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Question: If the distance between nuclei is \(2 \times {10^{ - 13}}cm\), the density of the nuclear material i...

If the distance between nuclei is 2×10−13cm2 \times {10^{ - 13}}cm, the density of the nuclear material is (mn=10−27kg)\left( {{m_n} = {{10}^{ - 27}}kg} \right)
A. 3.21×10−12 kg/m33.21 \times {10^{ - 12}}{\text{ kg/}}{{\text{m}}^3}
B. 1.6×10−3 kg/m31.6 \times {10^{ - 3}}{\text{ kg/}}{{\text{m}}^3}
C. 2×109 kg/m32 \times {10^9}{\text{ kg/}}{{\text{m}}^3}
D. 3×1016 kg/m33 \times {10^{16}}{\text{ kg/}}{{\text{m}}^3}

Explanation

Solution

In this question, we need to determine the density of the nuclear material such that the distance between nuclei is 2×10−13cm2 \times {10^{ - 13}}cm as the mass of the nuclei is mn=10−27kg{m_n} = {10^{ - 27}}kg. For this, we will use the relation between the density, mass and volume of the material.

Complete step by step answer:
The ratio of the mass of the element to the volume occupied by the element in the atmosphere results in the density of the element. Mathematically, ρ=mV\rho = \dfrac{m}{V} where, ρ\rho is the density of the material of mass ‘m’ and ‘V’ be the volume of the material in the surrounding. The SI unit of the density of the material is kilograms per cubic meters where mass should be in kilograms and volume should be cubic meters.
Here, in the question, the distance between the nuclei is 2×10−13cm2 \times {10^{ - 13}}cm so, we need to convert it into the SI units. To convert centimetres into meters, we need to divide the absolute value by the factor 100. So,
2×10−13cm=2×10−13100m =2×10−15m  2 \times {10^{ - 13}}cm = \dfrac{{2 \times {{10}^{ - 13}}}}{{100}}m \\\ = 2 \times {10^{ - 15}}m \\\
As, the nuclei are surrounded by nuclei so, this distance will act as the radius of the sphere in the nucleus. So, r=2×10−15mr = 2 \times {10^{ - 15}}m
Now, the volume of the sphere is given as 4π3\dfrac{{4\pi }}{3} times the cube of the radius. Mathematically, V=4πr33V = \dfrac{{4\pi {r^3}}}{3}.
Substitute r=2×10−15mr = 2 \times {10^{ - 15}}m in the equation V=4πr33V = \dfrac{{4\pi {r^3}}}{3} to determine the volume of the spherical (acting) nuclei.
V=4πr33 =4π(2×10−15)33m3 =4π×8×10−453m3 =32π×10−453m3  V = \dfrac{{4\pi {r^3}}}{3} \\\ = \dfrac{{4\pi {{\left( {2 \times {{10}^{ - 15}}} \right)}^3}}}{3}{m^3} \\\ = \dfrac{{4\pi \times 8 \times {{10}^{ - 45}}}}{3}{m^3} \\\ = \dfrac{{32\pi \times {{10}^{ - 45}}}}{3}{m^3} \\\
Now, substitute mass of the nuclei as 10−27kg{10^{ - 27}}kg and volume as 32π×10−453m3\dfrac{{32\pi \times {{10}^{ - 45}}}}{3}{m^3} in the formula ρ=mV\rho = \dfrac{m}{V} to determine the density of the nuclei.
ρ=mV =10−27kg(32π×10−453)m3 =3×10−2732π×10−45kgm3 =3×732×22×10(−27+45)kgm3 =0.0298×1018kgm3 =0.298×1017kgm3 =2.98×1016kgm3 ≈3×1016kgm3  \rho = \dfrac{m}{V} \\\ = \dfrac{{{{10}^{ - 27}}kg}}{{\left( {\dfrac{{32\pi \times {{10}^{ - 45}}}}{3}} \right){m^3}}} \\\ = \dfrac{{3 \times {{10}^{ - 27}}}}{{32\pi \times {{10}^{ - 45}}}}\dfrac{{kg}}{{{m^3}}} \\\ = \dfrac{{3 \times 7}}{{32 \times 22}} \times {10^{\left( { - 27 + 45} \right)}}\dfrac{{kg}}{{{m^3}}} \\\ = 0.0298 \times {10^{18}}\dfrac{{kg}}{{{m^3}}} \\\ = 0.298 \times {10^{17}}\dfrac{{kg}}{{{m^3}}} \\\ = 2.98 \times {10^{16}}\dfrac{{kg}}{{{m^3}}} \\\ \approx 3 \times {10^{16}}\dfrac{{kg}}{{{m^3}}} \\\
Hence, the density of the material approximately equals 3×1016kgm33 \times {10^{16}}\dfrac{{kg}}{{{m^3}}}.

So, the correct answer is “Option D”.

Note:
All the data should be converted into the SI units or in the same units before substituting them in the desired equation.
Moreover, the combination of the nuclei all together will act as a sphere.