Two-Body Recoil and Energy Sharing
In a two-body decay of a parent initially at rest, p1 = -p2. Under non-relativistic product motion, K1/K2 = m2/m1 and K1 + K2 = Q, giving K1 = Q m2/(m1 + m2).
Why this shows up in the exam
Alpha-particle recoil energy · Fragment speed and size comparisons · Gamma recoil corrections
Learn the idea
For a parent initially at rest, two products have equal opposite momenta and kinetic energies inversely proportional to mass. The light fragment must move faster so that its momentum cancels the heavy fragment's momentum. With equal momentum magnitude, the lighter product receives the larger kinetic-energy share.
🧠 Memory hook: Equal momentum means the lighter product gets more kinetic energy.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- |p1| = |p2| — momentum conservation for two products from a stationary parent
- K1/K2 = m2/m1 — non-relativistic kinetic-energy ratio at equal momentum
- K1 = Q m2/(m1 + m2) — energy carried by product 1 in a non-relativistic two-body decay
How to approach it
- 1Confirm the parent is initially at rest
- 2Write equal and opposite momenta
- 3Use K = p²/(2m) only when non-relativistic
Common slip-ups that cost marks
- •Giving equal kinetic energies to unequal masses
- •Ignoring daughter recoil
- •Using the non-relativistic ratio for photons
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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