DAT General Chemistry: Nuclear Stability and the Belt of Stability

DAT High-Yield Concept

Neutron-to-Proton Ratio ( $N/Z$ ): Stable light isotopes ( $Z \le 20$ ) have an $N/Z$ ratio of $\approx 1.0$ . For heavier elements, the ratio increases to $\approx 1.5$ at $Z = 83$ .
Above the Belt (Neutron-Rich): High $N/Z$ ratio. Nuclides undergo beta minus ( $\beta^-$ ) emission to convert a neutron into a proton, lowering the $N/Z$ ratio toward stability.
Below the Belt (Proton-Rich): Low $N/Z$ ratio. Nuclides undergo positron ( $\beta^+$ ) emission or electron capture to convert a proton into a neutron, raising the $N/Z$ ratio toward stability.
Beyond the Belt ( $Z > 83$ ): All isotopes with $Z > 83$ are inherently unstable and primarily decay via alpha emission to reduce overall nuclear mass.

The Neutron-to-Proton Ratio and Nuclear Stability

An atomic nucleus is held together by the strong nuclear force, an attractive force acting over short distances between nucleons (protons and neutrons). Simultaneously, electrostatic repulsion acts between positively charged protons. Neutrons act as a nuclear "glue," diluting proton-proton repulsions without adding electrostatic repulsion.

Nuclear stability depends heavily on the ratio of neutrons ( $N$ ) to protons ( $Z$ ), written as $N/Z$ . For light nuclides (atomic number $Z \le 20$ ), an equal number of protons and neutrons ( $N/Z = 1.0$ ) provides optimal stability, as seen in $\ce{^{12}_{6}C}$ and $\ce{^{16}_{8}O}$ . As $Z$ increases above 20, electrostatic repulsion between protons grows rapidly, requiring a higher proportion of neutrons to stabilize the nucleus. Consequently, stable heavy nuclides have $N/Z$ ratios approaching $1.5$ (for example, $\ce{^{209}_{83}Bi}$ ).

Nuclear Stability Boundary Rule: Every element with an atomic number

Z > 83

(beyond Bismuth) has no stable isotopes. Heavy nuclei are too large for the strong nuclear force to counteract electrostatic repulsion, making them radioactive regardless of their

N/Z

ratio.

The Belt of Stability

When stable nuclides are plotted on a graph of neutron number ( $N$ ) versus proton number ( $Z$ ), they form a narrow region known as the belt of stability (or band of stability).

At low atomic numbers ( $Z \le 20$ ), the belt falls directly along the line $N = Z$ . As atomic number increases, the belt bends upward away from the $N = Z$ line, illustrating the demand for additional neutrons to maintain stability in heavier elements.

The belt of stability: where a nuclide sits tells you how it will decay — Plotting neutrons against protons, the stable nuclides fall in a narrow band called the belt of stability. The belt follows a one-to-one ratio for light elements and then bends above it, because proton-proton repulsion acts across the whole nucleus while the attractive strong force only reaches nearby neighbours, so a heavy nucleus needs surplus neutrons to hold together. Where an unstable nuclide sits relative to the belt predicts how it decays. Above the belt there are too many neutrons, so a neutron converts into a proton and the nuclide emits a beta particle, raising the atomic number by one. Below the belt there are too few neutrons, so a proton converts into a neutron by positron emission or electron capture, lowering the atomic number by one. Beyond atomic number 83 no nuclide is stable at all, and these heavy nuclei shed an alpha particle to lose four mass units and two protons at once.

Predicting Decay Modes Relative to the Belt

Unstable nuclides lie outside the belt of stability and spontaneously undergo radioactive decay to adjust their $N/Z$ ratio toward the stable region:

Above the Belt (Neutron-Rich, High $N/Z$ ): These nuclides have too many neutrons relative to protons. They decay via beta ( $\beta^-$ ) emission, in which a neutron converts into a proton and an electron (beta particle): $\ce{^{1}_{0}n -> ^{1}_{1}p + ^{0}_{-1}e}$ This process decreases $N$ by 1 and increases $Z$ by 1, lowering the $N/Z$ ratio back toward the belt.
Below the Belt (Proton-Rich, Low $N/Z$ ): These nuclides have too many protons relative to neutrons. They decay via positron ( $\beta^+$ ) emission or electron capture: $\text{Positron emission: } \ce{^{1}_{1}p -> ^{1}_{0}n + ^{0}_{1}e}$ $\text{Electron capture: } \ce{^{1}_{1}p + ^{0}_{-1}e -> ^{1}_{0}n}$ Both pathways increase $N$ by 1 and decrease $Z$ by 1, raising the $N/Z$ ratio up toward the belt.
Beyond Bismuth ( $Z > 83$ ): Large, heavy nuclei undergo alpha decay to reduce total mass (alpha decay is covered in detail in its dedicated lesson).

Position Relative to Belt	Nuclear Condition	Primary Decay Mode	Nuclear Conversion Reaction	Effect on $N/Z$ Ratio
Above Belt	Neutron-rich (high $N/Z$ )	$\beta^-$ emission	$\ce{^{1}_{0}n -> ^{1}_{1}p + ^{0}_{-1}e}$	Decreases $N/Z$
Below Belt	Proton-rich (low $N/Z$ )	Positron ( $\beta^+$ ) emission / Electron Capture	$\ce{^{1}_{1}p + ^{0}_{-1}e -> ^{1}_{0}n}$	Increases $N/Z$
Beyond Belt ( $Z > 83$ )	Excessive mass	Alpha ( $\alpha$ ) decay	Emits $\ce{^{4}_{2}He}$ nucleus	Reduces total nuclear mass

DAT Memory Trick:
• ABove the belt = Beta minus decay (Adds a proton, lowers

N/Z

).
• BElow the belt = Positron / Electron capture (Erases a proton, raises

N/Z

Worked Examples and DAT Exam Traps

Worked Example 1 (Above the Belt): Predict the mode of radioactive decay for Carbon-14 ( $\ce{^{14}_{6}C}$ ).

Step 1: Calculate $N$ and $Z$ . Atomic number $Z = 6$ (protons). Mass number $A = 14$ . Neutrons $N = 14 - 6 = 8$ .
Step 2: Calculate $N/Z$ . $N/Z = 8 / 6 = 1.33$ .
Step 3: Evaluate against the Belt. For light elements ( $Z = 6$ ), stable isotopes have $N/Z \approx 1.0$ (e.g., $\ce{^{12}_{6}C}$ ). Since $1.33 > 1.0$ , $\ce{^{14}_{6}C}$ lies above the belt.
Step 4: Determine decay mode & product. It undergoes $\beta^-$ decay: $\ce{^{14}_{6}C -> ^{14}_{7}N + ^{0}_{-1}e}$ Product: Nitrogen-14 ( $\ce{^{14}_{7}N}$ ), which has $N/Z = 7/7 = 1.0$ (stable).

Worked Example 2 (Below the Belt): Predict the mode of decay for Carbon-10 ( $\ce{^{10}_{6}C}$ ).

Step 1: Calculate $N$ and $Z$ . Protons $Z = 6$ , Neutrons $N = 10 - 6 = 4$ .
Step 2: Calculate $N/Z$ . $N/Z = 4 / 6 = 0.67$ .
Step 3: Evaluate against the Belt. $0.67 < 1.0$ , so $\ce{^{10}_{6}C}$ lies below the belt.
Step 4: Determine decay mode & product. It undergoes positron emission: $\ce{^{10}_{6}C -> ^{10}_{5}B + ^{0}_{1}e}$ Product: Boron-10 ( $\ce{^{10}_{5}B}$ ), which has $N/Z = 5/5 = 1.0$ (stable).

Common DAT Exam Trap: Students often calculate

N/Z

using mass number

A

instead of neutron number

N

(writing

A/Z

instead of

N/Z

). Remember:

N = A - Z

. Always subtract the atomic number from the mass number before evaluating position relative to the belt!

Critical Thinking Questions

Why does the belt of stability curve upward away from the $N = Z$ line as atomic number increases beyond $Z = 20$ ?
An unknown radioisotope has $Z = 19$ and mass number $A = 40$ . Calculate its $N/Z$ ratio, determine whether it lies above or below the belt of stability, and write the balanced nuclear equation for its decay.
Both positron emission and electron capture achieve the same net change in atomic composition. How does the position relative to the belt of stability indicate that one of these processes will occur, and what determines which process takes place?

References

Sources returned by Google Search grounding during lesson verification.

21.2: Patterns of Nuclear Stability - Chemistry LibreTexts — chem.libretexts.org
2.5 The Belt of Stability - Predicting the Type of Radioactivity - Chemistry LibreTexts — chem.libretexts.org

Nuclear Stability and the Belt of Stability

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