krit.club logo

Nuclei - Size of the Nucleus

Grade 12CBSEPhysics

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

🔑Concepts

•

The size of the nucleus was first estimated using Rutherford's α\alpha-particle scattering experiment, specifically through the concept of the 'distance of closest approach'.

•

It has been found experimentally that the volume of a nucleus is directly proportional to its mass number AA. Therefore, V=43πR3∝AV = \frac{4}{3} \pi R^3 \propto A.

•

The nuclear radius RR is related to the mass number AA by the empirical formula R=R0A1/3R = R_0 A^{1/3}, where R0R_0 is a constant approximately equal to 1.2×10−15 m1.2 \times 10^{-15} \text{ m} (or 1.2 fm1.2 \text{ fm}).

•

Nuclear density is the mass per unit volume of the nucleus. Because both mass and volume are proportional to AA, the density ρ\rho is independent of the mass number AA.

•

The density of nuclear matter is extremely high, approximately 2.3×1017 kg m−32.3 \times 10^{17} \text{ kg m}^{-3}, which is nearly constant for all nuclei regardless of their size.

📐Formulae

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}

V=43πR3=43πR03AV = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R_0^3 A

ρ=Mass of nucleusVolume of nucleus=mA43πR03A=3m4πR03\rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{m A}{\frac{4}{3} \pi R_0^3 A} = \frac{3m}{4 \pi R_0^3}

💡Examples

Problem 1:

Calculate the radius of the Aluminium nucleus 1327Al^{27}_{13}\text{Al}. Take R0=1.2 fmR_0 = 1.2 \text{ fm}.

Solution:

Given mass number A=27A = 27 and R0=1.2×10−15 mR_0 = 1.2 \times 10^{-15} \text{ m}. Using the formula R=R0A1/3R = R_0 A^{1/3}: R=1.2×10−15×(27)1/3R = 1.2 \times 10^{-15} \times (27)^{1/3} R=1.2×10−15×3R = 1.2 \times 10^{-15} \times 3 R=3.6×10−15 m=3.6 fmR = 3.6 \times 10^{-15} \text{ m} = 3.6 \text{ fm}

Explanation:

The radius is determined by substituting the cube root of the mass number into the nuclear radius empirical relation.

Problem 2:

Two nuclei have mass numbers in the ratio 1:81:8. What is the ratio of their nuclear densities?

Solution:

Nuclear density ρ\rho is given by: ρ=3m4πR03\rho = \frac{3m}{4 \pi R_0^3} Since this expression does not contain the mass number AA, the density is independent of AA. Therefore, the ratio of their densities is: 1:11:1

Explanation:

Even though the volume and mass change with AA, they change proportionally, leaving the density constant for all nuclei.

Problem 3:

Calculate the difference in mass numbers between two nuclei if the radius of the first is 4.8 fm4.8 \text{ fm} and the second is 3.6 fm3.6 \text{ fm}. Use the vertical arithmetic method to find the difference in mass numbers if A1=64A_1 = 64 and A2=27A_2 = 27.

Solution:

Given A1=64A_1 = 64 and A2=27A_2 = 27. The difference in mass numbers is: 64−2737\begin{array}{r} 64 \\ - 27 \\ \hline 37 \end{array}

Explanation:

The mass number difference is calculated by simple subtraction using the provided values.