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Nuclei - Atomic Masses and Composition of Nucleus

Grade 12CBSEPhysics

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

🔑Concepts

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Atomic Mass Unit (uu): Defined as 112\frac{1}{12}th of the mass of an atom of the carbon isotope 12C^{12}\text{C}. 1u≈1.660539×10−27 kg1 u \approx 1.660539 \times 10^{-27} \text{ kg}.

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Nuclear Composition: A nucleus contains 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).

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Neutron Number (NN): Calculated as the difference between the mass number and the atomic number: N=A−ZN = A - Z.

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Isotopes: Nuclei with the same atomic number ZZ but different mass numbers AA (e.g., 11H^{1}_{1}\text{H}, 12H^{2}_{1}\text{H}, and 13H^{3}_{1}\text{H}).

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Isobars: Nuclei with the same mass number AA but different atomic numbers ZZ (e.g., 13H^{3}_{1}\text{H} and 23He^{3}_{2}\text{He}).

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Isotones: Nuclei with the same number of neutrons (NN) but different atomic numbers (ZZ).

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Nuclear Radius (RR): The radius of a nucleus is proportional to the cube root of its mass number AA. R=R0A1/3R = R_0 A^{1/3}, where R0≈1.2×10−15 mR_0 \approx 1.2 \times 10^{-15} \text{ m}.

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Nuclear Density: The density of nuclear matter is independent of the mass number AA and is approximately constant for all nuclei, roughly 2.3×1017 kg m−32.3 \times 10^{17} \text{ kg m}^{-3}.

📐Formulae

1u=1.660539×10−27 kg1 u = 1.660539 \times 10^{-27} \text{ kg}

A=Z+NA = Z + N

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

R0≈1.2×10−15 m=1.2 fmR_0 \approx 1.2 \times 10^{-15} \text{ m} = 1.2 \text{ fm}

Density (ρ)=MassVolume=mA43πR3=3m4πR03\text{Density } (\rho) = \frac{\text{Mass}}{\text{Volume}} = \frac{m A}{\frac{4}{3} \pi R^3} = \frac{3m}{4 \pi R_0^3}

E=mc2E = mc^2

1u≈931.5 MeV/c21 u \approx 931.5 \text{ MeV/c}^2

💡Examples

Problem 1:

Calculate the radius of a nucleus with mass number A=64A = 64. Given R0=1.2 fmR_0 = 1.2 \text{ fm}.

Solution:

R=R0A1/3R = R_0 A^{1/3} R=1.2×(64)1/3R = 1.2 \times (64)^{1/3} R=1.2×4R = 1.2 \times 4 R=4.8 fmR = 4.8 \text{ fm}

Explanation:

Using the nuclear radius formula, we take the cube root of the mass number and multiply by the constant R0R_0.

Problem 2:

Determine the number of protons and neutrons in the nucleus of Gold represented by 79197Au^{197}_{79}\text{Au}.

Solution:

Z=79Z = 79 A=197A = 197 N=A−ZN = A - Z 197−79118\begin{array}{r} 197 \\ -79 \\ \hline 118 \end{array}

Explanation:

The atomic number ZZ represents protons (7979). The mass number AA is 197197. Subtracting ZZ from AA gives the neutron count (118118).

Problem 3:

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

Solution:

ρ=Mass of NucleusVolume of Nucleus\rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}} ρ=m⋅A43π(R0A1/3)3\rho = \frac{m \cdot A}{\frac{4}{3} \pi (R_0 A^{1/3})^3} ρ=m⋅A43πR03A\rho = \frac{m \cdot A}{\frac{4}{3} \pi R_0^3 A} ρ=3m4πR03\rho = \frac{3m}{4 \pi R_0^3}

Explanation:

By substituting R=R0A1/3R = R_0 A^{1/3} into the density formula, the term AA cancels out in the numerator and denominator, proving density is constant for all nuclei.