Homework 7 Solutions
Stellar Structure I — Lifetimes, Equilibrium, and Fusion
Student note: These are model solutions written to show reasoning, units, and checks. Your own work can be shorter if your setup, logic, and interpretation are clear, but you should still show the reasoning steps that make your answer trustworthy.
Grade memo note: Use these solutions to identify what you understood, what broke, and what you will do differently next time.
How to study from this page: Before reading each final answer, stop at the equation or key insight and predict the next step yourself. Then check three things: Did I choose the right physical model? Did my units behave honestly? Does my final result make physical sense?
Part A — Stellar Clocks and Main-Sequence Lifetimes
Problem 1 — Fuel Is Not Lifetime
Restatement: Explain why more stellar mass does not automatically mean a longer main-sequence lifetime.
Key insight: A main-sequence lifetime is set by fuel divided by burn rate. More massive stars do have more hydrogen available, but their luminosities rise so steeply that they spend that fuel much faster.
Answer:
For part (a), the student’s statement is incomplete because it pays attention only to the fuel supply. A star’s lifetime depends on both how much fuel it has and how quickly it uses that fuel. More massive main-sequence stars have more hydrogen, but they also shine much more intensely, so they consume their available fuel at a much faster rate.
For part (b), the term that grows faster with stellar mass is the burn rate. On the main sequence, the burn rate is reflected by the luminosity \(L\), and \(L\) increases more steeply with mass than the fuel supply does.
For part (c), the scaling \[ \tau_{\text{nuc}} \propto \frac{M}{L} \] shows the logic directly. If mass increases by some factor but luminosity increases by a much larger factor, then the ratio \(M/L\) becomes smaller. That means the nuclear lifetime becomes shorter, not longer. Massive stars die young because their extra fuel does not keep up with their much larger energy output.
Sanity check: If a star has only a few times more fuel but burns it at tens or hundreds of times the Sun’s rate, it must run out sooner. \(\checkmark\)
Common misconception: “More fuel” sounds like “longer lifetime,” but for stars the burn rate is just as important as the size of the fuel tank.
Problem 2 — Solar-Normalized Nuclear Lifetime
Given: \(M = 2\,M_\odot\), \(L = 10\,L_\odot\), and \(\tau_{\text{nuc},\odot} \approx 10~\text{Gyr}\).
Find: The ratio \(\tau_{\text{nuc}}/\tau_{\text{nuc},\odot}\) and the star’s main-sequence lifetime.
Equation: \[ \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx \left(\frac{M}{M_\odot}\right) \left(\frac{L}{L_\odot}\right)^{-1} \]
Assumption: This scaling is for similar hydrogen-burning main-sequence stars, so comparing to the Sun is reasonable.
Steps:
For part (a), substitute the given mass and luminosity: \[ \frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx \left(\frac{2\,M_\odot}{M_\odot}\right) \left(\frac{10\,L_\odot}{L_\odot}\right)^{-1} \] \[ = 2 \times 10^{-1} = 0.20 \]
So the star’s lifetime is one-fifth of the Sun’s lifetime: \[ \boxed{\frac{\tau_{\text{nuc}}}{\tau_{\text{nuc},\odot}} \approx 0.20} \]
For part (b), multiply this ratio by the Sun’s nuclear lifetime: \[ \tau_{\text{nuc}} = \left(0.20\right)\left(10~\text{Gyr}\right) = 2.0~\text{Gyr} \]
\[ \boxed{\tau_{\text{nuc}} \approx 2.0~\text{Gyr}} \]
For part (c), the physical reason is that the star has only twice the Sun’s fuel supply in this rough picture, but it is radiating energy at ten times the Sun’s rate, so it burns through its fuel much faster.
Unit check: The ratios \(M/M_\odot\) and \(L/L_\odot\) are dimensionless, so \(\tau_{\text{nuc}}/\tau_{\text{nuc},\odot}\) is dimensionless. Multiplying by \(10~\text{Gyr}\) correctly gives a lifetime in Gyr. \(\checkmark\)
Sanity check: A star with \(10\times\) the Sun’s luminosity should not outlive the Sun if it has only \(2\times\) the fuel. A lifetime shorter than \(10~\text{Gyr}\) is exactly what we expect. \(\checkmark\)
Problem 3 — Turnoff as a Clock
Restatement: Use a \(10\,M_\odot\) main-sequence turnoff star to estimate the age of a cluster and explain the observable → model → inference chain.
Key elements: A complete answer must estimate the lifetime numerically, identify the observable, name the model used to convert that observable into an age, and explain why a massive turnoff implies youth.
Assumption: This method assumes the cluster is roughly coeval, so the turnoff mass really does trace the cluster age.
Sample response:
For part (a), start from the scaling \[ \tau_{\text{nuc}} = \tau_{\text{nuc},\odot} \left(\frac{M}{M_\odot}\right)^{-2.5}. \]
Using \(\tau_{\text{nuc},\odot} \approx 10~\text{Gyr}\) and \(M = 10\,M_\odot\) gives \[ \tau_{\text{nuc}} = 10~\text{Gyr} \left(\frac{10\,M_\odot}{M_\odot}\right)^{-2.5} \] \[ = 10~\text{Gyr}\left(10^{-2.5}\right). \]
Now evaluate the power of ten: \[ 10^{-2.5} = 10^{-2}10^{-0.5} = \left(0.01\right)\left(0.316\right) \approx 3.16 \times 10^{-3}. \]
So \[ \tau_{\text{nuc}} \approx \left(10~\text{Gyr}\right)\left(3.16 \times 10^{-3}\right) = 3.16 \times 10^{-2}~\text{Gyr}. \]
Convert to Myr: \[ 3.16 \times 10^{-2}~\text{Gyr} \left( \frac{10^3~\text{Myr}}{1~\text{Gyr}} \right) = 31.6~\text{Myr}. \]
So the cluster age scale is about \[ \boxed{\tau_{\text{cluster}} \sim 3 \times 10^7~\text{yr} \approx 32~\text{Myr}}. \]
For part (b), the observable is the main-sequence turnoff: the hottest or most massive star that is still on the main sequence. The model is the main-sequence lifetime scaling, which tells us how lifetime depends on stellar mass. The inference is that the cluster age is roughly the lifetime of the turnoff star, because stars more massive than the turnoff have already left the main sequence.
For part (c), a hot B2 turnoff star means the cluster cannot be old because a \(10\,M_\odot\) star has a very short main-sequence lifetime. If the cluster were hundreds of Myr or billions of years old, stars that massive would already have evolved away from the main sequence.
Sanity check: A \(10\,M_\odot\) star should live for only tens of Myr, not for billions of years, so an age near \(32~\text{Myr}\) is in the right ballpark. \(\checkmark\)
Grading guidance: Full credit requires the numerical age scale, explicit observable/model/inference language, and a sentence explaining why a massive turnoff rules out old ages.
Part B — Hydrostatic Support and Core Conditions
Problem 4 — Pressure Does Not Hold a Star Up
Restatement: Explain why hydrostatic support requires a pressure gradient rather than pressure by itself.
Key insight: A shell feels a net pressure force only if the pressure below it differs from the pressure above it.
Answer:
For part (a), if \[ \frac{dP}{dr} = 0 \] everywhere, then the pressure is the same on both sides of each fluid layer. That means the inward and outward pressure forces cancel, leaving no net pressure force to oppose gravity. The star would therefore collapse inward.
For part (b), if \[ \frac{dP}{dr} > 0 \] at some radius, then pressure increases outward. In that case the pressure is larger on the outside of the shell than on the inside, so the net pressure force points inward. Instead of supporting the star, that pressure pattern would reinforce gravity.
For part (c), hydrostatic equilibrium requires \[ \frac{dP}{dr} < 0 \] because pressure must decrease outward. Then the pressure is larger below a shell than above it, producing an outward pressure-gradient force that can balance the inward pull of gravity.
Sanity check: The only way pressure can push a shell outward is if the pressure below the shell is larger than the pressure above it. \(\checkmark\)
Common misconception: “High pressure” sounds supportive, but support comes from the difference in pressure across a layer, not from the absolute pressure value alone.
Problem 5 — Central Pressure from Hydrostatic Scaling
Given: \[ P_c \sim \frac{G M^2}{R^4}, \qquad G = 6.674 \times 10^{-8}~\text{cm}^3~\text{g}^{-1}~\text{s}^{-2}, \] \[ M_\odot = 2.0 \times 10^{33}~\text{g}, \qquad R_\odot = 7.0 \times 10^{10}~\text{cm}, \qquad 1~\text{atm} = 1.013 \times 10^6~\text{dyn/cm}^2. \]
Find: The pressure-change factor for \(R \rightarrow R/2\), the Sun’s central pressure in \(\text{dyn/cm}^2\), and that pressure in atmospheres.
Equation meaning: This scaling says that at fixed mass, a more compact star needs a much larger central pressure to support itself against gravity.
Steps:
For part (a), hold the mass fixed and compare the new pressure to the old pressure: \[ P_c \propto R^{-4}. \]
So \[ \frac{P_{c,\text{new}}}{P_{c,\text{old}}} = \left(\frac{R_{\text{new}}}{R_{\text{old}}}\right)^{-4} = \left(\frac{R/2}{R}\right)^{-4} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16. \]
Thus the required central pressure increases by a factor of \[ \boxed{16}. \]
For part (b), estimate the Sun’s central pressure: \[ P_{c,\odot} \sim \frac{G M_\odot^2}{R_\odot^4} = \frac{\left(6.674 \times 10^{-8}\right)\left(2.0 \times 10^{33}~\text{g}\right)^2}{\left(7.0 \times 10^{10}~\text{cm}\right)^4}. \]
Now evaluate the powers: \[ \left(2.0 \times 10^{33}~\text{g}\right)^2 = 4.0 \times 10^{66}~\text{g}^2, \] \[ \left(7.0 \times 10^{10}~\text{cm}\right)^4 = 7.0^4 \times 10^{40}~\text{cm}^4 = 2401 \times 10^{40}~\text{cm}^4 = 2.401 \times 10^{43}~\text{cm}^4. \]
Substitute these back in: \[ P_{c,\odot} \sim \frac{\left(6.674 \times 10^{-8}\right)\left(4.0 \times 10^{66}~\text{g}^2\right)}{2.401 \times 10^{43}~\text{cm}^4} \] \[ = \left(\frac{6.674 \times 4.0}{2.401}\right) \times 10^{-8+66-43} ~\text{g}~\text{cm}^{-1}~\text{s}^{-2} \] \[ \approx 11.1 \times 10^{15} ~\text{g}~\text{cm}^{-1}~\text{s}^{-2} = 1.11 \times 10^{16} ~\text{g}~\text{cm}^{-1}~\text{s}^{-2}. \]
Since \[ 1~\text{g}~\text{cm}^{-1}~\text{s}^{-2} = 1~\text{dyn/cm}^2, \] we get \[ \boxed{P_{c,\odot} \sim 1.1 \times 10^{16}~\text{dyn/cm}^2}. \]
For part (c), convert to atmospheres by multiplying by an identity equal to 1: \[ P_{c,\odot} \approx \left(1.1 \times 10^{16}~\frac{\text{dyn}}{\text{cm}^2}\right) \left( \frac{1~\text{atm}}{1.013 \times 10^6~\text{dyn/cm}^2} \right) \] \[ \approx 1.09 \times 10^{10}~\text{atm}. \]
So \[ \boxed{P_{c,\odot} \sim 1.1 \times 10^{10}~\text{atm}}. \]
For part (d), compactness matters so strongly because pressure scales as \(R^{-4}\). Halving the radius does not just double the required pressure; it multiplies it by \(16\). That steep dependence is why even moderate contraction has a dramatic effect on stellar interior conditions.
Unit check: \[ \left[\frac{G M^2}{R^4}\right] = \frac{ \left(\text{cm}^3~\text{g}^{-1}~\text{s}^{-2}\right) \left(\text{g}^2\right) }{ \text{cm}^4 } = \text{g}~\text{cm}^{-1}~\text{s}^{-2} = \text{dyn/cm}^2. \] The dimensions are correct for pressure. \(\checkmark\)
Sanity check: We expect stellar central pressures to be enormous, and the \(R^{-4}\) scaling tells us compact stars should be dramatically harder to support. A result of roughly ten billion atmospheres is therefore physically reasonable. \(\checkmark\)
Problem 6 — Core Temperature from Gravity
Given: \[ T_c \sim \frac{\mu G M m_p}{k_B R}, \qquad \mu = 0.6, \] \[ G = 6.674 \times 10^{-8}~\text{cm}^3~\text{g}^{-1}~\text{s}^{-2}, \qquad M_\odot = 2.0 \times 10^{33}~\text{g}, \] \[ m_p = 1.67 \times 10^{-24}~\text{g}, \qquad k_B = 1.38 \times 10^{-16}~\text{erg}~\text{K}^{-1}, \qquad R_\odot = 7.0 \times 10^{10}~\text{cm}. \]
Find: The Sun’s core temperature in K and MK, then the relative and absolute core temperature of a \(10\,M_\odot\) main-sequence star.
Equation meaning: This scaling compares the gravitational energy per particle to the thermal energy scale \(k_B T\).
Assumptions: We are using a rough hydrostatic/ideal-gas scaling, keeping \(\mu\) fixed, and using the main-sequence mass-radius relation \(R \propto M^{0.8}\).
Steps:
For part (a), estimate the Sun’s core temperature: \[ T_{c,\odot} \sim \frac{\mu G M_\odot m_p}{k_B R_\odot} = \frac{ \left(0.6\right) \left(6.674 \times 10^{-8}\right) \left(2.0 \times 10^{33}~\text{g}\right) \left(1.67 \times 10^{-24}~\text{g}\right) }{ \left(1.38 \times 10^{-16}~\text{erg}~\text{K}^{-1}\right) \left(7.0 \times 10^{10}~\text{cm}\right) }. \]
Evaluate the numerator: \[ \left(0.6\right)\left(6.674 \times 10^{-8}\right)\left(2.0 \times 10^{33}~\text{g}\right)\left(1.67 \times 10^{-24}~\text{g}\right) \approx 1.34 \times 10^2~\text{g}~\text{cm}^3~\text{s}^{-2}. \]
Evaluate the denominator: \[ \left(1.38 \times 10^{-16}~\text{erg}~\text{K}^{-1}\right)\left(7.0 \times 10^{10}~\text{cm}\right) = 9.66 \times 10^{-6}~\text{erg}~\text{cm}~\text{K}^{-1} = 9.66 \times 10^{-6}~\text{g}~\text{cm}^3~\text{s}^{-2}~\text{K}^{-1}. \]
Now divide: \[ T_{c,\odot} \approx \frac{1.34 \times 10^2}{9.66 \times 10^{-6}}~\text{K} \approx 1.39 \times 10^7~\text{K}. \]
Convert to MK: \[ 1.39 \times 10^7~\text{K} \left( \frac{1~\text{MK}}{10^6~\text{K}} \right) = 13.9~\text{MK}. \]
So \[ \boxed{T_{c,\odot} \approx 1.4 \times 10^7~\text{K} \approx 14~\text{MK}}. \]
This is reassuringly close to the reference solar-core value of about \(1.5 \times 10^7~\text{K}\) given in the homework, so the scaling is behaving the way it should.
For part (b), use the scaling form: \[ T_c \propto \frac{M}{R}. \]
Therefore \[ \frac{T_c}{T_{c,\odot}} = \left(\frac{M}{M_\odot}\right) \left(\frac{R}{R_\odot}\right)^{-1}. \]
Using the main-sequence mass-radius relation \[ \frac{R}{R_\odot} = \left(\frac{M}{M_\odot}\right)^{0.8}, \] we get \[ \frac{T_c}{T_{c,\odot}} = \left(\frac{M}{M_\odot}\right) \left[ \left(\frac{M}{M_\odot}\right)^{0.8} \right]^{-1} = \left(\frac{M}{M_\odot}\right)^{0.2}. \]
For a \(10\,M_\odot\) star, \[ \frac{T_{c,10}}{T_{c,\odot}} = \left(10\right)^{0.2} \approx 1.58. \]
So \[ \boxed{\frac{T_{c,10}}{T_{c,\odot}} \approx 1.6}. \]
For part (c), multiply by the Sun’s core temperature: \[ T_{c,10} \approx \left(1.58\right)\left(13.9~\text{MK}\right) \approx 22.0~\text{MK}. \]
Thus \[ \boxed{T_{c,10} \approx 2.2 \times 10^7~\text{K} \approx 22~\text{MK}}. \]
For part (d), the result shows that even a \(10\,M_\odot\) main-sequence star has a core temperature only modestly larger than the Sun’s. Main-sequence core temperatures do increase with mass, but the scaling is weak: \(T_c \propto M^{0.2}\). So the enormous luminosity differences along the main sequence do not come from core temperatures that differ by orders of magnitude.
Unit check: Using \(1~\text{erg} = 1~\text{g}~\text{cm}^2~\text{s}^{-2}\), \[ \left[\frac{G M m_p}{k_B R}\right] = \frac{ \left(\text{cm}^3~\text{g}^{-1}~\text{s}^{-2}\right) \left(\text{g}\right) \left(\text{g}\right) }{ \left(\text{g}~\text{cm}^2~\text{s}^{-2}~\text{K}^{-1}\right) \left(\text{cm}\right) } = \text{K}. \] The equation does indeed return a temperature. \(\checkmark\)
Sanity check: A temperature of order \(10^7~\text{K}\) is exactly the range required for stellar fusion arguments in this module, so the solar estimate passes the order-of-magnitude test. The \(10\,M_\odot\) star coming out only about \(1.6\) times hotter also matches the weak \(M^{0.2}\) scaling. \(\checkmark\)
Problem 7 — Losing Energy, Getting Hotter
Restatement: Explain why a protostar gets hotter when it radiates energy away.
Key elements: A complete answer should use the virial theorem, connect energy loss to contraction, explain why the thermal energy rises, and name the behavior as negative heat capacity.
Sample response:
For part (a), the virial theorem for a self-gravitating object says \[ 2K_{\text{th}} + U_{\text{grav}} = 0, \] so \[ K_{\text{th}} = -\frac{1}{2}U_{\text{grav}}. \]
The total energy is \[ E = K_{\text{th}} + U_{\text{grav}} = \frac{1}{2}U_{\text{grav}} = -K_{\text{th}}. \]
If the protostar radiates energy away, then \(E\) becomes more negative. To reach a more negative total energy, the star must contract into a more tightly bound state, which makes \(U_{\text{grav}}\) more negative.
For part (b), once \(U_{\text{grav}}\) becomes more negative, the virial theorem says \(K_{\text{th}}\) must increase. Larger thermal energy means larger average particle speeds, so the core temperature rises rather than falls.
For part (c), this is called negative heat capacity because a self-gravitating object behaves opposite to an everyday object. An ordinary object cools when it loses energy, but a bound gravitating object can get hotter as it loses energy. This matters for fusion ignition because gravitational contraction can keep raising the core temperature until nuclear reactions become possible.
Grading guidance: Full credit requires the chain “radiate energy \(\rightarrow\) total energy becomes more negative \(\rightarrow\) contraction \(\rightarrow\) larger thermal energy \(\rightarrow\) higher core temperature,” plus the phrase negative heat capacity.
Part C — Fusion Ignition and Energy Release
Problem 8 — Tunneling Makes Fusion Possible; the Weak Interaction Makes It Slow
Restatement: Distinguish what tunneling does for proton-proton fusion from what the weak interaction does.
Key insight: Tunneling lets protons reach classically forbidden separations, but the first successful pp reaction still requires a weak-interaction proton-to-neutron conversion.
Answer:
For part (a), quantum tunneling solves the Coulomb-barrier problem. At solar-core temperatures, typical protons do not have enough classical thermal energy to climb over the electrostatic repulsion between two positively charged protons. Tunneling makes it possible for a tiny fraction of collisions to reach the small separations where the strong interaction can act.
For part (b), even after tunneling allows two protons to get close together, the first step of the pp chain is still extremely slow because one of those protons must change into a neutron: \[ p + p \rightarrow d + e^+ + \nu_e. \] That proton-to-neutron conversion is a weak-interaction process, and weak processes are rare compared with ordinary electromagnetic collisions.
For part (c), the weak interaction largely controls the Sun’s hydrogen-burning rate because every successful pp chain must pass through that slow first step. The bottleneck is not the existence of proton collisions by itself; it is the rare weak conversion that creates deuterium.
Sanity check: If tunneling were the only bottleneck, the Sun would burn much faster than it actually does. Its multi-billion-year lifetime tells us another severe rate limiter must be present. \(\checkmark\)
Common misconception: The Sun does not live a long time because protons rarely collide. Protons collide often; what is rare is a collision that also produces the needed weak-interaction conversion.
Problem 9 — Classical Fusion Should Fail
Given: \[ \frac{3}{2}k_B T \approx 1.44~\text{MeV}, \qquad k_B = 8.617 \times 10^{-5}~\text{eV/K}, \qquad T_{c,\odot} \approx 1.5 \times 10^7~\text{K}. \]
Find: The temperature required for classical proton-proton barrier crossing, the comparison to the solar core, and the factor by which the classical temperature is larger.
Equation meaning: We are setting the typical thermal energy \(\frac{3}{2}k_B T\) equal to the Coulomb-barrier energy and asking what temperature would be needed if tunneling did not exist.
Steps:
For part (a), first convert the barrier energy from MeV to eV using the identity trick: \[ 1.44~\text{MeV} \left( \frac{10^6~\text{eV}}{1~\text{MeV}} \right) = 1.44 \times 10^6~\text{eV}. \]
Now solve for \(T\): \[ \frac{3}{2}k_B T = 1.44 \times 10^6~\text{eV} \] \[ T = \frac{2}{3} \frac{1.44 \times 10^6~\text{eV}}{8.617 \times 10^{-5}~\text{eV/K}}. \]
Compute the numerical value: \[ T \approx \frac{2}{3}\left(1.671 \times 10^{10}~\text{K}\right) \approx 1.11 \times 10^{10}~\text{K}. \]
So the required classical temperature is \[ \boxed{T_{\text{classical}} \approx 1.1 \times 10^{10}~\text{K}}. \]
For part (b), compare this to the solar-core temperature: \[ T_{\text{classical}} \approx 1.1 \times 10^{10}~\text{K}, \qquad T_{c,\odot} \approx 1.5 \times 10^7~\text{K}. \]
The classical requirement is therefore vastly larger than the actual solar-core temperature.
For part (c), compute the factor: \[ \frac{T_{\text{classical}}}{T_{c,\odot}} = \frac{1.11 \times 10^{10}~\text{K}}{1.5 \times 10^7~\text{K}} = 7.4 \times 10^2. \]
So the classical temperature is about \[ \boxed{7.4 \times 10^2 \approx 740\ \text{times larger}} \] than the actual solar-core temperature.
For part (d), this comparison shows that the Sun is nowhere near hot enough for classical barrier crossing. Therefore stars need quantum tunneling: fusion cannot be explained by ordinary thermal motion alone.
Unit check: \[ \left[\frac{\text{eV}}{\text{eV/K}}\right] = \text{K}, \] so the temperature calculation has the correct units. The ratio in part (c) is dimensionless because the kelvin units cancel. \(\checkmark\)
Sanity check: The answer should come out far above the solar-core temperature, not just a factor of two or three above it. A factor of about \(740\) confirms that classical fusion would fail badly in the Sun. \(\checkmark\)
Problem 10 — From Mass Deficit to the Iron Limit
Restatement: Compute the energy associated with the hydrogen-to-helium mass deficit, then explain why fusion releases energy below the iron peak but not beyond it.
Key elements: A complete answer should calculate the MeV release from the given mass deficit, interpret that result in terms of binding energy per nucleon, and explain why carbon fusion can still be exothermic while iron fusion cannot.
Sample response:
For part (a), start from \[ E = \Delta m c^2 \] and use \[ 1~\text{amu} = 931.5~\text{MeV}/c^2. \]
Then \[ E = \left(0.02872~\text{amu}\right)c^2 \left( \frac{931.5~\text{MeV}}{1~\text{amu}~c^2} \right). \]
Now cancel the units: \[ E = \left(0.02872\right)\left(931.5\right)~\text{MeV} \approx 26.75~\text{MeV}. \]
So the net pp-chain energy release is \[ \boxed{E_{\text{pp,net}} \approx 26.7~\text{MeV}}. \]
This is the total mass-energy released by the chain. In a real star, not all of that energy stays behind as thermal energy, because neutrinos carry some away.
For part (b), hydrogen-to-helium fusion is exothermic because the helium nucleus is more tightly bound per nucleon than the original hydrogen nuclei were. That means the final state has lower total mass-energy. The missing mass appears as released energy.
For part (c), carbon fusion can still release energy because carbon lies below the iron/nickel peak in binding energy per nucleon. Fusing carbon into heavier nuclei can still move the products upward on the binding-energy-per-nucleon curve, which lowers the total mass-energy and releases energy. Iron is different because nuclei near the iron peak are already among the most tightly bound. Fusing iron into even heavier nuclei would move the products downward in binding energy per nucleon, so it would require an input of energy instead of releasing it.
Grading guidance: Full credit requires the numerical value \(\approx 26.7~\text{MeV}\) and an explanation built around the binding-energy-per-nucleon curve, not just the phrase “iron is special.”