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Question

An unknown nucleus has a nuclear density of $2.29 \times 10^{17} \text{ kg/m}^3$ and mass of $19.926 \times 10^{-27} \text{ kg}$. Its mass number A is approximately :
(Take $R_0 = 1.2 \times 10^{-15} \text{ m}$, $4\pi = 12.56$)

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
12

Nuclear Density Calculation Steps

To find the approximate mass number (A) of the nucleus, we use the relationship between nuclear density ($\rho$), the mass of the nucleus ($m$), and the nuclear radius ($R$).

Nuclear Density Formula

Nuclear density is defined as mass per unit volume:

$ \rho = \frac{m}{V} $

The volume ($V$) of a nucleus is approximated as a sphere with radius $R$:

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

The nuclear radius ($R$) is related to the mass number ($A$) by the empirical formula:

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

Where $R_0$ is a constant ($1.2 \times 10^{-15} \text{ m}$).

Combining Formulas

Substituting the expression for $R$ into the volume formula:

$ V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A $

Now, substitute this volume into the density formula:

$ \rho = \frac{m}{\frac{4}{3}\pi R_0^3 A} $

Calculating Mass Number A

Rearrange the formula to solve for the mass number ($A$):

$ A = \frac{m}{\rho \frac{4}{3}\pi R_0^3} $

Given values:

  • Mass ($m$) = $19.926 \times 10^{-27} \text{ kg}$
  • Nuclear density ($\rho$) = $2.29 \times 10^{17} \text{ kg/m}^3$
  • $R_0 = 1.2 \times 10^{-15} \text{ m}$
  • $4\pi \approx 12.56$

Calculate the term $\frac{4}{3}\pi R_0^3$:

$ \frac{4}{3}\pi R_0^3 \approx \frac{1}{3} \times (12.56) \times (1.2 \times 10^{-15} \text{ m})^3 $ $ \frac{4}{3}\pi R_0^3 \approx \frac{12.56}{3} \times (1.728 \times 10^{-45} \text{ m}^3) $ $ \frac{4}{3}\pi R_0^3 \approx 4.1867 \times 1.728 \times 10^{-45} \text{ m}^3 \approx 7.235 \times 10^{-45} \text{ m}^3 $

Now substitute all values into the formula for $A$:

$ A = \frac{19.926 \times 10^{-27} \text{ kg}}{(2.29 \times 10^{17} \text{ kg/m}^3) \times (7.235 \times 10^{-45} \text{ m}^3)} $ $ A = \frac{19.926 \times 10^{-27}}{2.29 \times 7.235 \times 10^{17 - 45}} $ $ A = \frac{19.926 \times 10^{-27}}{16.568 \times 10^{-28}} $ $ A = \frac{19.926}{1.6568} \approx 12.026 $

Conclusion

The calculated mass number $A$ is approximately 12.026. Rounding to the nearest integer gives $A = 12$. This corresponds to Option 1.

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