Unit B.2: Greenhouse Effect单元 B.2：温室效应
A super-topic of Theme B "The particulate nature of matter" that applies thermal-radiation physics to a whole planet. The solar constant and inverse-square law, the incident power intercepted by a planet's cross-section, albedo and emissivity, the Stefan-Boltzmann law applied to a grey body, the planetary energy-balance model and its equilibrium temperature, the molecular resonance that lets greenhouse gases absorb outgoing infrared, and the enhanced greenhouse effect. This unit links radiation physics to climate, and is examined on both Paper 1 (MC + data) and Paper 2.主题 B"物质的粒子本质"下的一个超级专题，把热辐射物理应用到整颗行星。太阳常数与平方反比定律、行星横截面截获的入射功率、反照率与发射率、应用于灰体的斯特藩-玻尔兹曼定律、行星能量平衡模型及其平衡温度、使温室气体能够吸收外逸红外的分子共振，以及增强温室效应。本单元把辐射物理与气候联系起来，Paper 1（选择题+数据题）与 Paper 2 都会考。
How to use this guide本指南使用说明
B.2 is a "one model, many numbers" unit. Almost every quantitative question is a variation of a single energy-balance equation: power in (from the Sun, reduced by albedo) equals power out (radiated by the planet). The marks come from getting the geometry right — a disc of area $\pi r^{2}$ intercepts sunlight, but the whole sphere of area $4 \pi r^{2}$ radiates — and from quoting the data-booklet relations cleanly. The conceptual marks come from explaining molecular resonance and the enhanced greenhouse effect in words.B.2 是"一个模型、许多数字"的单元。几乎每道定量题都是同一条能量平衡方程的变形：输入功率（来自太阳，被反照率削减）等于输出功率（行星辐射出去）。分数来自把几何关系弄对——面积 $\pi r^{2}$ 的圆盘截获阳光，而整个面积 $4 \pi r^{2}$ 的球面向外辐射——以及干净地引用数据手册中的关系式。概念分来自用文字解释分子共振与增强温室效应。
Memorise $I = \frac{P}{4 \pi d^{2}}$ (intensity), the absorbed-power expression $(1 - \alpha) S \pi r^{2}$, and the emitted-power expression $e \sigma 4 \pi r^{2} T^{4}$. Set them equal for the equilibrium temperature. Know that $\alpha$ = albedo (fraction reflected) and $e$ = emissivity (between $0$ and $1$). Greenhouse gases absorb outgoing infrared and re-emit it, warming the surface.
背熟 $I = \frac{P}{4 \pi d^{2}}$（强度）、吸收功率表达式 $(1 - \alpha) S \pi r^{2}$ 和辐射功率表达式 $e \sigma 4 \pi r^{2} T^{4}$。令两者相等即得平衡温度。记住 $\alpha$ 是反照率（反射比例）、$e$ 是发射率（介于 $0$ 与 $1$）。温室气体吸收外逸红外并再辐射，使地表升温。
Be able to derive the equilibrium temperature from scratch, including the factor of $4$ that comes from disc-intercepts-versus-sphere-radiates. Explain emissivity as a grey-body correction to the ideal black body. Explain molecular resonance: a greenhouse gas absorbs an IR photon whose frequency matches a vibrational mode of the molecule. Distinguish the natural greenhouse effect (essential for life) from the enhanced greenhouse effect (the warming caused by rising gas concentrations).
能够从头推导平衡温度，包括"圆盘截获、球面辐射"带来的因子 $4$。把发射率理解为对理想黑体的灰体修正。解释分子共振：温室气体吸收频率与其某个振动模式匹配的红外光子。区分自然温室效应（生命所必需）与增强温室效应（气体浓度上升所致的额外升温）。
Solar Constant, Intensity, and Incident Power太阳常数、强度与入射功率 B.2 SL+HL
I = P / (4πd²):
$$ I = \frac{P}{4 \pi d^{2}}. $$
Inverse-square law. Intensity falls as $1/d^{2}$: double the distance, quarter the intensity. Units $\mathrm{W\,m^{-2}}$.
Solar constant $S$. The intensity of solar radiation at the top of Earth's atmosphere (mean Sun–Earth distance), $S \approx 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$.
Incident power on a planet. A planet of radius $r$ presents a circular cross-section (a disc) of area $\pi r^{2}$ to the incoming parallel rays, so the power it intercepts is
$$ P_{\text{in}} = S \, \pi r^{2}. $$
I = P / (4πd²)：
$$ I = \frac{P}{4 \pi d^{2}}. $$
平方反比定律。强度随 $1/d^{2}$ 衰减：距离加倍，强度变四分之一。单位 $\mathrm{W\,m^{-2}}$。
太阳常数 $S$。地球大气层顶（日地平均距离处）的太阳辐射强度，$S \approx 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$。
行星上的入射功率。半径为 $r$ 的行星，对平行入射光线呈现一个面积为 $\pi r^{2}$ 的圆形横截面（圆盘），因此它截获的功率为
$$ P_{\text{in}} = S \, \pi r^{2}. $$
The Sun radiates with total power (luminosity) $P = 3.85 \times 10^{26}\ \mathrm{W}$. The mean Sun–Earth distance is $d = 1.50 \times 10^{11}\ \mathrm{m}$. Find the solar constant $S$, and then the total power intercepted by Earth (radius $r = 6.37 \times 10^{6}\ \mathrm{m}$).太阳以总功率（光度）$P = 3.85 \times 10^{26}\ \mathrm{W}$ 辐射。日地平均距离 $d = 1.50 \times 10^{11}\ \mathrm{m}$。求太阳常数 $S$，再求地球（半径 $r = 6.37 \times 10^{6}\ \mathrm{m}$）截获的总功率。
Identify. Treat the Sun as a point source; the relevant relation is the inverse-square intensity I = P / (4πd²).
识别。把太阳视为点源；相关关系是平方反比强度 I = P / (4πd²)。
Set up. The solar constant is just the intensity at $d$:
列式。太阳常数即 $d$ 处的强度：
$$ S = \frac{P}{4 \pi d^{2}} = \frac{3.85 \times 10^{26}}{4 \pi (1.50 \times 10^{11})^{2}}. $$Execute. The denominator is $4 \pi (2.25 \times 10^{22}) = 2.83 \times 10^{23}$, so $S = 3.85 \times 10^{26} / 2.83 \times 10^{23} \approx 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$.
计算。分母为 $4 \pi (2.25 \times 10^{22}) = 2.83 \times 10^{23}$，故 $S = 3.85 \times 10^{26} / 2.83 \times 10^{23} \approx 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$。
Intercepted power. Earth intercepts over its disc $\pi r^{2}$: $P_{\text{in}} = S \pi r^{2} = (1.36 \times 10^{3}) \pi (6.37 \times 10^{6})^{2} \approx 1.73 \times 10^{17}\ \mathrm{W}$.
截获功率。地球以圆盘 $\pi r^{2}$ 截获：$P_{\text{in}} = S \pi r^{2} = (1.36 \times 10^{3}) \pi (6.37 \times 10^{6})^{2} \approx 1.73 \times 10^{17}\ \mathrm{W}$。
Evaluate. The recovered $S \approx 1.36\ \mathrm{kW\,m^{-2}}$ matches the accepted value. Note we used the disc area $\pi r^{2}$, not the surface area $4 \pi r^{2}$ — only the cross-section faces the Sun.
评估。所得 $S \approx 1.36\ \mathrm{kW\,m^{-2}}$ 与公认值吻合。注意此处用的是圆盘面积 $\pi r^{2}$，而非表面积 $4 \pi r^{2}$——只有横截面正对太阳。
Going deeper: why $\pi r^{2}$ and not $4 \pi r^{2}$ for the incident power深入：入射功率为何用 $\pi r^{2}$ 而非 $4 \pi r^{2}$
Sunlight arrives as a beam of effectively parallel rays. The amount of that beam a sphere captures equals the area of the sphere's shadow — its silhouette — which is a disc of area $\pi r^{2}$. The illuminated hemisphere has area $2 \pi r^{2}$, but rays hit it at varying angles; the perpendicular projection of that lit hemisphere onto a plane is exactly the disc $\pi r^{2}$. So:
阳光以一束近似平行的光线到达。球体捕获其中多少，等于球体阴影（轮廓）的面积，即面积为 $\pi r^{2}$ 的圆盘。被照亮的半球面积为 $2 \pi r^{2}$，但光线以不同角度入射；被照半球向平面的垂直投影恰好是圆盘 $\pi r^{2}$。所以：
$$ P_{\text{in}} = S \times (\text{cross-section}) = S \, \pi r^{2}. $$By contrast, the planet radiates from its entire surface $4 \pi r^{2}$ (it is warm all over, day and night). The ratio of these two areas, $\dfrac{\pi r^{2}}{4 \pi r^{2}} = \dfrac{1}{4}$, is the famous factor of $4$ that appears in the equilibrium-temperature derivation in Section B2.4.
相比之下，行星从它整个表面 $4 \pi r^{2}$ 辐射（昼夜各处都有温度）。这两个面积之比 $\dfrac{\pi r^{2}}{4 \pi r^{2}} = \dfrac{1}{4}$，正是 B2.4 节平衡温度推导中出现的著名因子 $4$。
Albedo: the Fraction Reflected反照率：被反射的比例 B.2 SL+HL
albedo = (total scattered power) / (total incident power):
$$ \alpha = \frac{P_{\text{reflected}}}{P_{\text{incident}}}, \qquad 0 \le \alpha \le 1. $$
Consequences. A fraction $\alpha$ is reflected; the remaining fraction $(1 - \alpha)$ is absorbed. So absorbed power is
$$ P_{\text{abs}} = (1 - \alpha) \, P_{\text{in}} = (1 - \alpha) S \, \pi r^{2}. $$
Typical values. Fresh snow/ice $\approx 0.8$–$0.9$; thick cloud $\approx 0.6$–$0.7$; desert sand $\approx 0.4$; ocean (overhead Sun) $\approx 0.06$; Earth as a whole $\approx 0.30$. High albedo = bright/reflective; low albedo = dark/absorbing.
albedo = (total scattered power) / (total incident power)：
$$ \alpha = \frac{P_{\text{reflected}}}{P_{\text{incident}}}, \qquad 0 \le \alpha \le 1. $$
推论。比例 $\alpha$ 被反射；剩下的 $(1 - \alpha)$ 被吸收。故吸收功率为
$$ P_{\text{abs}} = (1 - \alpha) \, P_{\text{in}} = (1 - \alpha) S \, \pi r^{2}. $$
典型值。新鲜雪/冰 $\approx 0.8$–$0.9$；厚云 $\approx 0.6$–$0.7$；沙漠沙 $\approx 0.4$；海洋（太阳当头）$\approx 0.06$；整个地球 $\approx 0.30$。高反照率＝明亮/反射强；低反照率＝深色/吸收强。
Earth has albedo $\alpha = 0.30$ and intercepts solar power $P_{\text{in}} = 1.73 \times 10^{17}\ \mathrm{W}$ (from B2.1). What power is reflected, and what power is absorbed?地球反照率 $\alpha = 0.30$，截获太阳功率 $P_{\text{in}} = 1.73 \times 10^{17}\ \mathrm{W}$（来自 B2.1）。求被反射的功率与被吸收的功率。
Identify. Albedo splits incident power into a reflected fraction $\alpha$ and an absorbed fraction $(1 - \alpha)$.
识别。反照率把入射功率分成被反射比例 $\alpha$ 与被吸收比例 $(1 - \alpha)$。
Reflected. $P_{\text{ref}} = \alpha P_{\text{in}} = 0.30 \times 1.73 \times 10^{17} \approx 5.2 \times 10^{16}\ \mathrm{W}$.
反射。$P_{\text{ref}} = \alpha P_{\text{in}} = 0.30 \times 1.73 \times 10^{17} \approx 5.2 \times 10^{16}\ \mathrm{W}$。
Absorbed. $P_{\text{abs}} = (1 - \alpha) P_{\text{in}} = 0.70 \times 1.73 \times 10^{17} \approx 1.2 \times 10^{17}\ \mathrm{W}$.
吸收。$P_{\text{abs}} = (1 - \alpha) P_{\text{in}} = 0.70 \times 1.73 \times 10^{17} \approx 1.2 \times 10^{17}\ \mathrm{W}$。
Evaluate. Roughly $30\%$ of intercepted sunlight is bounced straight back to space and never heats the planet; only the absorbed $70\%$ enters the energy balance of Section B2.4.
评估。约 $30\%$ 的截获阳光直接被弹回太空、根本不加热行星；只有被吸收的 $70\%$ 进入 B2.4 节的能量平衡。
Going deeper: the ice–albedo feedback深入：冰-反照率反馈
Albedo is not a fixed constant — it depends on the planet's surface, which itself depends on temperature. This creates a positive feedback loop:
反照率不是固定常数——它取决于行星表面，而表面又取决于温度。这构成一个正反馈循环：
Warming melts ice and snow (high $\alpha \approx 0.85$), exposing darker ocean and land (low $\alpha \approx 0.1$). Lower planetary albedo means more sunlight absorbed, which causes more warming, which melts more ice. The reverse runs during glaciation. The IB syllabus expects you to describe this qualitatively: a change in $\alpha$ shifts the energy balance and hence the equilibrium temperature.
升温融化冰雪（高 $\alpha \approx 0.85$），露出较暗的海洋与陆地（低 $\alpha \approx 0.1$）。行星反照率降低意味着吸收更多阳光，导致进一步升温，融化更多冰。冰期时该过程反向进行。IB 大纲要求你能定性描述这一点：$\alpha$ 的变化会改变能量平衡，从而改变平衡温度。
Emissivity, Black Bodies, and Stefan–Boltzmann发射率、黑体与斯特藩-玻尔兹曼 B.2 SL+HL
L = σ A T⁴:
$$ P_{\text{bb}} = \sigma A T^{4}, \qquad \sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}. $$
Emissivity $e$. The dimensionless ratio of the power radiated by a real ("grey") body to that radiated by a black body at the same temperature, emissivity = (power radiated per unit area) / (σT⁴):
$$ e = \frac{P_{\text{real}}}{P_{\text{bb}}}, \qquad 0 \le e \le 1. $$
A perfect black body has $e = 1$; real surfaces have $e < 1$.
Grey body radiated power.
$$ P = e \, \sigma A \, T^{4}. $$
For a planet ($A = 4 \pi r^{2}$):
$$ P_{\text{out}} = e \, \sigma \, 4 \pi r^{2} \, T^{4}. $$
L = σ A T⁴：
$$ P_{\text{bb}} = \sigma A T^{4}, \qquad \sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}. $$
发射率 $e$。真实（"灰"）体辐射功率与同温度黑体辐射功率之比，无量纲，emissivity = (power radiated per unit area) / (σT⁴)：
$$ e = \frac{P_{\text{real}}}{P_{\text{bb}}}, \qquad 0 \le e \le 1. $$
理想黑体 $e = 1$；真实表面 $e < 1$。
灰体辐射功率。
$$ P = e \, \sigma A \, T^{4}. $$
对行星而言（$A = 4 \pi r^{2}$）：
$$ P_{\text{out}} = e \, \sigma \, 4 \pi r^{2} \, T^{4}. $$
A planet of radius $r = 6.37 \times 10^{6}\ \mathrm{m}$ has surface temperature $T = 288\ \mathrm{K}$ and emissivity $e = 0.62$. Find the total power it radiates. Take $\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.半径 $r = 6.37 \times 10^{6}\ \mathrm{m}$ 的行星，表面温度 $T = 288\ \mathrm{K}$，发射率 $e = 0.62$。求它辐射的总功率。取 $\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$。
Identify. Use the grey-body form of Stefan–Boltzmann over the full surface area $A = 4 \pi r^{2}$.
识别。对整个表面积 $A = 4 \pi r^{2}$ 使用斯特藩-玻尔兹曼的灰体形式。
Surface area. $A = 4 \pi (6.37 \times 10^{6})^{2} \approx 5.10 \times 10^{14}\ \mathrm{m^{2}}$.
表面积。$A = 4 \pi (6.37 \times 10^{6})^{2} \approx 5.10 \times 10^{14}\ \mathrm{m^{2}}$。
Execute. $T^{4} = 288^{4} \approx 6.88 \times 10^{9}\ \mathrm{K^{4}}$. Then
计算。$T^{4} = 288^{4} \approx 6.88 \times 10^{9}\ \mathrm{K^{4}}$。于是
$$ P_{\text{out}} = e \sigma A T^{4} = (0.62)(5.67 \times 10^{-8})(5.10 \times 10^{14})(6.88 \times 10^{9}). $$This gives $P_{\text{out}} \approx 1.2 \times 10^{17}\ \mathrm{W}$.
得 $P_{\text{out}} \approx 1.2 \times 10^{17}\ \mathrm{W}$。
Evaluate. This output power matches the absorbed power of Section B2.2 ($\approx 1.2 \times 10^{17}\ \mathrm{W}$) — exactly the balance condition that fixes $T$. The emissivity $e = 0.62$ (below $1$) reflects the atmosphere's reduced efficiency at radiating to space.
评估。该输出功率与 B2.2 节的吸收功率（$\approx 1.2 \times 10^{17}\ \mathrm{W}$）相符——正是确定 $T$ 的平衡条件。发射率 $e = 0.62$（小于 $1$）反映了大气向太空辐射的效率被削弱。
Going deeper: emissivity, absorptivity, and Kirchhoff's law深入：发射率、吸收率与基尔霍夫定律
Kirchhoff's law of thermal radiation states that, at a given wavelength and temperature, a body's emissivity equals its absorptivity: a good absorber is a good emitter. A perfect black body absorbs everything ($\alpha_{\text{abs}} = 1$) and so also emits maximally ($e = 1$). A "grey body" has $e < 1$ and the same factor applies across all wavelengths, which is why we can pull a single constant $e$ out of the Stefan–Boltzmann integral:
基尔霍夫热辐射定律指出：在给定波长与温度下，物体的发射率等于其吸收率——好的吸收体也是好的辐射体。理想黑体吸收一切（$\alpha_{\text{abs}} = 1$），故也以最大功率辐射（$e = 1$）。"灰体"的 $e < 1$ 且对所有波长取同一因子，因此我们可以把单一常数 $e$ 从斯特藩-玻尔兹曼积分中提出来：
$$ P = e \int_{0}^{\infty} (\text{black-body spectral power})\, d\lambda = e \, \sigma A T^{4}. $$Real planets are not grey at all wavelengths — they absorb strongly in the visible (where the Sun peaks) yet emit in the infrared (where they peak). This wavelength selectivity is precisely what makes the greenhouse effect possible, and is treated in Section B2.5.
真实行星并非在所有波长都是灰体——它们在可见光（太阳峰值处）强烈吸收，却在红外（自身峰值处）发射。正是这种波长选择性使温室效应成为可能，详见 B2.5 节。
Energy Balance and Equilibrium Temperature能量平衡与平衡温度 B.2 SL+HL
Take $S = 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$, $\alpha = 0.30$, $\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$. (a) Find the equilibrium temperature treating Earth as a black body ($e = 1$). (b) The observed mean surface temperature is $288\ \mathrm{K}$. Comment.取 $S = 1.36 \times 10^{3}\ \mathrm{W\,m^{-2}}$、$\alpha = 0.30$、$\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$。(a) 把地球当作黑体（$e = 1$）求平衡温度。(b) 实测平均地表温度为 $288\ \mathrm{K}$，请评论。
Identify. Use the equilibrium-temperature result $T = \left[ (1 - \alpha) S / (4 e \sigma) \right]^{1/4}$ with $e = 1$.
识别。用平衡温度结果 $T = \left[ (1 - \alpha) S / (4 e \sigma) \right]^{1/4}$，取 $e = 1$。
Set up the bracket.
列出方括号内的式子。
$$ \frac{(1 - 0.30)(1.36 \times 10^{3})}{4 (1)(5.67 \times 10^{-8})} = \frac{952}{2.268 \times 10^{-7}} \approx 4.20 \times 10^{9}. $$Execute. $T = (4.20 \times 10^{9})^{1/4} \approx 255\ \mathrm{K}$ (about $-18\ ^\circ\mathrm{C}$).
计算。$T = (4.20 \times 10^{9})^{1/4} \approx 255\ \mathrm{K}$（约 $-18\ ^\circ\mathrm{C}$）。
(b) Comment. The black-body model predicts $255\ \mathrm{K}$, but the real surface is $288\ \mathrm{K}$ — about $33\ \mathrm{K}$ warmer. The gap is the natural greenhouse effect: the atmosphere traps outgoing infrared, raising the surface temperature above the bare radiative-balance value.
(b) 评论。黑体模型预测 $255\ \mathrm{K}$，但真实地表为 $288\ \mathrm{K}$——约高 $33\ \mathrm{K}$。这一差距即自然温室效应：大气捕获外逸红外，使地表温度高于纯辐射平衡值。
Evaluate. Equivalently, modelling Earth as a grey body with $e \approx 0.62$ instead reproduces $288\ \mathrm{K}$ directly — the lowered emissivity is how the greenhouse effect enters the energy-balance equation.
评估。等价地，把地球建模为发射率 $e \approx 0.62$ 的灰体，可直接得到 $288\ \mathrm{K}$——降低的发射率正是温室效应进入能量平衡方程的方式。
Going deeper: deriving $T$ from $(1-\alpha) S \pi r^{2} = e \sigma 4 \pi r^{2} T^{4}$深入：从 $(1-\alpha) S \pi r^{2} = e \sigma 4 \pi r^{2} T^{4}$ 推导 $T$
Start from the balance of intercepted-and-absorbed power against emitted power:
从"截获并吸收的功率"与"发射功率"的平衡出发：
$$ \underbrace{(1 - \alpha) S \, \pi r^{2}}_{\text{absorbed}} = \underbrace{e \, \sigma \, 4 \pi r^{2} \, T^{4}}_{\text{emitted}}. $$Both sides carry a factor $\pi r^{2}$, so the planet's radius cancels — equilibrium temperature is independent of size:
两边都含因子 $\pi r^{2}$，故行星半径相消——平衡温度与大小无关：
$$ (1 - \alpha) S = 4 \, e \, \sigma \, T^{4}. $$Isolate $T^{4}$ and take the fourth root:
解出 $T^{4}$ 再开四次方：
$$ T^{4} = \frac{(1 - \alpha) S}{4 e \sigma} \quad \Longrightarrow \quad T = \left[ \frac{(1 - \alpha) S}{4 e \sigma} \right]^{1/4}. $$The lone factor of $4$ in the denominator is the disc-to-sphere area ratio. A common exam slip is to forget it, which overestimates $T$ by a factor of $4^{1/4} \approx 1.41$ — turning $255\ \mathrm{K}$ into a wrong $360\ \mathrm{K}$. Always keep the $4$.
分母中孤立的因子 $4$ 即圆盘与球面的面积之比。常见考试失误是把它漏掉，这会把 $T$ 高估 $4^{1/4} \approx 1.41$ 倍——把 $255\ \mathrm{K}$ 算成错误的 $360\ \mathrm{K}$。务必保留这个 $4$。
Greenhouse Gases and Molecular Resonance温室气体与分子共振 B.2 SL+HL
A vibrational mode of a $\mathrm{CO_2}$ molecule absorbs infrared at wavelength $\lambda = 15\ \mathrm{\mu m}$. Find the photon frequency and energy, and explain what "resonance" means here. Take $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$, $h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$.$\mathrm{CO_2}$ 分子的某振动模式在波长 $\lambda = 15\ \mathrm{\mu m}$ 处吸收红外。求光子频率与能量，并解释此处"共振"的含义。取 $c = 3.0 \times 10^{8}\ \mathrm{m\,s^{-1}}$、$h = 6.63 \times 10^{-34}\ \mathrm{J\,s}$。
Identify. Use $c = f \lambda$ for the frequency and $E = h f$ for the photon energy.
识别。用 $c = f \lambda$ 求频率，用 $E = h f$ 求光子能量。
Frequency. $f = c / \lambda = (3.0 \times 10^{8}) / (15 \times 10^{-6}) = 2.0 \times 10^{13}\ \mathrm{Hz}$.
频率。$f = c / \lambda = (3.0 \times 10^{8}) / (15 \times 10^{-6}) = 2.0 \times 10^{13}\ \mathrm{Hz}$。
Energy. $E = h f = (6.63 \times 10^{-34})(2.0 \times 10^{13}) \approx 1.3 \times 10^{-20}\ \mathrm{J}$ (about $0.083\ \mathrm{eV}$).
能量。$E = h f = (6.63 \times 10^{-34})(2.0 \times 10^{13}) \approx 1.3 \times 10^{-20}\ \mathrm{J}$（约 $0.083\ \mathrm{eV}$）。
Resonance. Absorption is strong because $h f$ exactly matches the energy gap $\Delta E_{\text{vib}}$ between vibrational states of the $\mathrm{CO_2}$ molecule. Photons of nearby but mismatched frequencies pass through; only resonant photons are efficiently absorbed.
共振。吸收之所以强，是因为 $h f$ 恰好匹配 $\mathrm{CO_2}$ 分子振动态之间的能隙 $\Delta E_{\text{vib}}$。频率邻近但不匹配的光子会穿过；只有共振光子被高效吸收。
Evaluate. Earth's thermal emission peaks near $10\ \mathrm{\mu m}$, squarely in the IR band where $\mathrm{CO_2}$ and $\mathrm{H_2O}$ absorb — which is exactly why these gases intercept so much of the outgoing radiation.
评估。地球热辐射峰值约在 $10\ \mathrm{\mu m}$，正落在 $\mathrm{CO_2}$ 与 $\mathrm{H_2O}$ 吸收的红外波段——这正是这些气体能拦截大量外逸辐射的原因。
Going deeper: vibrational modes and the dipole rule深入：振动模式与偶极规则
A molecule can absorb an IR photon only if the vibration it excites changes the molecule's electric dipole moment, because an oscillating dipole is what couples to the oscillating electric field of the light. Compare:
分子只有当被激发的振动会改变其电偶极矩时，才能吸收红外光子，因为正是振荡的偶极子与光的振荡电场耦合。比较：
- $\mathrm{N_2}$, $\mathrm{O_2}$: symmetric diatomics. Their single stretch keeps zero dipole at all times — IR-inactive, not greenhouse gases.
- $\mathrm{N_2}$、$\mathrm{O_2}$：对称双原子分子。它们唯一的伸缩始终保持零偶极——红外非活性，不是温室气体。
- $\mathrm{CO_2}$: linear and symmetric, yet its bending mode and asymmetric stretch do create a transient dipole — IR-active at those modes.
- $\mathrm{CO_2}$：线形且对称，但其弯曲模式与不对称伸缩确实产生瞬时偶极——在这些模式上红外活性。
- $\mathrm{H_2O}$, $\mathrm{CH_4}$, $\mathrm{N_2O}$: bent or low-symmetry; multiple IR-active modes, strong absorbers.
- $\mathrm{H_2O}$、$\mathrm{CH_4}$、$\mathrm{N_2O}$：弯曲或低对称；多个红外活性模式，强吸收体。
Molecule for molecule, $\mathrm{CH_4}$ and $\mathrm{N_2O}$ are far stronger absorbers than $\mathrm{CO_2}$, but their atmospheric concentrations are much lower, so $\mathrm{CO_2}$ still dominates the human-caused forcing.
就单个分子而言，$\mathrm{CH_4}$ 与 $\mathrm{N_2O}$ 是远强于 $\mathrm{CO_2}$ 的吸收体，但它们的大气浓度低得多，故 $\mathrm{CO_2}$ 仍主导人为强迫。
The Enhanced Greenhouse Effect增强温室效应 B.2 SL+HL
Model Earth as a grey body. With emissivity $e_{1} = 0.62$ the equilibrium temperature is $288\ \mathrm{K}$. A rise in greenhouse-gas concentration lowers the effective emissivity to $e_{2} = 0.60$. Estimate the new equilibrium temperature (hold $\alpha$ and $S$ fixed).把地球建模为灰体。发射率 $e_{1} = 0.62$ 时平衡温度为 $288\ \mathrm{K}$。温室气体浓度上升使有效发射率降到 $e_{2} = 0.60$。估计新的平衡温度（$\alpha$ 与 $S$ 保持不变）。
Identify. Since $T \propto e^{-1/4}$ when $\alpha$ and $S$ are fixed, take a ratio rather than recomputing from scratch.
识别。当 $\alpha$ 与 $S$ 固定时 $T \propto e^{-1/4}$，故取比值而非从头重算。
Set up the ratio.
列出比值。
$$ \frac{T_{2}}{T_{1}} = \left( \frac{e_{1}}{e_{2}} \right)^{1/4} = \left( \frac{0.62}{0.60} \right)^{1/4}. $$Execute. $0.62 / 0.60 = 1.0333$, and $1.0333^{1/4} \approx 1.0082$. So $T_{2} = 288 \times 1.0082 \approx 290.4\ \mathrm{K}$.
计算。$0.62 / 0.60 = 1.0333$，$1.0333^{1/4} \approx 1.0082$。故 $T_{2} = 288 \times 1.0082 \approx 290.4\ \mathrm{K}$。
Evaluate. A small drop in emissivity ($0.62 \to 0.60$) raises the equilibrium temperature by about $2.4\ \mathrm{K}$ — illustrating how a modest change in atmospheric composition produces a measurable, persistent warming. During the adjustment $P_{\text{out}} < P_{\text{in}}$ until the surface reaches the new $T_{2}$.
评估。发射率小幅下降（$0.62 \to 0.60$）使平衡温度升高约 $2.4\ \mathrm{K}$——说明大气成分的适度变化即可产生可测、持续的升温。调整期间 $P_{\text{out}} < P_{\text{in}}$，直到地表达到新的 $T_{2}$。
Going deeper: how the energy balance shifts and re-settles深入：能量平衡如何移动并重新稳定
Before the change, the planet sits in balance: $(1-\alpha)S\pi r^{2} = e_{1}\sigma 4\pi r^{2}T_{1}^{4}$. Now add greenhouse gas, lowering the effective emissivity to $e_{2} < e_{1}$ while $T$ is momentarily still $T_{1}$. The instantaneous outgoing power drops to
变化前行星处于平衡：$(1-\alpha)S\pi r^{2} = e_{1}\sigma 4\pi r^{2}T_{1}^{4}$。现在加入温室气体，把有效发射率降到 $e_{2} < e_{1}$，而 $T$ 瞬间仍为 $T_{1}$。此刻外逸功率降为
$$ P_{\text{out}} = e_{2}\,\sigma\,4\pi r^{2}\,T_{1}^{4} \;<\; P_{\text{in}}. $$There is now an energy imbalance $P_{\text{in}} - P_{\text{out}} > 0$. The surplus is absorbed, the surface warms, and because $P_{\text{out}} \propto T^{4}$, the outgoing power climbs steeply as $T$ rises. Warming continues until $P_{\text{out}}$ once again equals $P_{\text{in}}$, at the new equilibrium $T_{2} = T_{1}(e_{1}/e_{2})^{1/4} > T_{1}$.
此时存在能量失衡 $P_{\text{in}} - P_{\text{out}} > 0$。盈余被吸收，地表升温，又因 $P_{\text{out}} \propto T^{4}$，随 $T$ 上升外逸功率陡增。升温持续到 $P_{\text{out}}$ 再次等于 $P_{\text{in}}$，即新平衡 $T_{2} = T_{1}(e_{1}/e_{2})^{1/4} > T_{1}$。
This is the essential physics of the enhanced greenhouse effect: a forced reduction in the planet's emissivity to space, met by a compensating rise in temperature. Feedbacks (ice–albedo, water-vapour amplification) can make the real shift larger than this single-step estimate.
这就是增强温室效应的核心物理：行星向太空发射率被强迫降低，靠温度的补偿性升高来抵消。反馈机制（冰-反照率、水汽放大）可能使真实变化大于这一单步估计。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Power intercepted from the Sun uses the cross-section $\pi r^{2}$. The planet only blocks parallel rays over its silhouette.
- 从太阳截获的功率用横截面 $\pi r^{2}$。行星只在其轮廓范围内挡住平行光线。
- Power radiated uses the full surface $4 \pi r^{2}$. The whole warm sphere emits. This mismatch is the factor of $4$.
- 辐射功率用整个表面 $4 \pi r^{2}$。整个温暖球面都发射。这一不匹配即因子 $4$。
- Equilibrium temperature is $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$. Forgetting the $4$ over-predicts $T$ by $\approx 41\%$.
- 平衡温度为 $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$。漏掉 $4$ 会把 $T$ 高估约 $41\%$。
- Always work in kelvin. Stefan–Boltzmann uses absolute temperature; a celsius slip is an instant zero on the calculation.
- 始终用开尔文。斯特藩-玻尔兹曼用绝对温度；用摄氏度立刻丢掉该计算分。
- Albedo $\alpha$ acts on incoming sunlight (visible); absorbed fraction is $1 - \alpha$. Emissivity $e$ acts on outgoing radiation (IR).
- 反照率 $\alpha$ 作用于入射阳光（可见光）；被吸收比例为 $1 - \alpha$。发射率 $e$ 作用于外逸辐射（红外）。
- Higher $\alpha$ cools; higher $e$ also cools. They enter the balance on opposite sides — keep them straight.
- $\alpha$ 越高越凉；$e$ 越高也越凉。它们分别进入平衡的两边——别弄混。
- "Explain the greenhouse effect" wants the chain: visible sunlight transmitted to surface $\to$ surface re-emits IR $\to$ greenhouse gases resonantly absorb IR $\to$ re-emit downward $\to$ surface warms.
- "解释温室效应"要写出链条：可见阳光透射到地表 $\to$ 地表再辐射红外 $\to$ 温室气体共振吸收红外 $\to$ 向下再辐射 $\to$ 地表升温。
- "Why is $\mathrm{N_2}$ not a greenhouse gas?" Because its symmetric vibration produces no changing dipole moment, so it cannot absorb IR.
- "为何 $\mathrm{N_2}$ 不是温室气体？"因为其对称振动不产生变化的偶极矩，故不能吸收红外。
Flashcards闪卡
Unit B.2 Practice Quiz单元 B.2 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Apply $I = P/(4\pi d^{2})$ and the inverse-square law to compute intensity at any distance用 $I = P/(4\pi d^{2})$ 与平方反比定律求任意距离处的强度
- State the solar constant $S$ and find the power a planet intercepts via $S \pi r^{2}$说出太阳常数 $S$ 并用 $S \pi r^{2}$ 求行星截获的功率
- Explain why incoming power uses $\pi r^{2}$ but outgoing power uses $4 \pi r^{2}$解释入射功率为何用 $\pi r^{2}$、出射功率为何用 $4 \pi r^{2}$
- Define albedo $\alpha$ and compute the absorbed power $(1-\alpha) S \pi r^{2}$定义反照率 $\alpha$ 并计算吸收功率 $(1-\alpha) S \pi r^{2}$
- Distinguish a black body ($e = 1$) from a grey body ($e < 1$) and define emissivity区分黑体（$e = 1$）与灰体（$e < 1$）并定义发射率
- Apply Stefan–Boltzmann $P = e \sigma A T^{4}$ to a planet's full surface in kelvin对行星整个表面用开尔文应用斯特藩-玻尔兹曼 $P = e \sigma A T^{4}$
- Set up and solve the energy balance $(1-\alpha) S \pi r^{2} = e \sigma 4 \pi r^{2} T^{4}$列出并求解能量平衡 $(1-\alpha) S \pi r^{2} = e \sigma 4 \pi r^{2} T^{4}$
- Derive and use $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$ without dropping the $4$推导并使用 $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$，且不漏掉 $4$
- Name the four greenhouse gases ($\mathrm{CO_2}, \mathrm{CH_4}, \mathrm{H_2O}, \mathrm{N_2O}$)说出四种温室气体（$\mathrm{CO_2}, \mathrm{CH_4}, \mathrm{H_2O}, \mathrm{N_2O}$）
- Explain molecular resonance ($hf = \Delta E_{\text{vib}}$) and why symmetric diatomics are IR-inactive解释分子共振（$hf = \Delta E_{\text{vib}}$）以及对称双原子为何红外非活性
- Explain the greenhouse mechanism: visible in, IR out, gases absorb and re-emit IR downward解释温室机理：可见光进、红外出、气体吸收红外并向下再辐射
- Distinguish natural from enhanced greenhouse effect and describe the energy-balance shift区分自然与增强温室效应，并描述能量平衡的移动
IB Paper-Style PracticeIB 试卷风格练习
B.2 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_B2_*.html with the bilingual built-in pattern.
B.2 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_B2_*.html。