An unknown nucleus has a nuclear density of 2.29 × 10¹ kg/m³ and mass of 19.926 × 10² kg. Its mass number A is approximately: (Take R₀ =1.2×10⁻¹⁵ m, 4π=12.56)

**Step 1 – Write the relation between mass, density and volume**

The mass \(M\) of a nucleus is related to its density \(\rho\) and its geometric volume \(V\) by

$M\; =\; \backslash rho\; \backslash ,\; V$

where the volume of a spherical nucleus of radius \(R\) is

$V\; =\; \backslash frac\{4\}\{3\}\backslash ,\backslash pi\; R^\{3\}$

**Step 2 – Express the radius through the mass number \(A\)**

For nuclei the empirical radius law is

$R\; =\; R\_\{0\}\backslash ,A^\{1/3\}$

with the constant \(R_{0}=1.2\times10^{-15}\,\text{m}\).

**Step 3 – Substitute the radius into the volume**

Insert \(R = R_{0}A^{1/3}\) into the volume formula:

$V\; =\; \backslash frac\{4\}\{3\}\backslash ,\backslash pi\; (R\_\{0\}A^\{1/3\})^\{3\}\; =\; \backslash frac\{4\}\{3\}\backslash ,\backslash pi\; R\_\{0\}^\{3\}\backslash ,A$

**Step 4 – Relate the mass to the mass number**

Now the mass becomes

$M\; =\; \backslash rho\; \backslash ,\backslash frac\{4\}\{3\}\backslash ,\backslash pi\; R\_\{0\}^\{3\}\backslash ,A$

Solving for \(A\) gives

$A\; =\; \backslash frac\{M\}\{\backslash rho\; \backslash ,\; \backslash dfrac\{4\}\{3\}\backslash ,\backslash pi\; R\_\{0\}^\{3\}\}$

**Step 5 – Insert the numerical values**

Given data (written in standard scientific notation for nuclear scales):

• \(\rho = 2.29 \times 10^{17}\ \text{kg·m}^{-3}\)
• \(M = 19.926 \times 10^{-27}\ \text{kg}\)
• \(R_{0}=1.2 \times 10^{-15}\ \text{m}\)
• \(4\pi = 12.56\) (so \(\dfrac{4}{3}\pi = \dfrac{12.56}{3}=4.1867\))

First compute the denominator:

$\backslash rho\; \backslash ,\backslash frac\{4\}\{3\}\backslash ,\backslash pi\; R\_\{0\}^\{3\}\; =\; (2.29\backslash times10^\{17\})\backslash times(4.1867)\backslash times(1.2\backslash times10^\{-15\})^\{3\}$

Calculate \(R_{0}^{3}\):

$(1.2\backslash times10^\{-15\})^\{3\}=1.728\backslash times10^\{-45\}\backslash \; \backslash text\{m\}^\{3\}$

Now multiply:

$2.29\backslash times10^\{17\}\backslash times4.1867\backslash times1.728\backslash times10^\{-45\}\; =\; 1.658\backslash times10^\{-27\}\backslash \; \backslash text\{kg\}$

**Step 6 – Obtain the mass number**

Insert the mass of the nucleus:

$A\; =\; \backslash frac\{19.926\backslash times10^\{-27\}\backslash \; \backslash text\{kg\}\}\{1.658\backslash times10^\{-27\}\backslash \; \backslash text\{kg\}\}\; \backslash approx\; 12.0$

**Conclusion**

The calculated mass number is approximately 12, which matches the given correct option.

The correct answer is 12.