Nuclear Structure

The atomic nucleus is a bound system of protons and neutrons held together by the strong nuclear force. Nuclear binding energy, measured by the semi-empirical mass formula, determines nuclear stability and energy release in reactions.

Key Concepts

Nucleon numbers: Z (protons), N (neutrons), A = Z+N (mass number)
Nuclear radius: R = R₀ A^{1/3}, R₀ ≈ 1.2 fm
Binding energy: BE = (Zm_p + Nm_n - M_nucleus)c²
Semi-empirical mass formula (Bethe-Weizsäcker)
Magic numbers: 2, 8, 20, 28, 50, 82, 126

Key Equations

Nuclear radius

R = R_0 A^{1/3},\quad R_0 \approx 1.2\text{ fm}

Binding energy per nucleon

B/A = a_v - a_s A^{-1/3} - a_c \frac{Z^2}{A^{4/3}} - a_a\frac{(N-Z)^2}{A^2} \pm \delta

Q-value

Q = (m_i - m_f)c^2

Mass excess

\Delta = (M - A)u\cdot c^2\text{ in MeV}

Worked Example

Example Problem

Problem

Find the radius of ⁵⁶Fe (A=56).

Solution

R = 1.2 × 56^{1/3} fm = 1.2 × 3.826 = 4.59 fm.

Practice

Exercises

7 problems

1 of 7

Find the radius of ²³⁸U (A=238) in fm.

Your answer

2 of 7

The binding energy of ⁴He is 28.3 MeV. Find BE/A in MeV/nucleon.

Your answer

MeV/nucleon

3 of 7

The mass excess of ¹²C is 0.000 MeV (by definition), and ¹H is 7.289 MeV. For ⁴He: Δ=-2.425 MeV. Find the Q-value (in MeV) for ³α→¹²C: Q = 3Δ(He) - Δ(C).

Unlock Exercise 3