Atoms & Nuclei

JEE Main level, numerical answer, medium

171 of 239

Question 171 of 239

Assume that protons and neutrons have equal masses. Mass of a nucleon is 1.6×10−27kg1.6\times {10}^{-27}kg and radius of nucleus is 1.5×10−15A13m1.5\times {10}^{-15}{A}^{\dfrac{1}{3}}m. The approximate ratio of the nuclear density and water density is n×1013n\times {10}^{13}. The value of nn is _____.

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Answer: 11

As we know, density of nuclei=mass of nucleivolume of nuclei\text{density of nuclei}=\dfrac{\text{mass of nuclei}}{\text{volume of nuclei}}
⇒ρ=1.6×10−27A43π(1.5×10−15)3A\Rightarrow \rho =\dfrac{1.6\times {10}^{-27}A}{\dfrac{4}{3}\pi {\left(1.5\times {10}^{-15}\right)}^{3}A}
=1.6×10−2714.14×10−45=0.113×1018kgm−3=\dfrac{1.6\times {10}^{-27}}{14.14\times {10}^{-45}} =0.113\times {10}^{18}kg{m}^{-3}
We also know that, density of water is ρw=103kgm−3{\rho }_{w}={10}^{3}kg{m}^{-3}.
Hence,
ρρw=11.31×1013\dfrac{\rho }{{\rho }_{w}}=11.31\times {10}^{13}.
Therefore, n≈11n\approx 11

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