Question

An unknown nucleus has a nuclear density of 2.29 $\times$ 10¹⁷ kg/m³ and mass of 19.926 $\times$ 10⁻²⁷ kg. Its mass number A is approximately: (Take $R_0 = 1.2 \times 10^{-15}$ m, $4\pi = 12.56$)

• A. 12
• B. 16
• C. 19
• D. 20

Hint:

The nuclear density is constant for all nuclei. If you are given the total mass, simply divide by $1.66 \times 10^{-27}$ kg to get the mass number immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The mass number $A$ represents the total number of protons and neutrons. We can find it by dividing the total mass of the nucleus by the average mass of a single nucleon (approx. $1.66 \times 10^{-27}$ kg).

Step 2: Key Formula or Approach:
\[ A = \frac{\text{Total Mass of Nucleus}}{\text{Mass of one nucleon } (m_n)} \]

Step 3: Detailed Explanation:
1. Given total mass $M = 19.926 \times 10^{-27}$ kg. 2. Standard mass of one nucleon ($1 \text{ amu}$) $\approx 1.66 \times 10^{-27}$ kg. 3. Calculate $A$: \[ A = \frac{19.926 \times 10^{-27}}{1.66 \times 10^{-27}} \approx 12.003 \] 4. Alternatively, using density $\rho = M/V$: \[ V = \frac{M}{\rho} = \frac{19.926 \times 10^{-27}}{2.29 \times 10^{17}} \approx 8.7 \times 10^{-45} \text{ m}^3 \] 5. Since $V = \frac{4}{3}\pi R_0^3 A$: \[ A = \frac{3V}{4\pi R_0^3} = \frac{3 \times 8.7 \times 10^{-45}}{12.56 \times (1.2 \times 10^{-15})^3} \approx 12 \]

Step 4: Final Answer:
The mass number $A$ is approximately 12.