Know that stars are powered by nuclear reactions that release energy and that in stable stars the nuclear reactions involve the fusion of hydrogen into helium
6.2.1 The Sun as a Star
Learning objectives (AO 1‑3)
Explain why stars, including the Sun, shine – nuclear fusion in the core.
Identify the dominant fusion reaction in a stable, Sun‑like star (the proton‑proton chain).
Describe the CNO cycle and state when it becomes the main energy source.
Understand hydrostatic equilibrium, the two main energy‑transport zones, main‑sequence status and how a star’s lifetime is estimated.
Carry out simple quantitative calculations (mass defect, energy released, lifetime) to demonstrate understanding.
1. Core conditions
Quantity
Typical value (Sun)
Temperature, Tc
≈ 1.5 × 107 K
Pressure, Pc
≈ 2 × 1011 Pa
Density, ρc
≈ 1.5 × 105 kg m⁻³
These extreme values give protons enough kinetic energy to overcome their electrostatic repulsion, allowing nuclear fusion to occur.
2. Nuclear fusion in stable stars
2.1 Proton‑proton (p‑p) chain – dominant in stars ≤ 1.3 M☉ (including the Sun)
The p‑p chain converts four hydrogen nuclei (protons) into one helium‑4 nucleus. It proceeds in three main steps.
Step 1 – Formation of deuterium
\[
p + p \;\rightarrow\; ^2\!H + e^{+} + \nu_e
\]
A positron (e⁺) and an electron‑neutrino (νₑ) are emitted.
Step 2 – Production of helium‑3
\[
^2\!H + p \;\rightarrow\; ^3\!He + \gamma
\]
A gamma‑ray photon (γ) carries away the released energy.
Step 3 – Formation of helium‑4
\[
^3\!He + ^3\!He \;\rightarrow\; ^4\!He + 2p
\]
Two protons are released to start the cycle again.
In hotter, more massive stars carbon, nitrogen and oxygen act as catalysts. The overall conversion is the same as the p‑p chain (4 p → ⁴He + energy) but proceeds through a series of rapid reactions.
The CNO rate is extremely temperature‑sensitive (≈ T15), so it overtakes the p‑p chain when the core temperature exceeds ≈ 1.8 × 107 K.
3. Energy released – mass defect (worked example)
Quantity
Value
Mass of 4 protons
4 × 1.007825 u = 4.03130 u
Mass of a ⁴He nucleus
4.002603 u
Mass defect Δm
0.028697 u
1 u = 931.5 MeV c⁻²
Energy released per reaction
Δm c² ≈ 0.028697 u × 931.5 MeV ≈ 26.7 MeV
Convert to joules
26.7 MeV × 1.602 × 10⁻¹³ J MeV⁻¹ ≈ 4.3 × 10⁻¹² J
Number of reactions per second in the Sun ≈ 1056. Multiplying gives a luminosity of ≈ 3.8 × 1026 W, matching the observed solar output.
4. Hydrostatic equilibrium – why the Sun is stable
Outward pressure: gas pressure from the hot plasma + radiation pressure from photons.
Inward force: gravity pulling all mass toward the centre.
When these forces balance at every radius, the star is in hydrostatic equilibrium and its size remains constant.
Self‑regulating feedback: a slight rise in core temperature ↑ pressure → core expands → temperature falls → fusion rate drops, restoring equilibrium.
5. Energy transport inside the Sun
Radiative zone (≈ 0.25 R☉ – 0.70 R☉) – photons are repeatedly scattered by ions; energy moves outward by diffusion.
Convective zone (≈ 0.70 R☉ – 1.00 R☉) – the temperature gradient becomes steep enough that bulk motion of plasma (convection) carries the energy.
Photosphere – the visible surface (≈ 5 × 10⁻⁴ R☉ thick) from which photons finally escape into space; it defines the Sun’s “surface” temperature (~5 800 K) and the light we see.
Suggested diagram: cross‑section of the Sun showing core, radiative zone, convective zone and photosphere with arrows indicating radiative diffusion and convective currents.
6. Main‑sequence status and stellar lifetime
The Sun is a main‑sequence star: it is fusing hydrogen in its core and lies on the stable band of the Hertzsprung–Russell diagram.
Mass‑luminosity relation (approximate, for main‑sequence stars):
\[
L \;\approx\; L{\odot}\, \left(\frac{M}{M{\odot}}\right)^{3.5}
(≈ 10 Gyr for the Sun; 0.1 M☉ is the fraction of the Sun’s mass that ever reaches the core.)
7. Solar neutrinos – direct evidence of core fusion
Each complete p‑p chain produces two electron‑neutrinos (νₑ). Because neutrinos interact only via the weak force, they escape the Sun’s interior essentially unhindered.
Neutrino detectors on Earth (e.g., the Sudbury Neutrino Observatory, Super‑Kamiokande) measure a flux that matches the rate predicted by the p‑p chain, confirming that nuclear fusion powers the Sun.
Low‑mass stars (M < 0.5 M☉): core temperatures too low for the CNO cycle; they rely entirely on the p‑p chain and can live > 100 Gyr because their fuel consumption is extremely slow.
9. Classroom activity – estimating the Sun’s energy output
Use the mass‑defect result (26.7 MeV per 4 p) and the fact that the Sun converts ≈ 4 × 10⁹ kg of hydrogen to helium each second.
Calculate the energy released per second:
\[
E = \frac{4\times10^{9}\ \text{kg}}{4m_{\text{p}}}\times 26.7\ \text{MeV}
\]
(where \(m_{\text{p}} = 1.67\times10^{-27}\) kg). Show that the answer is ≈ 3.8 × 1026 J s⁻¹, i.e. the Sun’s luminosity.
Discuss sources of error (e.g., not all core mass is fused, neutrino energy loss) and relate the result to the solar constant measured at Earth.
Key points to remember
The Sun’s energy comes from nuclear fusion; the dominant reaction in Sun‑like stars is the proton‑proton chain.
In more massive stars the CNO cycle replaces the p‑p chain as the main energy source.
Mass is converted to energy: 1 kg → 9 × 10¹⁶ J (E = mc²).
Hydrostatic equilibrium (balance of outward pressure and inward gravity) keeps a star stable for billions of years.
Energy produced in the core travels outward by radiation (radiative zone) then by convection (convective zone) before escaping from the photosphere, the visible surface.
Solar neutrino detection provides direct experimental proof of core fusion.
Stellar mass determines which fusion cycle dominates, how luminous a star is, and how long it will live.
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