NUC-15

Mass defect

Δ = Z m_H + (A−Z) m_n − M. Binding energy B = Δ c².

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NuclearMass defect

Governing equation

\Delta=Z m_H+(A-Z)m_n-M

where

Z: Atomic number (—)
A: Mass number (—)
M: Atomic mass (u)
\Delta: Mass defect (u)
B: Binding energy (MeV)

Lecture brief

Historical brief

Rutherford, Chadwick, the semi-empirical mass formula, fission (Hahn–Strassmann 1938) and fusion Q-values, then Compton and Bethe–Bloch stopping, are the nuclear toolkit. Decay, binding and dose start here. This sheet (NUC-15 — Mass defect) is the form associated with Mass defect. Working symbols: $Z$ , $A$ , $M$ $\rightarrow$ $\Delta$ , $B$ . m_H = 1.007825 u, m_n = 1.008665 u. The iron region maximises B/A.

Purpose

Purpose: compute

\Delta

B

from

Z

A

M

in Nuclear physics via

\Delta=Z m_H+(A-Z)m_n-M

Δ = Z m_H + (A−Z) m_n − M. Binding energy B = Δ c². Use it when a real nuclear physics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

Z = 26.000\,\mathrm{—}

A = 56.000\,\mathrm{—}

M = 55.935\,\mathrm{u}

, the governing relation

\Delta=Z m_H+(A-Z)m_n-M

yields

\Delta = 0.52850\,\mathrm{u}

B = 492.29\,\mathrm{MeV}

. Free nucleons versus a bound mass, a Δ gap. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Atomic number Z26 —Mass number A56 —Atomic mass M55.9349 u

Outputs

Mass defect \Delta0.52850 u
Binding energy B492.29 MeV

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Narration of this film

Free nucleons versus a bound mass, a Δ gap.

m_H = 1.007825 u, m_n = 1.008665 u. The iron region maximises B/A.

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