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Grade 10 Science · Problem type
The mass defect: where the energy comes from
Use this for “U-238 gives off an alpha particle. How much mass is lost, and how much energy is released?”
Diagram
Before: 238.0508 u
After: 234.0436 + 4.0026 = 238.0462 u
Δm = 238.0508 − 238.0462
Δm = 0.004584 u
E = 0.004584 u × 931.5 MeV/u
E = 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)
=
m (u)
−
(
m₁ (u)
+
m₂ (u)
)
E
Energy released
worked out (not typed) · MeV
Formulas
Δm
=
m
−
(
m₁
+
m₂
)
0.00458 = 238.050788 − (234.043601 + 4.002603)
E
=
Δm
×
931.5
4.27 = 0.00458 × 931.5
Assumptions
•
The masses are of the nuclei with their electrons (atomic masses), in unified atomic mass units.
•
The mass lost becomes energy, E = mc²: 931.5 MeV for each unit of mass.
Step-by-step
WE KNOW
m = 238.0508 u, m₁ = 234.0436 u, m₂ = 4.0026 u
FIND
mass lost (Δm), energy released (E)
Step 1 · Find mass lost (Δm)
Use Δm = m − (m₁ + m₂)
The mass before less the mass after.
Δm = 238.050788 − (234.043601 + 4.002603)
Δm = 0.00458 u
Step 2 · Find energy released (E)
Use E = Δm × 931.5
Each unit of mass lost becomes 931.5 MeV of energy.
E
=
0.00458
×
931.5
E
=
4.27
MeV
Check
Put the numbers back in. Both sides should match.
0.00458 ≈ 238.050788 − (234.043601 + 4.002603) ✓
4.27 ≈ 0.00458 × 931.5 ✓
Topic check
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Nuclear chemistry: radioactive decay, half-life, fission and fusion
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