The radius of ^27_13Al is 3.6 fm. The radius of ^125_52Te is about

Class 12 Physics · NCERT notes for NEET

Nuclei: NCERT Notes, Formulas and Tricks for NEET

What a nucleus is made of, how big it is, why it holds together (binding energy) and how fission and fusion release energy. NEET questions here are usually on binding energy per nucleon, nuclear radius or a Q-value.

1. Nuclear size, mass defect and binding energy
2. Nuclear fission and fusion 🔒

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Topic 1

Nuclear size, mass defect and binding energy

A teaspoon of pure nuclear matter would weigh about a billion tonnes, and every nucleus, from hydrogen to uranium, has this same density.

A nucleus has $Z$ protons and $N$ neutrons (together called nucleons). Mass number $A = Z + N$ ; we write $_{Z}^{A} X$ , e.g. $_{8}^{16} O$ .
Atomic mass unit: $1 u = \frac{1}{12}$ of the mass of a $^{12} C$ atom $= 1.660539 \times 1 0^{- 27} kg$ , and its energy equivalent is $931.5 MeV$ .
Radius $R = R_{0} A^{1/3}$ with $R_{0} = 1.2 fm$ . So volume $\propto A$ and nuclear density $\approx 2.3 \times 1 0^{17} kg m^{- 3}$ is the same for all nuclei.
Mass defect: a nucleus weighs less than its separate nucleons, $Δ M = Z m_{p} + (A - Z) m_{n} - M$ . Binding energy $E_{b} = Δ M c^{2}$ is the energy needed to pull it fully apart.
NCERT example: $_{8}^{16} O$ has $Δ M = 0.13691 u$ , so $E_{b} = 127.5 MeV$ , i.e. about $7.97 MeV$ per nucleon.
Binding energy per nucleon $E_{bn} = E_{b} / A$ measures stability. It is small for light nuclei, nearly constant ( $\approx 8 MeV$ ) for $30 < A < 170$ , maximum $\approx 8.75 MeV$ at $A \approx 56$ and about $7.6 MeV$ at $A = 238$ .
Nuclear force: strongest force, short range (a few fm), attractive beyond about $0.8 fm$ and repulsive closer, nearly the same for p-p, n-n and n-p, and saturating (each nucleon pulls only its neighbours).
Saturation is why $E_{bn}$ stays flat in the middle; Coulomb repulsion between many protons is why it drops for heavy nuclei.

E_b/A climbs fast for light nuclei, peaks near iron (A approx 56) and falls slowly to 7.6 MeV for uranium. ^4He sits well above its neighbours. — $E_{b} / A$ climbs fast for light nuclei, peaks near iron ( $A \approx 56$ ) and falls slowly to 7.6 MeV for uranium. $^{4} He$ sits well above its neighbours.

Potential energy of two nucleons: the minimum at r_0 approx 0.8 fm is where force is zero. Closer: strong repulsion; farther: attraction that vanishes within a few fm. — Potential energy of two nucleons: the minimum at $r_{0} \approx 0.8$ fm is where force is zero. Closer: strong repulsion; farther: attraction that vanishes within a few fm.

Formulas

$R = R_{0} A^{1/3}, R_{0} = 1.2 \times 1 0^{- 15} m$
Radius depends only on mass number
$ρ = \frac{3 m _{N}}{4 π R _{0}^{3}} \approx 2.3 \times 1 0^{17} kg m^{- 3}$
Nuclear density, independent of $A$ ( $m_{N}$ = nucleon mass)
$Δ M = Z m_{p} + (A - Z) m_{n} - M$
Mass defect; $M$ = nuclear mass
$E_{b} = Δ M c^{2} = Δ M (in u) \times 931.5 MeV$
Binding energy
$E_{bn} = \frac{E _{b}}{A}$
Binding energy per nucleon; higher means more stable
$E = m c^{2}$
Mass-energy equivalence; $1 u \leftrightarrow 931.5 MeV$

Tips, tricks and hacks

Cube-root radius ratio

R_{1} / R_{2} = (A_{1} / A_{2})^{1/3}

. Choose nuclei with cube ratios:

^{27} Al

^{125} Te

gives

(125/27)^{1/3} = 5/3

. Volume ratio is simply

A_{1} / A_{2}

931.5 shortcutMass defect in u times 931.5 gives MeV directly; no need for

c^{2}

in SI. Deuteron:

Δ M = 0.00239 u

gives

\approx 2.22 MeV

Density is a constantAny question asking density ratio of two nuclei: answer

1 : 1

. Mass ratio

= A_{1} : A_{2}

, radius ratio

= (A_{1} : A_{2})^{1/3}

Stability = height on curveCompare stability with

E_{b} / A

, not

E_{b}

^{238} U

has

E_{b} \approx 1800

MeV (far more than Fe's 490 MeV) but is less stable since

7.6 < 8.75

Common mistakes

✗ Larger total binding energy means a more stable nucleus.

✓ Stability depends on binding energy per nucleon; heavy nuclei have big $E_{b}$ but lower $E_{b} / A$ .

✗ Nuclear radius is proportional to $A$ , so a nucleus 8 times heavier is 8 times larger.

✓ $R \propto A^{1/3}$ : 8 times the mass gives only 2 times the radius.

✗ Mass defect means the nucleons lose some mass when they are inside.

✓ The bound system as a whole has less energy; the missing mass equals the binding energy released when it formed.

✗ Nuclear force is stronger between two protons because they are charged.

✓ Nuclear force is charge independent; p-p, n-n and n-p are about equal. Only the Coulomb part differs.

NEET traps

"Nuclear density increases with mass number": FALSE. Density is the same ( $\sim 1 0^{17} kg m^{- 3}$ ) for all nuclei.

"Binding energy is the energy released when nucleons combine" and "energy needed to separate them" are both correct statements of the same quantity.

NEET may give atomic masses (with electrons); when $m_{H}$ is used in place of $m_{p}$ , the electron masses cancel and the answer is the same.

Quick check: The radius of

_{13}^{27} Al

is 3.6 fm. The radius of

_{52}^{125} Te

is about

6.0 fm
16.7 fm
4.6 fm
8.0 fm

Answer: A. $R \propto A^{1/3}$ : $(125/27)^{1/3} = 5/3$ , so $R = 3.6 \times 5/3 = 6.0$ fm.

Topic 2 · 🔒

Nuclear fission and fusion

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