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Nuclei - Nuclear Force

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The nuclear force is the strong attractive force that binds protons and neutrons (nucleons) together inside a nucleus, overcoming the strong electrostatic repulsion between protons.

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It is the strongest known fundamental force in nature, approximately 100100 times stronger than the electromagnetic force.

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The nuclear force is a short-range force, effective only over distances of about 22 to 3 fm3 \text{ fm} (1 fm=10−15 m1 \text{ fm} = 10^{-15} \text{ m}). Its strength drops to zero rapidly beyond this range.

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The force is charge-independent; the nuclear force between (n−n)(n-n), (p−p)(p-p), and (n−p)(n-p) is approximately the same, provided the nucleons are in the same state.

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The potential energy of a pair of nucleons is a function of their separation rr. It is minimum at r0≈0.8 fmr_0 \approx 0.8 \text{ fm}. The force is attractive for r>r0r > r_0 and becomes strongly repulsive for r<r0r < r_0.

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Nuclear forces show saturation: a nucleon interacts only with its immediate neighbors rather than with all nucleons in the nucleus.

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Unlike gravitational or electrostatic forces, the nuclear force is a non-central force and depends on the orientation of the spins of the nucleons.

📐Formulae

R=R0A1/3 (where R0≈1.2×10−15 m)R = R_0 A^{1/3} \text{ (where } R_0 \approx 1.2 \times 10^{-15} \text{ m)}

ρ=m⋅A43π(R0A1/3)3=3m4πR03≈2.3×1017 kg/m3\rho = \frac{m \cdot A}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{3m}{4\pi R_0^3} \approx 2.3 \times 10^{17} \text{ kg/m}^3

Δm=[Zmp+(A−Z)mn]−Mnucleus\Delta m = [Z m_p + (A - Z) m_n] - M_{nucleus}

BE=Δm⋅c2=Δm (in amu)×931.5 MeVBE = \Delta m \cdot c^2 = \Delta m \text{ (in amu)} \times 931.5 \text{ MeV}

💡Examples

Problem 1:

Two nuclei have mass numbers in the ratio 1:271:27. What is the ratio of their (i) nuclear radii and (ii) nuclear densities?

Solution:

(i) Since R=R0A1/3R = R_0 A^{1/3}, the ratio of radii is R1R2=(A1A2)1/3=(127)1/3=13\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3}. (ii) Nuclear density ρ\rho is independent of mass number AA, so the ratio is 1:11:1.

Explanation:

The radius of a nucleus scales with the cube root of the mass number, while the volume scales linearly with the mass number, resulting in a constant density across all nuclei.

Problem 2:

Explain why the nuclear force must be repulsive at very short distances (r<0.8 fmr < 0.8 \text{ fm}).

Solution:

If the nuclear force were only attractive at all distances, the nucleons would collapse into each other, leading to a nucleus of infinite density. The repulsion at r<0.8 fmr < 0.8 \text{ fm} ensures the stability and finite size of the nucleus.

Explanation:

This repulsion is represented by the steep rise in the potential energy curve as the distance rr approaches zero.

Nuclear Force - Revision Notes & Key Formulas | CBSE Class 12 Physics