Q10 (2024)
An unknown nucleus has a nuclear density of $2.29 \times 10^{17} \, \mathrm{kg/m}^{3}$ and mass of $19.926 \times 10^{-27} \, \mathrm{kg}$ . Its mass number A is approximately: (Take $R_{0} = 1.2 \times 10^{-15} \, \mathrm{m}, 4\pi = 12.56$ )
- 12
- 19
- 20
- 16
### Related Formula
$$\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$$
$R = R_0 A^{1/3}$
### Core Logic
$$A = \frac{M}{\rho} \times \frac{3}{4\pi R_0^3}$$
$$A = \frac{19.926 \times 10^{-27} \times 3}{2.29 \times 10^{17} \times 12.56 \times (1.2 \times 10^{-15})^3} \approx 12$$
### Pattern Recognition
Relation between mass, volume, nuclear radius, and mass number.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
