SME
University Physics III: Waves, Optics & Modern · Problem type
The energy released in a decay (Q value)
Use this for “Radium-226 (226.025410 u) gives off an alpha particle (4.002603 u) and becomes radon-222 (222.017578 u). How much energy does each decay release?”
Diagram
Before: 238.0508 u
After: 234.0436 + 4.0026 = 238.0462 u
Δm = 238.0508 − 238.0462
Δm = 0.004584 u
Q = 0.004584 u × 931.5 MeV/u
Q = 4.27 MeV
The axis is broken: the bars start near the masses, so the tiny gap shows.
The lost mass became energy.
This is an example. Type your own values to start a new problem.
Δm (u)
=
Mₚₐᵣₑₙₜ (u)
−
Mdaughter (u)
−
Mparticle (u)
Q (MeV)
=
Δm
×
931.5
MeV/u
Formulas
Δm = Mₚₐᵣₑₙₜ − Mdaughter − Mparticle
0.004584 = 238.050788 − 234.043601 − 4.002603
Q
=
Δm
×
931.5
4.27 = 0.004584 × 931.5
Assumptions
•
Atomic masses in u, electrons included; an alpha particle is a helium-4 atom, 4.002603 u.
•
For β⁻ decay the daughter atom already counts the new electron, so the particle’s mass is 0.
•
The missing mass comes out as kinetic energy: Q = Δm × 931.5 MeV for each u. Q > 0 means the decay can happen on its own.
Step-by-step
WE KNOW
Mₚₐᵣₑₙₜ = 238.0508 u, Mdaughter = 234.0436 u, Mparticle = 4.0026 u
FIND
mass that goes missing (Δm), energy released (Q)
Step 1 · Find mass that goes missing (Δm)
Use Δm = Mₚₐᵣₑₙₜ − Mdaughter − Mparticle
The mass before less the masses after.
Δm = 238.050788 − 234.043601 − 4.002603
Δm = 0.004584 u
Step 2 · Find energy released (Q)
Use Q = Δm × 931.5
E = mc²: each u of mass is 931.5 MeV, shared as kinetic energy by the daughter and the particle.
Q
=
0.004584
×
931.5
Q
=
4.27
MeV
Check
Put the numbers back in. Both sides should match.
0.004584 = 238.050788 − 234.043601 − 4.002603 ✓
4.27 = 0.004584 × 931.5 ✓
Topic check
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