撞，并释放光电子。在入射截止处不发射光电子。结果，光电效应立即发生。
光的粒子性质：光子
因此，光电效应提供了一个不寻常的证据，即光的行为就好像它是由量子或能量包形成的，每个能量为 hν。
能量的光量子要与粒子相连吗？爱因斯坦发现光量子也可能与动量hv有关。能量和动量的确定值强烈表明光量子可以与粒子连接。
这个粒子最终被命名为光子。 1924 年，AH Compton (1892-1962) 对电子散射 X 射线的实验确立了光的类粒子行为。爱因斯坦因对理论物理学和光电效应的贡献而于 1921 年获得诺贝尔物理学奖。密立根因发现电的基本电荷和光电效应而于 1923 年获得诺贝尔物理学奖。
Summarized Photon picture of electromagnetic radiation
- When radiation interacts with matter, it behaves as if it were made up of particles known as photons.
- Each photon has energy E (= hu) and momentum p (= hν / c) where c is the speed of light.
- Whatever the intensity of radiation, all photons of light of a specific frequency ν or wavelength λ, have the identical energy E( = hν = hc / λ) and momentum p (= hν / c = h/λ). By raising the intensity of light of a certain wavelength, the number of photons per second crossing a given area increases, with each photon having the same energy. As a result, photon energy is independent of radiation intensity.
- Photons are electrically neutral and are unaffected by electric or magnetic forces.
- Total energy and total momentum are preserved in a photon-particle collision (such as a photon electron collision). However, the number of photons in a collision may not be preserved. The photon might be absorbed or a new photon could be produced.
示例问题
问题 1：电子通过 500 伏的电位差从静止状态加速。求电子的速度。 （给定 e = 1.6 × 10 -19 C，m = 19 ×10 -31 Kg）
解决方案：
Given that,
Voltage = 500 V
e = 1.6 × 10-19 C
m = 19 × 10 Kg
KE of emitted electron = eV = 1/2 m v2
= 2 × (1.6 × 10-19) × (500) / (9.1 × 10-31 )
V = √1.758 × 1014
= 1.326 × 107 m/s
问题 2：功函数为 4.2 eV 的金属被波长为 2000 A°的辐射照射。求发射电子的最大动能。
解决方案：
Given that,
φ0 = 4.2eV = 6.72 × 10-19 J
λ = 2000 A°
1/2 m v2max = hν – φ0
KEmax = hν – φ0
= (hc / λ) – φ0
= (6.63 ×10-34 × 3 × 10) / (2 × 10-7) – (6.72 × 10-10)
= (9.945 – 6.72) × 10-19
= 2.015 eV
问题3：如果要以6×10 5 m/s 的速度从钾表面发射光电子，必须使用什么频率的辐射？钾的阈值频率为 4.22 ×10 14 Hz。
解决方案：
vmax = 6 × 105 m/s
ν0 = 4.22 × 1014 Hz
KEmax = 1/2 mv2max = h(ν – ν0)
ν = 1/2 × (mv2max / h) + ν0
= (1/2) × [(9.1 × 10-31 ×(6 × 105)2 / 6.63 × 10-34] +4.22 × 1014 Hz
= 2.47 × 1014 + 4.22 × 1014
= 6.69 × 1014 Hz
问题4：一个波长为3310A°的光子落在一个光电阴极上，一个能量为3 × 10 -19 J 的电子被射出。如果将入射光子的波长改为5000 A°，则射出电子的能量为9.72×10 -20 J。计算普朗克常数的值和光子的阈值波长。
解决方案：
λ1 = 3310 = 3.31 × 10-7 m
KEmax (1) = 3 × 10-19 J
λ2 = 5000 = 5 × 10-7 m
KEmax = 9.72 × 10-20 J= 0.972 × 10-19 J
KEmax (1) = hc/λ1 – φ0
KEmax (2) = hc/λ2 – φ0
KEmax (1) – KEmax (2) = hc (1/λ1 – 1/λ2)
3 × 10-19 – 0.972 × 10-19 = h × 3 × 108 [(1/3.31 × 10-7) – (1/5 × 10-7)]
2.028 × 10-19 = h × 3 × 1015 (1.69 / 16.55)
h = 2.028 × 10-19 × 16.55 / 3 × 1015 × 1.69
= 6.62 × 10-34 Js
Now,
KEmax (1) = hc/λ1 – φ0
φ0 = hc/λ1 – KEmax (1)
= (6.62 × 10-34 × 3 × 108 / 3.31 × 10-7 ) – 3 × 10-19
= 3 × 10-19 J
Also,
φ0 = hc / λ0
λ0 = hc / φ0
= 6.62 × 10-34 × 3 × 108 /3 × 10-19
= 6.62 × 10-7
= 6620 A°
问题 5：金属表面的光电函数为 2.4 eV。如果波长为5000 A°的光入射到金属表面，求阈值频率和入射频率，是否会发射光电子？
解决方案：
φ0 = 2.34eV = 3.84 × 10-19 J
λ = 5000A° = 5 × 10-7
ν = c / λ
= 3 × 108 / 5 × 10-7
= 6 × 1034 Hz
ν0 = φ0 / h
= 3.84 × 10-19 / 6.63 × 10-34
= 5.792 × 1014 Hz
As ν > ν0, photoelectric emission is possible.