Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus B2.1 to B2.6考纲 B2.1 至 B2.6PHYSICS HL
Sun luminosity $P = 3.85\times 10^{26}\ \mathrm{W}$; Jupiter at $d = 7.78\times 10^{11}\ \mathrm{m}$, radius $r = 6.99\times 10^{7}\ \mathrm{m}$. (a) intensity at Jupiter; (b) power Jupiter intercepts.太阳光度 $P = 3.85\times 10^{26}\ \mathrm{W}$；木星距 $d = 7.78\times 10^{11}\ \mathrm{m}$，半径 $r = 6.99\times 10^{7}\ \mathrm{m}$。(a) 木星处强度；(b) 木星截获功率。
The data-booklet relation for a point source is $I = \dfrac{P}{4 \pi d^{2}}$. (A1)
Substitute (M1):
$$ I = \frac{3.85\times 10^{26}}{4 \pi (7.78\times 10^{11})^{2}} = \frac{3.85\times 10^{26}}{7.61\times 10^{24}} \approx 51\ \mathrm{W\,m^{-2}}. $$(A1)
Jupiter intercepts parallel rays over its circular cross-section (silhouette), area $\pi r^{2}$, not the full surface. (M1)
$$ P_{\text{in}} = I \, \pi r^{2} = (50.6)\,\pi (6.99\times 10^{7})^{2}. $$(M1 for substitution)
$$ P_{\text{in}} = (50.6)(1.535\times 10^{16}) \approx 7.8\times 10^{17}\ \mathrm{W}. $$(A1)
点源的数据手册关系式为 $I = \dfrac{P}{4 \pi d^{2}}$。(A1)
代入（M1）：
$$ I = \frac{3.85\times 10^{26}}{4 \pi (7.78\times 10^{11})^{2}} = \frac{3.85\times 10^{26}}{7.61\times 10^{24}} \approx 51\ \mathrm{W\,m^{-2}}. $$(A1)
木星以其圆形横截面（轮廓）截获平行光线，面积为 $\pi r^{2}$，而非整个表面。(M1)
$$ P_{\text{in}} = I \, \pi r^{2} = (50.6)\,\pi (6.99\times 10^{7})^{2}. $$(代入得 M1)
$$ P_{\text{in}} = (50.6)(1.535\times 10^{16}) \approx 7.8\times 10^{17}\ \mathrm{W}. $$(A1)
$P_{\text{in}} = 2.0\times 10^{17}\ \mathrm{W}$, albedo $\alpha = 0.35$. (a) define albedo and find reflected power; (b) absorbed power and which one enters the energy balance.$P_{\text{in}} = 2.0\times 10^{17}\ \mathrm{W}$，反照率 $\alpha = 0.35$。(a) 定义反照率并求反射功率；(b) 吸收功率及哪个进入能量平衡。
Albedo is the ratio of the total power reflected (scattered back) to the total power incident: $\alpha = P_{\text{ref}} / P_{\text{in}}$, a dimensionless number between $0$ and $1$. (B1)
$$ P_{\text{ref}} = \alpha P_{\text{in}} = 0.35 \times 2.0\times 10^{17} = 7.0\times 10^{16}\ \mathrm{W}. $$(A1)
The absorbed fraction is $(1 - \alpha)$: (M1)
$$ P_{\text{abs}} = (1 - \alpha) P_{\text{in}} = 0.65 \times 2.0\times 10^{17} = 1.3\times 10^{17}\ \mathrm{W}. $$It is the absorbed power that enters the planetary energy balance; the reflected part never heats the planet. (A1)
反照率是反射（散射回去）总功率与入射总功率之比：$\alpha = P_{\text{ref}} / P_{\text{in}}$，是介于 $0$ 与 $1$ 之间的无量纲数。(B1)
$$ P_{\text{ref}} = \alpha P_{\text{in}} = 0.35 \times 2.0\times 10^{17} = 7.0\times 10^{16}\ \mathrm{W}. $$(A1)
被吸收比例为 $(1 - \alpha)$：(M1)
$$ P_{\text{abs}} = (1 - \alpha) P_{\text{in}} = 0.65 \times 2.0\times 10^{17} = 1.3\times 10^{17}\ \mathrm{W}. $$进入行星能量平衡的是吸收功率；被反射的部分根本不加热行星。(A1)
Grey planet $r = 3.4\times 10^{6}\ \mathrm{m}$, $T = 210\ \mathrm{K}$, $e = 0.90$. (a) define emissivity and give black-body value; (b) radiated power; (c) factor if $T$ doubles.灰体行星 $r = 3.4\times 10^{6}\ \mathrm{m}$，$T = 210\ \mathrm{K}$，$e = 0.90$。(a) 定义发射率并给黑体值；(b) 辐射功率；(c) $T$ 加倍时的倍数。
Emissivity is the ratio of the power radiated per unit area by a real (grey) body to that radiated by a black body at the same temperature, $e = P_{\text{real}} / P_{\text{bb}}$, with $0 \le e \le 1$. (B1)
A perfect black body has $e = 1$. (B1)
Use the grey-body Stefan-Boltzmann law over the full surface $A = 4 \pi r^{2}$: $P = e \sigma A T^{4}$. (M1)
$$ A = 4 \pi (3.4\times 10^{6})^{2} \approx 1.45\times 10^{14}\ \mathrm{m^{2}}, \qquad T^{4} = 210^{4} \approx 1.94\times 10^{9}\ \mathrm{K^{4}}. $$(M1 for substitution)
$$ P = (0.90)(5.67\times 10^{-8})(1.45\times 10^{14})(1.94\times 10^{9}) \approx 1.4\times 10^{16}\ \mathrm{W}. $$(A1)
Since $P \propto T^{4}$, doubling $T$ multiplies $P$ by $2^{4} = 16$. (B1)
发射率是真实（灰）体单位面积辐射功率与同温度黑体辐射功率之比，$e = P_{\text{real}} / P_{\text{bb}}$，且 $0 \le e \le 1$。(B1)
理想黑体 $e = 1$。(B1)
对整个表面 $A = 4 \pi r^{2}$ 使用灰体斯特藩-玻尔兹曼定律：$P = e \sigma A T^{4}$。(M1)
$$ A = 4 \pi (3.4\times 10^{6})^{2} \approx 1.45\times 10^{14}\ \mathrm{m^{2}}, \qquad T^{4} = 210^{4} \approx 1.94\times 10^{9}\ \mathrm{K^{4}}. $$(代入得 M1)
$$ P = (0.90)(5.67\times 10^{-8})(1.45\times 10^{14})(1.94\times 10^{9}) \approx 1.4\times 10^{16}\ \mathrm{W}. $$(A1)
因 $P \propto T^{4}$，$T$ 加倍使 $P$ 乘以 $2^{4} = 16$。(B1)
Airless planet: $S = 1.0\times 10^{3}\ \mathrm{W\,m^{-2}}$, $\alpha = 0.25$. (a) derive $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$ and explain the factor $4$; (b) black-body equilibrium temperature; (c) effect of raising albedo.无大气行星：$S = 1.0\times 10^{3}\ \mathrm{W\,m^{-2}}$，$\alpha = 0.25$。(a) 推导 $T = [(1-\alpha)S/(4e\sigma)]^{1/4}$ 并解释因子 $4$；(b) 黑体平衡温度；(c) 增大反照率的影响。
At equilibrium absorbed power equals radiated power. Absorbed: $(1 - \alpha) S \pi r^{2}$ (disc intercept, minus albedo). Radiated: $e \sigma 4 \pi r^{2} T^{4}$ (whole surface, grey body). (M1)
$$ (1 - \alpha) S \, \pi r^{2} = e \, \sigma \, 4 \pi r^{2} \, T^{4} \;\Rightarrow\; T = \left[ \frac{(1 - \alpha) S}{4 e \sigma} \right]^{1/4}. $$The $\pi r^{2}$ cancels. (A1) The factor of $4$ is the ratio of the radiating surface $4 \pi r^{2}$ to the intercepting cross-section $\pi r^{2}$. (R1)
Set $e = 1$: (M1)
$$ T = \left[ \frac{(1 - 0.25)(1.0\times 10^{3})}{4 (1)(5.67\times 10^{-8})} \right]^{1/4} = \left[ \frac{750}{2.268\times 10^{-7}} \right]^{1/4} = (3.31\times 10^{9})^{1/4}. $$ $$ T \approx 240\ \mathrm{K}. $$(A1)
$T \propto (1 - \alpha)^{1/4}$, so a larger $\alpha$ shrinks $(1 - \alpha)$, less power is absorbed, and the equilibrium temperature falls. (B1)
平衡时吸收功率等于辐射功率。吸收：$(1 - \alpha) S \pi r^{2}$（圆盘截获，扣除反照率）。辐射：$e \sigma 4 \pi r^{2} T^{4}$（整个表面，灰体）。(M1)
$$ (1 - \alpha) S \, \pi r^{2} = e \, \sigma \, 4 \pi r^{2} \, T^{4} \;\Rightarrow\; T = \left[ \frac{(1 - \alpha) S}{4 e \sigma} \right]^{1/4}. $$$\pi r^{2}$ 相消。(A1) 因子 $4$ 是辐射表面 $4 \pi r^{2}$ 与截获横截面 $\pi r^{2}$ 之比。(R1)
取 $e = 1$：(M1)
$$ T = \left[ \frac{(1 - 0.25)(1.0\times 10^{3})}{4 (1)(5.67\times 10^{-8})} \right]^{1/4} = \left[ \frac{750}{2.268\times 10^{-7}} \right]^{1/4} = (3.31\times 10^{9})^{1/4}. $$ $$ T \approx 240\ \mathrm{K}. $$(A1)
$T \propto (1 - \alpha)^{1/4}$，故 $\alpha$ 越大，$(1 - \alpha)$ 越小，吸收功率越少，平衡温度下降。(B1)
$\mathrm{CO_2}$ absorbs IR at $\lambda = 14\ \mathrm{\mu m}$; $c = 3.0\times 10^{8}$, $h = 6.63\times 10^{-34}$. (a) frequency and photon energy; (b) why strong absorption (resonance); (c) why $\mathrm{N_2}$ is not a greenhouse gas.$\mathrm{CO_2}$ 在 $\lambda = 14\ \mathrm{\mu m}$ 吸收红外；$c = 3.0\times 10^{8}$，$h = 6.63\times 10^{-34}$。(a) 频率与光子能量；(b) 为何强吸收（共振）；(c) 为何 $\mathrm{N_2}$ 不是温室气体。
From $c = f \lambda$: (M1)
$$ f = \frac{c}{\lambda} = \frac{3.0\times 10^{8}}{14\times 10^{-6}} \approx 2.1\times 10^{13}\ \mathrm{Hz}. $$(A1)
From $E = h f$:
$$ E = (6.63\times 10^{-34})(2.143\times 10^{13}) \approx 1.4\times 10^{-20}\ \mathrm{J}. $$(A1)
Absorption is strong because the photon energy $hf$ exactly matches the energy gap between two vibrational states of the molecule, $hf = \Delta E_{\text{vib}}$. (R1)
This is resonance: only photons whose frequency equals a natural vibrational frequency of the molecule are efficiently absorbed; mismatched frequencies pass through. (R1)
$\mathrm{N_2}$ is a symmetric diatomic whose stretch keeps zero dipole moment, so it has no IR-active vibrational mode and cannot absorb infrared. (B1)
由 $c = f \lambda$：(M1)
$$ f = \frac{c}{\lambda} = \frac{3.0\times 10^{8}}{14\times 10^{-6}} \approx 2.1\times 10^{13}\ \mathrm{Hz}. $$(A1)
由 $E = h f$：
$$ E = (6.63\times 10^{-34})(2.143\times 10^{13}) \approx 1.4\times 10^{-20}\ \mathrm{J}. $$(A1)
吸收之所以强，是因为光子能量 $hf$ 恰好匹配分子两个振动态之间的能隙，$hf = \Delta E_{\text{vib}}$。(R1)
这就是共振：只有频率等于分子某个固有振动频率的光子才被高效吸收；不匹配的频率会穿过。(R1)
$\mathrm{N_2}$ 是对称双原子分子，其伸缩始终保持零偶极矩，故没有红外活性振动模式，不能吸收红外。(B1)
Probe measures $I$ at distances $d$ (in $10^{11}\ \mathrm{m}$): $(1.0,3060)$, $(1.5,1360)$, $(2.0,766)$, $(3.0,340)$. (a) axes to linearise and meaning of gradient; (b) gradient from end points; (c) Sun's luminosity; (d) percentage uncertainty in $I$ at $d = 1.0\times 10^{11}\ \mathrm{m}$ ($\pm 60$).探测器在距离 $d$（以 $10^{11}\ \mathrm{m}$ 计）处测 $I$：$(1.0,3060)$、$(1.5,1360)$、$(2.0,766)$、$(3.0,340)$。(a) 线性化坐标轴与斜率含义；(b) 由端点求斜率；(c) 太阳光度；(d) $d = 1.0\times 10^{11}\ \mathrm{m}$ 处 $I$ 的百分比不确定度（$\pm 60$）。
Write $I = \dfrac{P}{4 \pi}\cdot \dfrac{1}{d^{2}}$. This has the form $I = (\text{gradient})\times \dfrac{1}{d^{2}}$. (M1)
So plot $I$ (vertical) against $\dfrac{1}{d^{2}}$ (horizontal); the data then lie on a straight line through the origin. (A1)
The gradient equals $\dfrac{P}{4 \pi}$. (A1)
At $d = 1.0\times 10^{11}\ \mathrm{m}$: $1/d^{2} = 1.0\times 10^{-22}\ \mathrm{m^{-2}}$, $I = 3060$. At $d = 3.0\times 10^{11}\ \mathrm{m}$: $1/d^{2} = 1.11\times 10^{-23}\ \mathrm{m^{-2}}$, $I = 340$. (M1 for both $1/d^{2}$ values) (M1 for the gradient setup)
$$ \text{gradient} = \frac{3060 - 340}{(1.0\times 10^{-22}) - (1.11\times 10^{-23})} = \frac{2720}{8.89\times 10^{-23}} \approx 3.06\times 10^{25}. $$(A1)
From gradient $= P/(4 \pi)$: (M1)
$$ P = 4 \pi \times (3.06\times 10^{25}) \approx 3.8\times 10^{26}\ \mathrm{W}. $$(A1)
At $d = 1.0\times 10^{11}\ \mathrm{m}$, $I = 3060\ \mathrm{W\,m^{-2}}$ with $\pm 60\ \mathrm{W\,m^{-2}}$: (M1)
$$ \frac{60}{3060}\times 100\% \approx 2.0\%. $$(A1)
写成 $I = \dfrac{P}{4 \pi}\cdot \dfrac{1}{d^{2}}$。此式形如 $I = (\text{斜率})\times \dfrac{1}{d^{2}}$。(M1)
故以 $I$（纵轴）对 $\dfrac{1}{d^{2}}$（横轴）作图；数据即落在过原点的直线上。(A1)
斜率等于 $\dfrac{P}{4 \pi}$。(A1)
$d = 1.0\times 10^{11}\ \mathrm{m}$ 处：$1/d^{2} = 1.0\times 10^{-22}\ \mathrm{m^{-2}}$，$I = 3060$。$d = 3.0\times 10^{11}\ \mathrm{m}$ 处：$1/d^{2} = 1.11\times 10^{-23}\ \mathrm{m^{-2}}$，$I = 340$。(两个 $1/d^{2}$ 值得 M1)（列出斜率式得 M1）
$$ \text{斜率} = \frac{3060 - 340}{(1.0\times 10^{-22}) - (1.11\times 10^{-23})} = \frac{2720}{8.89\times 10^{-23}} \approx 3.06\times 10^{25}. $$(A1)
由斜率 $= P/(4 \pi)$：(M1)
$$ P = 4 \pi \times (3.06\times 10^{25}) \approx 3.8\times 10^{26}\ \mathrm{W}. $$(A1)
$d = 1.0\times 10^{11}\ \mathrm{m}$ 处 $I = 3060\ \mathrm{W\,m^{-2}}$，不确定度 $\pm 60\ \mathrm{W\,m^{-2}}$：(M1)
$$ \frac{60}{3060}\times 100\% \approx 2.0\%. $$(A1)
Satellite measures incoming $S = 1.50\times 10^{3}\ \mathrm{W\,m^{-2}}$ and reflected $0.45\times 10^{3}\ \mathrm{W\,m^{-2}}$; planet is a black body in equilibrium. (a) albedo; (b) equilibrium temperature ($e=1$); (c) percentage uncertainty in $T$ given $4\%$ in $S$; (d) why a real surface temperature would differ.卫星测得入射 $S = 1.50\times 10^{3}\ \mathrm{W\,m^{-2}}$、反射 $0.45\times 10^{3}\ \mathrm{W\,m^{-2}}$；行星为平衡黑体。(a) 反照率；(b) 平衡温度（$e=1$）；(c) 由 $S$ 的 $4\%$ 求 $T$ 的百分比不确定度；(d) 真实地表温度为何不同。
Albedo is reflected over incident: (M1)
$$ \alpha = \frac{0.45\times 10^{3}}{1.50\times 10^{3}} = 0.30. $$(A1)
Black body, $e = 1$: $T = \left[ (1 - \alpha) S / (4 \sigma) \right]^{1/4}$. (M1)
$$ T = \left[ \frac{(0.70)(1.50\times 10^{3})}{4 (5.67\times 10^{-8})} \right]^{1/4} = \left[ \frac{1050}{2.268\times 10^{-7}} \right]^{1/4} = (4.63\times 10^{9})^{1/4}. $$(M1 for substitution)
$$ T \approx 261\ \mathrm{K}. $$(A1)
Since $T \propto S^{1/4}$, a power of $\tfrac{1}{4}$ scales the fractional uncertainty by $\tfrac{1}{4}$. (M1·R1)
$$ \frac{\Delta T}{T} = \frac{1}{4}\cdot\frac{\Delta S}{S} = \frac{1}{4}(4\%) = 1\%. $$(A1)
A real planet with an atmosphere has a greenhouse effect: greenhouse gases absorb outgoing infrared and re-emit part of it downward. (B1)
This traps energy, so the actual surface temperature is higher than the bare radiative-balance ("effective") temperature calculated here. (R1)
反照率为反射比入射：(M1)
$$ \alpha = \frac{0.45\times 10^{3}}{1.50\times 10^{3}} = 0.30. $$(A1)
黑体，$e = 1$：$T = \left[ (1 - \alpha) S / (4 \sigma) \right]^{1/4}$。(M1)
$$ T = \left[ \frac{(0.70)(1.50\times 10^{3})}{4 (5.67\times 10^{-8})} \right]^{1/4} = \left[ \frac{1050}{2.268\times 10^{-7}} \right]^{1/4} = (4.63\times 10^{9})^{1/4}. $$(代入得 M1)
$$ T \approx 261\ \mathrm{K}. $$(A1)
因 $T \propto S^{1/4}$，$\tfrac{1}{4}$ 次幂使分数不确定度乘以 $\tfrac{1}{4}$。(M1·R1)
$$ \frac{\Delta T}{T} = \frac{1}{4}\cdot\frac{\Delta S}{S} = \frac{1}{4}(4\%) = 1\%. $$(A1)
带大气的真实行星存在温室效应：温室气体吸收外逸红外并把一部分向下再辐射。(B1)
这会捕获能量，故实际地表温度高于此处算出的纯辐射平衡（"有效"）温度。(R1)
Earth: $S = 1.36\times 10^{3}\ \mathrm{W\,m^{-2}}$, $\alpha = 0.30$, $r = 6.37\times 10^{6}\ \mathrm{m}$. (a) intercepted power; (b) absorbed power; (c) black-body equilibrium temperature; (d) name the effect for the gap to $288\ \mathrm{K}$ and find the grey-body emissivity; (e) why $\pi r^{2}$ in, $4\pi r^{2}$ out.地球：$S = 1.36\times 10^{3}\ \mathrm{W\,m^{-2}}$，$\alpha = 0.30$，$r = 6.37\times 10^{6}\ \mathrm{m}$。(a) 截获功率；(b) 吸收功率；(c) 黑体平衡温度；(d) 命名到 $288\ \mathrm{K}$ 差距的效应并求灰体发射率；(e) 为何入射 $\pi r^{2}$、出射 $4\pi r^{2}$。
Earth intercepts over its cross-section $\pi r^{2}$: $P_{\text{in}} = S \pi r^{2}$. (M1)
$$ P_{\text{in}} = (1.36\times 10^{3})\,\pi (6.37\times 10^{6})^{2} \approx 1.7\times 10^{17}\ \mathrm{W}. $$(A1)
Absorb the fraction $(1 - \alpha)$: (M1)
$$ P_{\text{abs}} = (1 - 0.30)(1.73\times 10^{17}) \approx 1.2\times 10^{17}\ \mathrm{W}. $$(A1)
Set $P_{\text{abs}} = e \sigma 4 \pi r^{2} T^{4}$ with $e = 1$, equivalently $T = \left[ (1-\alpha)S/(4\sigma) \right]^{1/4}$. (M1)
$$ T = \left[ \frac{(0.70)(1.36\times 10^{3})}{4(5.67\times 10^{-8})} \right]^{1/4} = (4.20\times 10^{9})^{1/4}. $$(M1 for substitution)
$$ T \approx 255\ \mathrm{K}. $$(A1)
The model gives $255\ \mathrm{K}$ but the real surface is $288\ \mathrm{K}$, about $33\ \mathrm{K}$ warmer; this gap is the (natural) greenhouse effect. (B1)
Model Earth as a grey body that reaches $288\ \mathrm{K}$: $(1-\alpha)S = 4 e \sigma T^{4}$, so $e = \dfrac{(1-\alpha)S}{4 \sigma T^{4}}$. (M1)
$$ e = \frac{(0.70)(1.36\times 10^{3})}{4(5.67\times 10^{-8})(288)^{4}} = \frac{952}{4(5.67\times 10^{-8})(6.88\times 10^{9})}. $$(M1 for substitution)
$$ e = \frac{952}{1560} \approx 0.61. $$(A1)
Incoming sunlight is a parallel beam, intercepted only over the planet's silhouette $\pi r^{2}$; the warm planet radiates from its entire surface $4 \pi r^{2}$. (B1)
地球以其横截面 $\pi r^{2}$ 截获：$P_{\text{in}} = S \pi r^{2}$。(M1)
$$ P_{\text{in}} = (1.36\times 10^{3})\,\pi (6.37\times 10^{6})^{2} \approx 1.7\times 10^{17}\ \mathrm{W}. $$(A1)
吸收比例 $(1 - \alpha)$：(M1)
$$ P_{\text{abs}} = (1 - 0.30)(1.73\times 10^{17}) \approx 1.2\times 10^{17}\ \mathrm{W}. $$(A1)
令 $P_{\text{abs}} = e \sigma 4 \pi r^{2} T^{4}$，取 $e = 1$，等价于 $T = \left[ (1-\alpha)S/(4\sigma) \right]^{1/4}$。(M1)
$$ T = \left[ \frac{(0.70)(1.36\times 10^{3})}{4(5.67\times 10^{-8})} \right]^{1/4} = (4.20\times 10^{9})^{1/4}. $$(代入得 M1)
$$ T \approx 255\ \mathrm{K}. $$(A1)
模型给出 $255\ \mathrm{K}$，而真实地表为 $288\ \mathrm{K}$，约高 $33\ \mathrm{K}$；该差距即（自然）温室效应。(B1)
把地球建模为达到 $288\ \mathrm{K}$ 的灰体：$(1-\alpha)S = 4 e \sigma T^{4}$，故 $e = \dfrac{(1-\alpha)S}{4 \sigma T^{4}}$。(M1)
$$ e = \frac{(0.70)(1.36\times 10^{3})}{4(5.67\times 10^{-8})(288)^{4}} = \frac{952}{4(5.67\times 10^{-8})(6.88\times 10^{9})}. $$(代入得 M1)
$$ e = \frac{952}{1560} \approx 0.61. $$(A1)
入射阳光是平行光束，只在行星轮廓 $\pi r^{2}$ 范围内被截获；温暖的行星从整个表面 $4 \pi r^{2}$ 辐射。(B1)
$\mathrm{CH_4}$ absorbs IR at $\lambda = 7.7\ \mathrm{\mu m}$; $c = 3.0\times 10^{8}$, $h = 6.63\times 10^{-34}$. (a) photon energy; (b) why greenhouse gases warm the surface (Sun's vs Earth's spectrum); (c) natural vs enhanced effect; (d) new equilibrium $T$ if $e$ falls $0.61\to 0.58$ from $288\ \mathrm{K}$ using $T \propto e^{-1/4}$.$\mathrm{CH_4}$ 在 $\lambda = 7.7\ \mathrm{\mu m}$ 吸收红外；$c = 3.0\times 10^{8}$，$h = 6.63\times 10^{-34}$。(a) 光子能量；(b) 温室气体为何使地表变暖（太阳光谱与地球光谱）；(c) 自然与增强效应；(d) 由 $288\ \mathrm{K}$、用 $T \propto e^{-1/4}$ 求 $e$ 从 $0.61\to 0.58$ 时的新平衡温度。
Use $E = hf = hc/\lambda$: (M1)
$$ E = \frac{(6.63\times 10^{-34})(3.0\times 10^{8})}{7.7\times 10^{-6}} \approx 2.6\times 10^{-20}\ \mathrm{J}. $$(A1)
The Sun's radiation peaks in the visible; greenhouse gases are largely transparent to it, so sunlight passes through and is absorbed at the surface. (M1)
The warmed surface re-radiates in the infrared (its spectrum peaks in the IR because it is much cooler than the Sun). (M1)
Greenhouse gases resonantly absorb this outgoing IR, because the IR photon energies match their vibrational energy gaps. (R1)
They then re-emit in all directions, returning part of the energy to the surface, so the surface settles at a higher temperature than it would without the atmosphere. (R1)
The natural greenhouse effect is the baseline warming (about $33\ \mathrm{K}$ for Earth) that the existing atmosphere provides, making the planet habitable. (B1)
The enhanced greenhouse effect is the additional warming caused by human activity raising greenhouse-gas concentrations (mainly $\mathrm{CO_2}$, also $\mathrm{CH_4}$ and $\mathrm{N_2O}$). (B1)
With $\alpha$ and $S$ fixed, $T \propto e^{-1/4}$, so take a ratio: $\dfrac{T_{2}}{T_{1}} = \left( \dfrac{e_{1}}{e_{2}} \right)^{1/4}$. (M1)
$$ T_{2} = 288 \left( \frac{0.61}{0.58} \right)^{1/4} = 288 \,(1.0517)^{1/4}. $$(M1 for substitution)
$$ T_{2} = 288 \times 1.0127 \approx 292\ \mathrm{K}. $$(A1) The temperature rise is $T_{2} - T_{1} \approx 292 - 288 = 3.7\ \mathrm{K}$. (A1)
用 $E = hf = hc/\lambda$：(M1)
$$ E = \frac{(6.63\times 10^{-34})(3.0\times 10^{8})}{7.7\times 10^{-6}} \approx 2.6\times 10^{-20}\ \mathrm{J}. $$(A1)
太阳辐射峰值在可见光；温室气体对其大致透明，故阳光透过并被地表吸收。(M1)
被加热的地表以红外再辐射（因其温度远低于太阳，光谱峰值落在红外）。(M1)
温室气体共振吸收这些外逸红外，因为红外光子能量匹配其振动能隙。(R1)
随后它们向各方向再辐射，把一部分能量送回地表，故地表稳定在比无大气时更高的温度。(R1)
自然温室效应是现有大气提供的基线升温（地球约 $33\ \mathrm{K}$），使行星宜居。(B1)
增强温室效应是人类活动提高温室气体浓度（主要是 $\mathrm{CO_2}$，也有 $\mathrm{CH_4}$ 与 $\mathrm{N_2O}$）所致的额外升温。(B1)
$\alpha$ 与 $S$ 固定时 $T \propto e^{-1/4}$，故取比值：$\dfrac{T_{2}}{T_{1}} = \left( \dfrac{e_{1}}{e_{2}} \right)^{1/4}$。(M1)
$$ T_{2} = 288 \left( \frac{0.61}{0.58} \right)^{1/4} = 288 \,(1.0517)^{1/4}. $$(代入得 M1)
$$ T_{2} = 288 \times 1.0127 \approx 292\ \mathrm{K}. $$(A1) 升温为 $T_{2} - T_{1} \approx 292 - 288 = 3.7\ \mathrm{K}$。(A1)
A planet in equilibrium ($P_{\text{in}} = P_{\text{out}}$) gains greenhouse gas, lowering emissivity $e_{1} \to e_{2} < e_{1}$ while $T$ is momentarily $T_{1}$. (a) the instantaneous balance and warm-or-cool; (b) how a new equilibrium is reached using $P_{\text{out}} \propto T^{4}$, and is it higher or lower; (c) two observable consequences for Earth.处于平衡（$P_{\text{in}} = P_{\text{out}}$）的行星获得温室气体，使发射率 $e_{1} \to e_{2} < e_{1}$，而 $T$ 瞬间仍为 $T_{1}$。(a) 瞬时平衡与升降温；(b) 用 $P_{\text{out}} \propto T^{4}$ 说明如何达到新平衡及高低；(c) 对地球的两个可观测后果。
The outgoing power is $P_{\text{out}} = e \sigma 4 \pi r^{2} T^{4}$. Lowering $e$ to $e_{2}$ while $T = T_{1}$ reduces it below the unchanged absorbed power $P_{\text{in}}$. (M1)
$$ P_{\text{out}} = e_{2}\,\sigma\,4 \pi r^{2}\,T_{1}^{4} \;<\; e_{1}\,\sigma\,4 \pi r^{2}\,T_{1}^{4} = P_{\text{in}}. $$So $P_{\text{out}} < P_{\text{in}}$. (A1) There is a net energy gain, so the planet warms. (R1)
As the surface warms, the outgoing power rises steeply because $P_{\text{out}} \propto T^{4}$. (M1)
Warming continues until $P_{\text{out}}$ has climbed back to equal $P_{\text{in}}$, restoring balance. (R1)
Because a smaller $e$ needs a larger $T^{4}$ to deliver the same $P_{\text{out}}$, the new equilibrium temperature is higher than $T_{1}$. (A1)
Any two of: rising mean surface temperatures; melting ice caps and glaciers (with ice-albedo feedback); thermal expansion of the oceans and sea-level rise; shifts in climate and weather patterns. (B1·B1)
外逸功率为 $P_{\text{out}} = e \sigma 4 \pi r^{2} T^{4}$。在 $T = T_{1}$ 时把 $e$ 降到 $e_{2}$，使其低于不变的吸收功率 $P_{\text{in}}$。(M1)
$$ P_{\text{out}} = e_{2}\,\sigma\,4 \pi r^{2}\,T_{1}^{4} \;<\; e_{1}\,\sigma\,4 \pi r^{2}\,T_{1}^{4} = P_{\text{in}}. $$故 $P_{\text{out}} < P_{\text{in}}$。(A1) 存在净能量获得，故行星升温。(R1)
地表升温时，因 $P_{\text{out}} \propto T^{4}$，外逸功率陡增。(M1)
升温持续到 $P_{\text{out}}$ 回升至等于 $P_{\text{in}}$，恢复平衡。(R1)
由于较小的 $e$ 需要较大的 $T^{4}$ 才能给出相同的 $P_{\text{out}}$，新平衡温度高于 $T_{1}$。(A1)
以下任取两点：平均地表温度上升；冰盖与冰川融化（伴随冰-反照率反馈）；海洋热膨胀与海平面上升；气候与天气格局改变。(B1·B1)