含义Meaning
光照射金属时，只有频率高于阈值 ν₀ 的光才能打出电子，与光强无关；光电子最大动能随频率线性增大。爱因斯坦（1905）用光子解释，获 1921 年诺贝尔物理学奖。
Light ejects electrons from a metal only above a threshold frequency ν₀, regardless of intensity; the maximum kinetic energy rises linearly with frequency. Einstein’s photon explanation (1905) won the 1921 Nobel Prize in Physics.
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(\Phi\) | 逸出功（功函数）work function | eV |
| \(\nu_0\) | 截止频率threshold frequency | Hz |
| \(E_{k,\max}\) | 光电子最大动能maximum photoelectron kinetic energy | eV |
推导Derivation
一个光子把全部能量 hν 交给一个电子。
One photon gives all its energy hν to one electron.
电子逸出至少消耗 Φ，剩余为动能。
Escape costs at least Φ; the rest is kinetic energy.
用遏止电压测量：\(eU_s = h\nu - \Phi\)；直线斜率 h/e（密立根 1916 年据此测得 h）。
Stopping voltage: \(eU_s = h\nu - \Phi\); the slope h/e gave Millikan’s 1916 value of h.
适用条件与局限Assumptions & limitations
表面状况影响 Φ；温度效应和多光子过程略去。
Surface condition affects Φ; thermal and multiphoton effects ignored.
例题Worked example
钠 Φ = 2.28 eV（ν₀ = 5.5×10¹⁴ Hz，λ₀ = 544 nm）；用 400 nm 光照射时 E_k,max = 3.10 − 2.28 = 0.82 eV。
Sodium Φ = 2.28 eV; 400 nm light gives E_k,max = 0.82 eV.