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Atoms and Nuclei - Composition and Size of Nucleus

Grade 12ICSEPhysics

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

🔑Concepts

The nucleus is the central, positively charged part of an atom containing protons and neutrons, collectively known as nucleons. The number of protons is the atomic number ZZ, and the total number of nucleons is the mass number AA.

Atomic Mass Unit (uu): It is defined as 112\frac{1}{12} of the mass of a carbon-12 atom (612C^{12}_{6}C). 1u1.660539×1027 kg1 u \approx 1.660539 \times 10^{-27} \text{ kg}.

Nuclear Radius (RR): The volume of a nucleus is proportional to its mass number AA. The radius RR is given by the empirical relation R=R0A1/3R = R_0 A^{1/3}, where R01.2×1015 mR_0 \approx 1.2 \times 10^{-15} \text{ m} (or 1.2 fm1.2 \text{ fm}).

Nuclear Density (ρ\rho): Nuclear density is the mass per unit volume of the nucleus. It is approximately 2.3×1017 kg m32.3 \times 10^{17} \text{ kg m}^{-3} and is independent of the mass number AA. This implies all nuclei have nearly the same density.

Isotopes are atoms with the same ZZ but different AA. Isobars are atoms with the same AA but different ZZ. Isotones are atoms with the same number of neutrons (N=AZN = A - Z).

Einstein's Mass-Energy Equivalence: Mass and energy are interconvertible, expressed as E=mc2E = mc^2. The energy equivalent of 1u1 u is approximately 931.5 MeV931.5 \text{ MeV}.

📐Formulae

R=R0A1/3R = R_0 A^{1/3}

N=AZN = A - Z

ρ=Mass of NucleusVolume of Nucleus=mA43πR3=3m4πR03\rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}} = \frac{m A}{\frac{4}{3} \pi R^3} = \frac{3m}{4 \pi R_0^3}

1u=1.66×1027 kg1 u = 1.66 \times 10^{-27} \text{ kg}

E=mc2E = mc^2

1u931.5 MeV1 u \approx 931.5 \text{ MeV}

💡Examples

Problem 1:

Calculate the ratio of the radii of the nuclei of Aluminum (1327Al^{27}_{13}Al) and Tellurium (52125Te^{125}_{52}Te).

Solution:

Given A1=27A_1 = 27 and A2=125A_2 = 125. Using the formula R=R0A1/3R = R_0 A^{1/3}, the ratio is: R1R2=(A1A2)1/3=(27125)1/3=35\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3} = \left( \frac{27}{125} \right)^{1/3} = \frac{3}{5}.

Explanation:

The radius of a nucleus is directly proportional to the cube root of its mass number. By taking the ratio of A1/3A^{1/3} for both elements, we find the relative sizes.

Problem 2:

Show that the density of the nucleus is independent of the mass number AA.

Solution:

Density ρ=MassVolume\rho = \frac{\text{Mass}}{\text{Volume}}. Let mm be the average mass of a nucleon. Total mass mA\approx mA. Volume V=43πR3=43π(R0A1/3)3=43πR03AV = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A. Therefore, ρ=mA43πR03A=3m4πR03\rho = \frac{mA}{\frac{4}{3} \pi R_0^3 A} = \frac{3m}{4 \pi R_0^3}.

Explanation:

Since mm and R0R_0 are constants, the resulting expression for density does not contain AA, proving that nuclear density is constant for all elements.

Composition and Size of Nucleus - Revision Notes & Key Formulas | ICSE Class 12 Physics