Fusion and stars

Joining two or more **light** nuclei into a **heavier** one; the product is slightly lighter and the missing mass is released as energy.

The electrical **push** between two positive charges. Nuclei are positive, so they repel — fusion must overcome this to bring them together.

Very high **temperature** (fast-moving nuclei) and very high **density / pressure** (frequent collisions).

The series of reactions that fuses **hydrogen into helium** in stars like the Sun, releasing energy at each step.

$E = mc^{2}$ — mass-energy equivalence (given in the data booklet). Here m is the mass defect.

Multiply Δm (in u) by **931.5**, because 1 u = 931.5 MeV c⁻².

The **mass defect** — the product is slightly lighter than the nuclei that fused, and that missing mass becomes energy.

The state where the **outward** pressure from fusion's radiation and hot gas exactly **balances** gravity's **inward** pull, so the radius stays stable.

The **outward pressure** from the heat of fusion — radiation pressure plus the pressure of the hot gas. (Not the reactions themselves directly.)

If it shrinks → core heats → fusion speeds up → more pressure → it expands back. If it expands → cools → fusion slows → gravity pulls it back in.

E = 0.0265 × 931.5 ≈ **24.7 MeV**.

1) mass defect Δm = total mass of nuclei − mass of product; 2) E = mc² (or Δm × 931.5 for MeV); 3) keep the unit.

**4 ¹H + 2 e⁻ → ⁴He + 2ν + γ rays**, releasing about **26.7 MeV**.

Their nuclei have **more charge**, so the repulsion is bigger and a much **higher core temperature** is needed.

Advantage: **plentiful fuel**, no long-lived waste. Difficulty: about **10⁸ K** needed and the plasma must be **contained**.

How long the star spends steadily **fusing hydrogen into helium** — the long, stable middle of its life.

The total energy a star radiates **every second** — its power output, in watts (W = J s⁻¹).

Lifetime = energy the fusible hydrogen releases ÷ luminosity: **t = E ÷ L**. Then convert seconds to years.

**No** — you build it yourself from 'luminosity = energy used per second', so lifetime = energy available ÷ luminosity.

Only the **core's** hydrogen fuses (~10–12% of the mass), and only **~0.7%** of that mass becomes energy. Multiply by both.

$E = mc^{2}$ — mass-energy equivalence (given in the data booklet).

A high luminosity means it **burns through its fuel faster**, so even with lots of fuel it runs out sooner.

**Δm = E ÷ c²**, where E is the total energy it radiates. (Rearranged from E = mc².)

The **luminosity stays constant**; or only the core hydrogen fuses; or a fixed ~0.7% of the mass is converted; or the fusion rate is steady.

**Divide by about 3.16 × 10⁷** — the number of seconds in one year.

t = E ÷ L = 3.6 × 10¹⁷ s ≈ **1.1 × 10¹⁰ years** (÷ 3.16 × 10⁷).

Δm = E ÷ c² = 1.8×10⁴⁴ ÷ (3.00×10⁸)² ≈ **2.0 × 10²⁷ kg**.

Fuel mass ÷ (4 × proton mass) = number of p–p chains; × energy per chain = total energy E.

**Shorter**: they have more fuel, but their luminosity is far greater, so they use it up much faster.

About **10 %**: the hydrogen in its core.

The **total power** the star radiates in all directions (in watts, W). It is a property of the star itself and does **not** depend on distance.

The power we **receive per square metre** at Earth (in W m⁻²). It **depends on distance** — the same star looks dimmer farther away.

$b = \dfrac{L}{4\pi d^{2}}$ — the inverse-square law (given in the data booklet).

By distance d the light has spread over a **sphere** of radius d, whose surface area is 4π d². The power L is shared over that area.

The apparent brightness falls to a **quarter** (1/2² = 1/4): twice as far → 4× the area → ¼ the brightness.

The tiny apparent **shift** of a nearby star against distant background stars as Earth orbits the Sun. A bigger shift means a closer star.

$d\,(\text{parsec}) = \dfrac{1}{p\,(\text{arc-second})}$ — distance in parsecs is one over the parallax angle in arc-seconds.

The distance at which a star shows a parallax angle of exactly **1 arc-second**. 1 pc ≈ 3.26 light-years ≈ 3.1 × 10¹⁶ m.

d = 1/p = 1/0.020 = **50 parsec**.

Equal b ⇒ d ∝ √L, so √100 = **10 times farther** away.

Only its **apparent brightness** (it drops as 1/d²). The **luminosity is unchanged** — that's a fixed property of the star.

1 pc = **3.26 ly** = **3.09 × 10¹⁶ m**.

p = φ/2, so d = 1/p = **2/φ** parsecs (φ in arc-seconds).

For distant stars the parallax angle is **too small to measure**.

An ideal object that absorbs all radiation hitting it and re-radiates a spectrum set **only by its temperature**. A star is a good approximation.

The wavelength at which the star radiates **most intensely** — the top of its black-body curve. A shorter peak means a hotter star.

$\lambda_{max}T = 2.9\times10^{-3}$ m K (given). The peak wavelength and the absolute temperature are **inversely** related.

Read off the peak wavelength λ_{max} (in metres), then T = 2.9 × 10⁻³ ÷ λ_{max}.

$L = \sigma A T^{4}$ (given), with A = 4πR² for a sphere, so $L = \sigma(4\pi R^{2})T^{4}$ and L ∝ R²T⁴.

The **total power** a star radiates, in watts (W). It is set by the star's surface area and the fourth power of its temperature.

Because it appears as **T⁴**. Doubling the temperature multiplies the luminosity by 2⁴ = **16**, while doubling the radius gives only 4×.

R_B/R_A = √(L_B/L_A) ÷ (T_B/T_A)² — take the ratio of the two Stefan-Boltzmann equations so σ and 4π cancel.

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (a given data-booklet constant).

T = 2.9 × 10⁻³ ÷ (580 × 10⁻⁹) ≈ **5000 K** — a yellow star.

Wien first (peak → temperature T), then Stefan-Boltzmann (L and T → radius R, via L = σ(4πR²)T⁴).

At equal T, L ∝ R², so R ratio = √4 = **2**.

L from brightness and distance (L = 4πd²b); T from the peak (Wien); then **R = √(L ÷ 4πσT⁴)**.

**L/L☉ = (R/R☉)²(T/T☉)⁴** (σ and 4π cancel).

**10 R☉**: at equal T, L ∝ R², so R ∝ √L.

A star's **luminosity** (vertical, up = brighter) against its **surface temperature** (horizontal).

**Backwards** — **hot stars on the LEFT**, cool stars on the right. (A classic exam trap.)

The **total power** a star radiates, in watts. (Different from apparent brightness, which also depends on distance.)

On the **diagonal band** through the middle; they are fusing **hydrogen into helium**. The Sun is one.

**Top-right** — cool but very luminous. It is bright because it is **huge** (large radius), not because it is hot.

**Bottom-left** — **hot** surface but **very dim**, because it is **tiny** (small radius).

$L = \sigma A T^{4}$ (given). With A = 4πr² it becomes **L ∝ r²T⁴**.

$R_{\text{star}}/R_{\text{sun}} = (T_{\text{sun}}/T_{\text{star}})^{2}\sqrt{L_{\text{star}}/L_{\text{sun}}}$ — from L ∝ r²T⁴.

**Larger**. For fixed L, r ∝ 1/T², so a lower temperature means a bigger radius.

From its **position**: diagonal band = main sequence; top-right = red giant/supergiant; bottom-left = white dwarf.

Equal T makes the bracket 1, so R/R_{sun} = √16 = **4 R_{sun}**.

Because L ∝ r²T⁴ — a large enough **radius** makes up for the low temperature, so a big cool star (red giant) is still bright.

A near-vertical band above the main sequence where stars are unstable and **pulsate**, their luminosity varying.

From **upper left to lower right**; stars nearer the top right are **larger**.

From the **main sequence** up and to the **right** (cooler, more luminous), into the red-giant region.

Its **mass**. Low-mass stars end as white dwarfs; high-mass stars end in supernovae, leaving neutron stars or black holes.

main sequence → **red giant** → **planetary nebula** → **white dwarf**.

main sequence → **red supergiant** → **supernova** → **neutron star** (or **black hole** if heavy enough).

The glowing shell of gas a dying **low-mass** star gently puffs off (it has nothing to do with planets).

The small, hot, dense leftover core of a **low-mass** star after it sheds its outer layers; it just cools over time.

The violent explosion that ends a **massive** star's life, leaving a neutron star or a black hole.

The making of **heavier elements** by fusion inside stars (e.g. helium → carbon → ... up to iron in massive stars).

The Sun fuses only **hydrogen into helium**. A hotter, massive star fuses **heavier elements** (carbon, oxygen...) up to **iron**.

Heavier nuclei repel more strongly, so fusing them needs a **hotter** core — only a massive star's core gets that hot.

Fusing up TO iron releases energy, but fusing iron into heavier elements would **cost** energy — so even massive stars can go no further by fusion.

From its **absorption spectral lines** — each element absorbs its own wavelengths, leaving a unique pattern of dark lines (a fingerprint).

Its electrons only absorb photons whose energy exactly matches the gaps between its **energy levels**, which are unique to that element.

**Up and right** to the red giant, then **down and left** to the white dwarf.

A **white dwarf**: it is below the mass needed for a supernova (about 8 solar masses).

**Main sequence → red giant → white dwarf.**