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A nucleus at rest spits out an alpha particle. The energy released, the Q value, comes from the mass that disappears. But the alpha particle does not get all of it: the daughter nucleus recoils and takes a share. Exam questions ask for that share, or give the alpha's energy and ask for Q. This post shows the two-line method, works through polonium-210 with real masses, and lists the slips that cost marks.
Which exams ask this?
This is a syllabus question worth settling first. The JEE (Advanced) 2026 brochure lists "α, β and γ radiations" and "Energy calculation in these processes" under Modern Physics, so alpha-decay energetics is in scope there. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list mass-energy relation, mass defect, binding energy, fission and fusion, but not radioactive decay. The NCERT Class 12 chapter Nuclei (2026-27 reprint) keeps only a short section naming alpha, beta and gamma decay, with no decay equations.
So for JEE Main and NEET, alpha decay itself is out, but the method is not. The same momentum split is in NCERT's Laws of Motion chapter, where Exercise 4.17 asks you to show that a nucleus at rest breaking into two pieces sends them in opposite directions. And the Q value is the mass-energy calculation both syllabi do include. Check your own year's documents.
How is the energy shared?
Take a parent nucleus at rest, mass number A. It splits into an alpha particle (mass number 4) and a daughter (A − 4).
Momentum: the total was zero, so the two momenta are equal and opposite: pα = pd.
Kinetic energy: K = p²/2m. With equal p, Kd/Kα = mα/md ≈ 4/(A − 4).
Energy: Kα + Kd = Q.
Solving gives
Kα = Q × (A − 4)/A and Kd = Q × 4/A
The lighter fragment takes almost everything. For a heavy nucleus with A around 200, the alpha gets about 98% of Q.
Worked example: polonium-210
Polonium-210 decays to lead-206 by alpha emission. The NIST atomic mass tables give Po-210 = 209.9828741 u, Pb-206 = 205.9744657 u and He-4 = 4.00260325413 u.
Mass lost: 209.9828741 − 205.9744657 − 4.0026033 = 0.0058051 u.
Q value: using NCERT's 1 u = 931.5 MeV/c², Q = 0.0058051 × 931.5 ≈ 5.407 MeV.
Alpha energy: Kα = 5.407 × 206/210 ≈ 5.304 MeV.
Recoil energy: Kd = 5.407 × 4/210 ≈ 0.103 MeV, about 103 keV.
Why can you use atomic masses? The polonium atom has 84 electrons. The lead atom has 82 and the helium atom 2, so the electron masses cancel on both sides.
How fast does the lead nucleus move? From K = ½mv², 103 keV on a mass of 206 u gives about 3.1 × 10⁵ m/s. The alpha moves at about 1.6 × 10⁷ m/s. The speed ratio is 206/4 ≈ 51.5, as equal momentum demands.
Worked example: going backwards from the alpha energy
A question says (sample number) that radium-226 emits alpha particles of 4.78 MeV and asks for Q and the recoil energy of the radon-222 nucleus.
Kα = Q × 222/226, so Q = 4.78 × 226/222 ≈ 4.87 MeV.
Recoil = Q − Kα ≈ 4.866 − 4.78 ≈ 0.086 MeV, about 86 keV. As a check, Q × 4/226 gives the same.
Which mistakes cost marks?
Giving the whole Q to the alpha. The question usually exists to test this.
Swapping the ratio. The alpha's share is (A − 4)/A, not 4/A. If your alpha energy comes out as a few keV, you swapped them.
Using A − 4 for the parent. A in the formula is the parent's mass number: 210 for polonium, not 206.
Mixing atomic and nuclear masses. Use atomic masses throughout, or nuclear masses throughout, never one of each.
Forgetting excited states. If the daughter is left excited, part of Q leaves later as a gamma ray, and the alpha's energy is lower. Use the formula only for decay to the ground state unless told otherwise.
Assuming the parent was moving. The formula needs the parent at rest in your frame.
The same two-body split appears in fission and binding-energy questions; our post on nuclear binding energy problems works through Q values from mass defect. If you get stuck on where a momentum or energy step comes from, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit the momentum idea before returning to the nuclear version.
A practice routine
Day 1: Derive Kα = Q(A − 4)/A from momentum and energy without notes. Then do NCERT Exercise 4.17.
Day 2: Find Q, Kα and the recoil energy for americium-241 decaying to neptunium-237, using NIST masses (Am-241 = 241.0568293 u, Np-237 = 237.0481736 u). You should get Q ≈ 5.64 MeV.
Day 3: Two backwards problems: given Kα, find Q and the recoil.
Day 5: For JEE Advanced students, a mixed set with half-life questions from our radioactive decay and half-life guide.
Tonight, compute the recoil speed of the radon nucleus in the radium example and check that its momentum matches the alpha's.
