Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus E5.1 to E5.6考纲 E5.1 至 E5.6PHYSICS HL
Binding energy per nucleon peaks near iron. (a) why fusing two light nuclei releases energy; (b) the two conditions for fusion and the barrier each addresses.每核子结合能在铁附近达峰。(a) 为何两轻核聚变释放能量；(b) 聚变的两个条件及各自克服的障碍。
Light nuclei lie low on the binding-energy-per-nucleon curve, which rises steeply toward the iron peak. (M1)
Fusing two light nuclei moves the product up the curve, so it has a higher binding energy per nucleon and is more tightly bound than the reactants. (A1)
The product therefore has less mass than the reactants; this mass defect $\Delta m$ is released as energy via $E = \Delta m c^2$. (A1)
High temperature gives the nuclei enough kinetic energy to overcome the Coulomb barrier (the electrostatic repulsion between the positively charged nuclei). (A1)
High density provides a high collision rate, so that enough fusions occur per second despite each collision being unlikely to fuse. (A1)
轻核位于每核子结合能曲线的低处，该曲线向铁峰急升。(M1)
两轻核聚变使产物沿曲线上行，故其每核子结合能更高、比反应物结合得更紧。(A1)
因此产物质量小于反应物；该质量亏损 $\Delta m$ 经 $E = \Delta m c^2$ 以能量释放。(A1)
高温使核获得足够动能以克服库仑势垒（带正电的核之间的静电斥力）。(A1)
高密度提供高碰撞率，使尽管每次碰撞难以聚变，每秒仍发生足够多次聚变。(A1)
$^{2}\mathrm{H} + {}^{2}\mathrm{H} \to {}^{3}\mathrm{He} + \mathrm{n}$; $^{2}\mathrm{H} = 2.01410$, $^{3}\mathrm{He} = 3.01603$, $\mathrm{n} = 1.00867\ \mathrm{u}$. (a) mass defect; (b) energy in MeV; (c) in joules.$^{2}\mathrm{H} + {}^{2}\mathrm{H} \to {}^{3}\mathrm{He} + \mathrm{n}$；$^{2}\mathrm{H} = 2.01410$、$^{3}\mathrm{He} = 3.01603$、$\mathrm{n} = 1.00867\ \mathrm{u}$。(a) 质量亏损；(b) 能量（MeV）；(c) 焦耳。
Reactant mass: $2(2.01410) = 4.02820\ \mathrm{u}$. Product mass: $3.01603 + 1.00867 = 4.02470\ \mathrm{u}$. (M1)
$$ \Delta m = 4.02820 - 4.02470 = 0.00350\ \mathrm{u}. $$(A1)
Convert the mass defect with $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$: (M1)
$$ E = (0.00350)(931.5) \approx 3.26\ \mathrm{MeV}. $$(A1)
(A1)
反应物质量：$2(2.01410) = 4.02820\ \mathrm{u}$。产物质量：$3.01603 + 1.00867 = 4.02470\ \mathrm{u}$。(M1)
$$ \Delta m = 4.02820 - 4.02470 = 0.00350\ \mathrm{u}. $$(A1)
用 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$ 换算质量亏损：(M1)
$$ E = (0.00350)(931.5) \approx 3.26\ \mathrm{MeV}. $$(A1)
(A1)
Sun powered by the p-p chain; $L_\odot = 3.85\times 10^{26}\ \mathrm{W}$, $c = 3.00\times 10^{8}$. (a) net p-p reaction; (b) why the energy source is nuclear not chemical; (c) rate of mass-to-energy conversion.太阳由 p-p 链供能；$L_\odot = 3.85\times 10^{26}\ \mathrm{W}$、$c = 3.00\times 10^{8}$。(a) 净 p-p 反应；(b) 为何能源是核而非化学；(c) 质量转能率。
Over the whole chain, four protons become one helium-4 nucleus, emitting two positrons and two electron neutrinos: (A1 reactants/product)
$$ 4\,{}^{1}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + 2e^{+} + 2\nu_e. $$(A1 leptons correct and balanced)
The energy comes from fusing atomic nuclei: a mass defect is converted to energy via $E = \Delta m c^2$. (A1)
Chemical reactions only rearrange electron bonds and release a millionth as much energy per reaction, far too little to power a star. (R1)
The radiated power equals the rest-mass energy converted per second: $L = \dfrac{dm}{dt}c^2$, so $\dfrac{dm}{dt} = \dfrac{L}{c^2}$. (M1)
$$ \frac{dm}{dt} = \frac{3.85\times 10^{26}}{(3.00\times 10^{8})^2} = \frac{3.85\times 10^{26}}{9.00\times 10^{16}}. $$(M1 substitution)
$$ \frac{dm}{dt} \approx 4.28\times 10^{9}\ \mathrm{kg\,s^{-1}}. $$(A1)
整条链中，四个质子变成一个氦-4 核，放出两个正电子与两个电子中微子：(反应物/产物 A1)
$$ 4\,{}^{1}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + 2e^{+} + 2\nu_e. $$(轻子正确且守恒 A1)
能量来自原子核聚变：质量亏损经 $E = \Delta m c^2$ 转为能量。(A1)
化学反应只重组电子键，每次反应释放的能量约为百万分之一，远不足以为恒星供能。(R1)
辐射功率等于每秒转换的静质能：$L = \dfrac{dm}{dt}c^2$，故 $\dfrac{dm}{dt} = \dfrac{L}{c^2}$。(M1)
$$ \frac{dm}{dt} = \frac{3.85\times 10^{26}}{(3.00\times 10^{8})^2} = \frac{3.85\times 10^{26}}{9.00\times 10^{16}}. $$(代入 M1)
$$ \frac{dm}{dt} \approx 4.28\times 10^{9}\ \mathrm{kg\,s^{-1}}. $$(A1)
A main-sequence star holds a steady size. (a) name the two opposing agents and directions; (b) state hydrostatic equilibrium; (c) why a small contraction is self-correcting.主序星保持稳定尺寸。(a) 两个对抗主体及方向；(b) 流体静力平衡的含义；(c) 为何微小收缩能自我纠正。
Inward: gravitation, pulling all the mass toward the centre. (A1)
Outward: pressure from the hot core, namely radiation pressure together with gas (thermal) pressure. (A1)
At every layer of the star these two agents are equal in magnitude, so the net force on each layer is zero. (A1)
The star therefore neither collapses nor expands and holds a steady size; this balance is hydrostatic equilibrium. (A1)
If the core contracts slightly, it is compressed and heats up; the fusion rate is strongly temperature-dependent, so it rises, and the pressure rises with it. (M1)
The increased outward pressure exceeds gravity and pushes the layers back out, reversing the contraction: a negative-feedback loop that restores equilibrium. (R1)
向内：引力，把全部质量拉向中心。(A1)
向外：来自高温核心的压力，即辐射压连同气体（热）压。(A1)
在恒星的每一层，这两个主体大小相等，故每一层所受合力为零。(A1)
因此恒星既不坍缩也不膨胀，保持稳定尺寸；这一平衡即流体静力平衡。(A1)
若核心略微收缩，被压缩而升温；聚变率强烈依赖温度故上升，压力随之上升。(M1)
增大的向外压力超过引力，把各层推回外，逆转收缩：这是恢复平衡的负反馈回路。(R1)
Star A: twice the radius and three times the surface temperature of star B. (a) Stefan-Boltzmann law for a sphere; (b) ratio $L_A/L_B$; (c) which is more luminous and what dominates.A 星：半径为 B 星两倍、表面温度为其三倍。(a) 球体的斯特藩-玻尔兹曼定律；(b) 比值 $L_A/L_B$；(c) 哪颗更亮、何者主导。
With surface area $A = 4\pi R^2$: $L = \sigma A T^4 = 4\pi R^2 \sigma T^4$. (A1)
Forming the ratio, the constants $4\pi$ and $\sigma$ cancel: (M1)
$$ \frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4. $$Substitute $R_A/R_B = 2$ and $T_A/T_B = 3$: (M1)
$$ \frac{L_A}{L_B} = 2^2 \times 3^4 = 4 \times 81 = 324. $$(A1)
Star A is the more luminous (by a factor of $324$); the temperature term ($3^4 = 81$) dominates over the radius term ($2^2 = 4$). (A1)
表面积 $A = 4\pi R^2$：$L = \sigma A T^4 = 4\pi R^2 \sigma T^4$。(A1)
作比值时常数 $4\pi$ 与 $\sigma$ 相消：(M1)
$$ \frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4. $$代入 $R_A/R_B = 2$、$T_A/T_B = 3$：(M1)
$$ \frac{L_A}{L_B} = 2^2 \times 3^4 = 4 \times 81 = 324. $$(A1)
A 星更亮（$324$ 倍）；温度项（$3^4 = 81$）主导，超过半径项（$2^2 = 4$）。(A1)
$b$ vs $1/d^2$ data given. (a) show the plot is linear through the origin and state the gradient; (b) gradient and luminosity $L$; (c) brightness of an identical star three times farther; (d) percentage uncertainty at the last point.给出 $b$ 对 $1/d^2$ 的数据。(a) 证明该图为过原点直线并说明斜率；(b) 斜率与光度 $L$；(c) 相同恒星远三倍时的视亮度；(d) 末点的百分比不确定度。
From the data booklet $b = \dfrac{L}{4\pi d^2}$. For a fixed luminosity $L$, rewrite with $1/d^2$ as the variable: (M1)
$$ b = \frac{L}{4\pi}\cdot\frac{1}{d^2}. $$This has the form $b = (\text{gradient})\times (1/d^2)$ with no intercept, so a graph of $b$ against $1/d^2$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\dfrac{L}{4\pi}$. (A1)
Read the gradient from two well-separated points, $(1.0\times 10^{-34},\,2.0\times 10^{-9})$ and $(4.0\times 10^{-34},\,8.0\times 10^{-9})$: (M1)
$$ \text{gradient} = \frac{(8.0 - 2.0)\times 10^{-9}}{(4.0 - 1.0)\times 10^{-34}} = \frac{6.0\times 10^{-9}}{3.0\times 10^{-34}} = 2.0\times 10^{25}\ \mathrm{W}. $$(A1)
Since the gradient equals $\dfrac{L}{4\pi}$: $L = 4\pi(2.0\times 10^{25}) \approx 2.5\times 10^{26}\ \mathrm{W}$. (A1)
Apparent brightness obeys $b \propto 1/d^2$ at fixed $L$. Tripling the distance multiplies $b$ by $1/3^2$. (M1)
So the second star appears $\dfrac{1}{9}$ as bright as the first. (A1)
At $1/d^2 = 4.0\times 10^{-34}\ \mathrm{m^{-2}}$, $b = 8.0\times 10^{-9}\ \mathrm{W\,m^{-2}}$ with absolute uncertainty $\pm 0.2\times 10^{-9}\ \mathrm{W\,m^{-2}}$: (M1)
$$ \frac{0.2\times 10^{-9}}{8.0\times 10^{-9}}\times 100\% = 2.5\%. $$(A1)
数据手册 $b = \dfrac{L}{4\pi d^2}$。对固定光度 $L$，以 $1/d^2$ 为变量改写：(M1)
$$ b = \frac{L}{4\pi}\cdot\frac{1}{d^2}. $$此式形如 $b = (\text{斜率})\times (1/d^2)$，无截距，故 $b$ 对 $1/d^2$ 作图为过原点直线。(A1)
与 $y = mx$ 比较，斜率为 $\dfrac{L}{4\pi}$。(A1)
用相距较远的两点读斜率，$(1.0\times 10^{-34},\,2.0\times 10^{-9})$ 与 $(4.0\times 10^{-34},\,8.0\times 10^{-9})$：(M1)
$$ \text{斜率} = \frac{(8.0 - 2.0)\times 10^{-9}}{(4.0 - 1.0)\times 10^{-34}} = \frac{6.0\times 10^{-9}}{3.0\times 10^{-34}} = 2.0\times 10^{25}\ \mathrm{W}. $$(A1)
因斜率等于 $\dfrac{L}{4\pi}$：$L = 4\pi(2.0\times 10^{25}) \approx 2.5\times 10^{26}\ \mathrm{W}$。(A1)
固定 $L$ 时视亮度满足 $b \propto 1/d^2$。距离变三倍使 $b$ 乘以 $1/3^2$。(M1)
故第二颗恒星的视亮度为第一颗的 $\dfrac{1}{9}$。(A1)
在 $1/d^2 = 4.0\times 10^{-34}\ \mathrm{m^{-2}}$ 处，$b = 8.0\times 10^{-9}\ \mathrm{W\,m^{-2}}$，绝对不确定度 $\pm 0.2\times 10^{-9}\ \mathrm{W\,m^{-2}}$：(M1)
$$ \frac{0.2\times 10^{-9}}{8.0\times 10^{-9}}\times 100\% = 2.5\%. $$(A1)
Star peaks at $\lambda_{\max} = 5.0\times 10^{-7}\ \mathrm{m}$; Wien's law $\lambda_{\max}T = 2.9\times 10^{-3}$. HR diagram: $T$ left, $L$ up. (a) surface temperature; (b) second star peaks at $2.9\times 10^{-7}\ \mathrm{m}$, which is hotter/bluer; (c) where the main sequence sits and what mass determines; (d) why a same-$T$ red giant has larger radius; (e) end states of low- and high-mass stars.恒星峰值在 $\lambda_{\max} = 5.0\times 10^{-7}\ \mathrm{m}$；维恩定律 $\lambda_{\max}T = 2.9\times 10^{-3}$。HR 图：$T$ 向左、$L$ 向上。(a) 表面温度；(b) 第二颗峰值在 $2.9\times 10^{-7}\ \mathrm{m}$，哪颗更热/更蓝；(c) 主序的位置与质量决定什么；(d) 相同 $T$ 的红巨星为何半径更大；(e) 低、大质量恒星的归宿。
Rearrange Wien's law $\lambda_{\max}T = 2.9\times 10^{-3}$: (M1)
$$ T = \frac{2.9\times 10^{-3}}{5.0\times 10^{-7}} = 5800\ \mathrm{K}. $$(A1)
The second star: $T = \dfrac{2.9\times 10^{-3}}{2.9\times 10^{-7}} = 1.0\times 10^{4}\ \mathrm{K}$, higher than $5800\ \mathrm{K}$. (A1)
A shorter peak wavelength means a higher temperature (Wien's law is inverse), and a peak shifted toward the blue end means the star appears bluer. (R1)
The main sequence is a diagonal band running from hot, bright, blue stars at the top-left to cool, faint, red stars at the bottom-right. (A1)
It is where stars spend most of their lives fusing hydrogen to helium. (A1)
Position along it is set by mass: massive stars sit at the top-left (hot and luminous), low-mass stars at the bottom-right. (A1)
From $L = 4\pi R^2 \sigma T^4$, at fixed $T$ the luminosity depends only on $R^2$. (M1)
The red giant lies far higher on the diagram, so its luminosity is much greater than the first star's despite the same surface temperature. (R1)
Since $T$ is the same, the only way to be much more luminous is to have a much larger radius $R$. (A1)
A low-mass star (like the Sun) becomes a red giant and then a white dwarf. (A1)
A high-mass star ends in a supernova, leaving a neutron star or a black hole. (A1)
变形维恩定律 $\lambda_{\max}T = 2.9\times 10^{-3}$：(M1)
$$ T = \frac{2.9\times 10^{-3}}{5.0\times 10^{-7}} = 5800\ \mathrm{K}. $$(A1)
第二颗：$T = \dfrac{2.9\times 10^{-3}}{2.9\times 10^{-7}} = 1.0\times 10^{4}\ \mathrm{K}$，高于 $5800\ \mathrm{K}$。(A1)
峰值波长更短意味着温度更高（维恩定律为反比），峰值偏向蓝端意味着恒星看起来更蓝。(R1)
主序是一条对角带，从左上又热又亮的蓝星延伸到右下又冷又暗的红星。(A1)
这是恒星一生大部分时间把氢聚变为氦之处。(A1)
沿主序的位置由质量决定：大质量星在左上（又热又亮），低质量星在右下。(A1)
由 $L = 4\pi R^2 \sigma T^4$，$T$ 不变时光度只取决于 $R^2$。(M1)
红巨星在图上位置高得多，故尽管表面温度相同，其光度远大于第一颗恒星。(R1)
既然 $T$ 相同，要光度大得多，唯一途径是半径 $R$ 大得多。(A1)
低质量恒星（如太阳）变为红巨星再成白矮星。(A1)
大质量恒星以超新星告终，留下中子星或黑洞。(A1)
$L = 4.0\times 10^{27}\ \mathrm{W}$, $T = 6000\ \mathrm{K}$, $b = 8.0\times 10^{-10}\ \mathrm{W\,m^{-2}}$, $\sigma = 5.67\times 10^{-8}$, $1\ \mathrm{AU} = 1.5\times 10^{11}\ \mathrm{m}$. (a) radius from Stefan-Boltzmann; (b) distance from $b$; (c) distance in AU; (d) factor change in $L$ if $T$ doubles; (e) distinguish $L$ and $b$.$L = 4.0\times 10^{27}\ \mathrm{W}$、$T = 6000\ \mathrm{K}$、$b = 8.0\times 10^{-10}\ \mathrm{W\,m^{-2}}$、$\sigma = 5.67\times 10^{-8}$、$1\ \mathrm{AU} = 1.5\times 10^{11}\ \mathrm{m}$。(a) 由斯特藩-玻尔兹曼求半径；(b) 由 $b$ 求距离；(c) 距离用 AU；(d) $T$ 翻倍时 $L$ 变化倍数；(e) 区分 $L$ 与 $b$。
Rearrange $L = 4\pi R^2 \sigma T^4$ for $R$: $R = \sqrt{\dfrac{L}{4\pi \sigma T^4}}$. (M1)
Fourth power of $T$: $(6000)^4 = 1.296\times 10^{15}\ \mathrm{K^4}$. (M1)
$$ R = \sqrt{\frac{4.0\times 10^{27}}{4\pi (5.67\times 10^{-8})(1.296\times 10^{15})}} = \sqrt{\frac{4.0\times 10^{27}}{9.23\times 10^{8}}}. $$(M1 substitution)
$$ R = \sqrt{4.33\times 10^{18}} \approx 2.1\times 10^{9}\ \mathrm{m}. $$(A1)
Rearrange $b = \dfrac{L}{4\pi d^2}$ for $d$: $d = \sqrt{\dfrac{L}{4\pi b}}$. (M1)
$$ d = \sqrt{\frac{4.0\times 10^{27}}{4\pi (8.0\times 10^{-10})}} = \sqrt{\frac{4.0\times 10^{27}}{1.005\times 10^{-8}}}. $$(M1 substitution)
$$ d = \sqrt{3.98\times 10^{35}} \approx 6.3\times 10^{17}\ \mathrm{m}. $$(A1)
Divide by $1\ \mathrm{AU} = 1.5\times 10^{11}\ \mathrm{m}$: (M1)
$$ d = \frac{6.3\times 10^{17}}{1.5\times 10^{11}} \approx 4.2\times 10^{6}\ \mathrm{AU}. $$(A1)
At fixed radius $L \propto T^4$, so doubling $T$ multiplies $L$ by $2^4 = 16$. (A1)
The factor is $16$ because the temperature appears to the fourth power in the Stefan-Boltzmann law. (R1)
Luminosity $L$ is the intrinsic total power the star radiates (in W); apparent brightness $b$ is the power received per unit area at Earth (in $\mathrm{W\,m^{-2}}$) and depends on distance. (A1)
把 $L = 4\pi R^2 \sigma T^4$ 解出 $R$：$R = \sqrt{\dfrac{L}{4\pi \sigma T^4}}$。(M1)
$T$ 的四次方：$(6000)^4 = 1.296\times 10^{15}\ \mathrm{K^4}$。(M1)
$$ R = \sqrt{\frac{4.0\times 10^{27}}{4\pi (5.67\times 10^{-8})(1.296\times 10^{15})}} = \sqrt{\frac{4.0\times 10^{27}}{9.23\times 10^{8}}}. $$(代入 M1)
$$ R = \sqrt{4.33\times 10^{18}} \approx 2.1\times 10^{9}\ \mathrm{m}. $$(A1)
把 $b = \dfrac{L}{4\pi d^2}$ 解出 $d$：$d = \sqrt{\dfrac{L}{4\pi b}}$。(M1)
$$ d = \sqrt{\frac{4.0\times 10^{27}}{4\pi (8.0\times 10^{-10})}} = \sqrt{\frac{4.0\times 10^{27}}{1.005\times 10^{-8}}}. $$(代入 M1)
$$ d = \sqrt{3.98\times 10^{35}} \approx 6.3\times 10^{17}\ \mathrm{m}. $$(A1)
除以 $1\ \mathrm{AU} = 1.5\times 10^{11}\ \mathrm{m}$：(M1)
$$ d = \frac{6.3\times 10^{17}}{1.5\times 10^{11}} \approx 4.2\times 10^{6}\ \mathrm{AU}. $$(A1)
半径不变时 $L \propto T^4$，故 $T$ 翻倍使 $L$ 变 $2^4 = 16$ 倍。(A1)
倍数为 $16$，因为温度在斯特藩-玻尔兹曼定律中以四次方出现。(R1)
光度 $L$ 是恒星辐射的内禀总功率（W）；视亮度 $b$ 是地球处单位面积接收的功率（$\mathrm{W\,m^{-2}}$），且依赖距离。(A1)
Sun $L_\odot = 3.85\times 10^{26}\ \mathrm{W}$, p-p chain gives $26.7\ \mathrm{MeV}$ per He-4; $c = 3.00\times 10^{8}$, $1\ \mathrm{MeV} = 1.60\times 10^{-13}\ \mathrm{J}$. (a) mass-to-energy rate; (b) energy per He-4 in J; (c) He-4 nuclei per second; (d) why heating energy is slightly less; (e) why high $T$ and high density are both needed.太阳 $L_\odot = 3.85\times 10^{26}\ \mathrm{W}$，p-p 链每个氦-4 给出 $26.7\ \mathrm{MeV}$；$c = 3.00\times 10^{8}$、$1\ \mathrm{MeV} = 1.60\times 10^{-13}\ \mathrm{J}$。(a) 质量转能率；(b) 每个氦-4 的能量（J）；(c) 每秒氦-4 核数；(d) 为何加热能量略小；(e) 为何同时需要高温与高密度。
$L = \dfrac{dm}{dt}c^2 \Rightarrow \dfrac{dm}{dt} = \dfrac{L}{c^2}$: (M1)
$$ \frac{dm}{dt} = \frac{3.85\times 10^{26}}{9.00\times 10^{16}} \approx 4.28\times 10^{9}\ \mathrm{kg\,s^{-1}}. $$(A1)
Convert $26.7\ \mathrm{MeV}$ with $1\ \mathrm{MeV} = 1.60\times 10^{-13}\ \mathrm{J}$: (M1)
$$ E = (26.7)(1.60\times 10^{-13}) \approx 4.27\times 10^{-12}\ \mathrm{J}. $$(A1)
Each helium-4 releases $E \approx 4.27\times 10^{-12}\ \mathrm{J}$, and the Sun radiates $L_\odot$ joules per second, so the number per second is $L_\odot / E$. (M1)
$$ N = \frac{3.85\times 10^{26}}{4.27\times 10^{-12}}. $$(M1 substitution)
$$ N \approx 9.0\times 10^{37}\ \mathrm{s^{-1}}. $$(A1)
Each chain emits neutrinos, which barely interact and escape the Sun directly, carrying away a small part of the $26.7\ \mathrm{MeV}$, so slightly less remains to heat the Sun. (R1)
High temperature gives the nuclei enough kinetic energy to overcome the Coulomb barrier and approach within range of the strong force. (R1)
High density gives a high collision rate, so that enough fusions occur per second to sustain the Sun's power output. (R1)
$L = \dfrac{dm}{dt}c^2 \Rightarrow \dfrac{dm}{dt} = \dfrac{L}{c^2}$：(M1)
$$ \frac{dm}{dt} = \frac{3.85\times 10^{26}}{9.00\times 10^{16}} \approx 4.28\times 10^{9}\ \mathrm{kg\,s^{-1}}. $$(A1)
用 $1\ \mathrm{MeV} = 1.60\times 10^{-13}\ \mathrm{J}$ 换算 $26.7\ \mathrm{MeV}$：(M1)
$$ E = (26.7)(1.60\times 10^{-13}) \approx 4.27\times 10^{-12}\ \mathrm{J}. $$(A1)
每个氦-4 释放 $E \approx 4.27\times 10^{-12}\ \mathrm{J}$，太阳每秒辐射 $L_\odot$ 焦耳，故每秒数目为 $L_\odot / E$。(M1)
$$ N = \frac{3.85\times 10^{26}}{4.27\times 10^{-12}}. $$(代入 M1)
$$ N \approx 9.0\times 10^{37}\ \mathrm{s^{-1}}. $$(A1)
每条链放出中微子，它们几乎不相互作用、直接逃离太阳，带走 $26.7\ \mathrm{MeV}$ 的一小部分，故留下加热太阳的能量略少。(R1)
高温使核获得足够动能以克服库仑势垒，接近到强核力作用范围内。(R1)
高密度提供高碰撞率，使每秒发生足够多次聚变以维持太阳的功率输出。(R1)
$M = 5.0\,M_\odot$; $L \propto M^{3.5}$, $t \propto M/L$; Sun's lifetime $\approx 1.0\times 10^{10}$ yr. (a) $L/L_\odot$; (b) show $t \propto M^{-2.5}$ and find $t/t_\odot$; (c) lifetime in years; (d) why more massive means shorter-lived despite more fuel.$M = 5.0\,M_\odot$；$L \propto M^{3.5}$、$t \propto M/L$；太阳寿命 $\approx 1.0\times 10^{10}$ 年。(a) $L/L_\odot$；(b) 证明 $t \propto M^{-2.5}$ 并求 $t/t_\odot$；(c) 寿命（年）；(d) 为何质量更大尽管燃料更多却寿命更短。
$\dfrac{L}{L_\odot} = \left(\dfrac{M}{M_\odot}\right)^{3.5} = (5.0)^{3.5}$. (M1)
$$ (5.0)^{3.5} = 5^3 \times 5^{0.5} = 125 \times 2.236 \approx 280. $$(A1)
Lifetime is fuel divided by burn rate: $t \propto \dfrac{M}{L} \propto \dfrac{M}{M^{3.5}} = M^{-2.5}$. (M1)
$$ \frac{t}{t_\odot} = (5.0)^{-2.5} = \frac{1}{5^{2.5}} = \frac{1}{55.9} \approx 1.8\times 10^{-2}. $$(A1)
Multiply the ratio by the Sun's lifetime: (M1)
$$ t \approx (1.8\times 10^{-2})(1.0\times 10^{10}) \approx 1.8\times 10^{8}\ \mathrm{yr}. $$(A1)
A more massive star starts with more fuel (fuel $\propto M$), which alone would lengthen its life. (R1)
But its luminosity rises far faster than its mass ($L \propto M^{3.5}$), so it burns the fuel disproportionately quickly; the net effect $t \propto M^{-2.5}$ is a much shorter life. (R1)
$\dfrac{L}{L_\odot} = \left(\dfrac{M}{M_\odot}\right)^{3.5} = (5.0)^{3.5}$。(M1)
$$ (5.0)^{3.5} = 5^3 \times 5^{0.5} = 125 \times 2.236 \approx 280. $$(A1)
寿命是燃料除以消耗率：$t \propto \dfrac{M}{L} \propto \dfrac{M}{M^{3.5}} = M^{-2.5}$。(M1)
$$ \frac{t}{t_\odot} = (5.0)^{-2.5} = \frac{1}{5^{2.5}} = \frac{1}{55.9} \approx 1.8\times 10^{-2}. $$(A1)
把比值乘以太阳寿命：(M1)
$$ t \approx (1.8\times 10^{-2})(1.0\times 10^{10}) \approx 1.8\times 10^{8}\ \mathrm{yr}. $$(A1)
质量更大的恒星起初燃料更多（燃料 $\propto M$），仅此会延长寿命。(R1)
但其光度上升远快于质量（$L \propto M^{3.5}$），故消耗燃料快得不成比例；净效应 $t \propto M^{-2.5}$ 是寿命短得多。(R1)