Question

A black hole begins to form as an aging star abruptly collapses when it ceases
emitting radiation from fusion reactions
. This occurs after the star has consumed its component elements in this order:
hydrogen, helium, then through iron
.

Answer 1

You're basically on the right track, but a few points need sharpening.

Key corrections and clarifications
- Massive stars do fuse elements in roughly this progression: hydrogen → helium → carbon/oxygen → neon → oxygen → silicon → iron-group. Iron (Fe) and nearby nuclei are the most tightly bound, so fusion beyond iron is endothermic and cannot produce net energy.
- It’s not simply that the star “stops emitting radiation.” The relevant event is that the iron core can no longer generate pressure (via fusion or electron degeneracy support) to resist gravity. When the iron core grows near the Chandrasekhar mass (~1.4 M☉) it catastrophically collapses in seconds.
- Core collapse usually produces a core‑collapse supernova (huge neutrino burst, outer layers expelled). The compact remnant depends on the core mass and physics:
- If the proto‑neutron star mass is below the neutron-star limit (Tolman–Oppenheimer–Volkoff limit, roughly ~2–3 M☉, uncertain) it becomes a neutron star.
- If it’s heavier (or if enough fallback mass accretes), it becomes a black hole.
- Whether a star ends as a black hole depends on its initial mass, mass loss, metallicity, and details of the explosion. Roughly, very massive stars (tens of solar masses) are the typical progenitors of stellar‑mass black holes, but there are exceptions (pair‑instability supernovae can leave no remnant; some very massive stars may collapse directly with little visible supernova).

Concise corrected statement
- In a sufficiently massive star, successive fusion builds an iron core that can no longer produce pressure to support itself; when that core exceeds the support limit it collapses, and if the remnant is massive enough the collapse produces a black hole.