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JEE Main 2020 (3 Sep, Shift 2) · Nucleus: Composition, Size & Nuclear Force

Nucleus: Composition, Size & Nuclear Force question from JEE Main 2020 (3 Sep, Shift 2)

Q24JEE Main 2020 · 3 SEP · SHIFT 2Medium

Nucleus: Composition, Size & Nuclear Force

The radius R of a nucleus of mass number A can be estimated by the formula

R = (1.3 ×\times 10–15)A1/3 m.

It follows that the mass density of a nucleus is of the order of:

(Mprot. ≅\cong Mneut ≃\simeq 1.67 ×\times 10–27 kg)

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Answer: (c)

  1. ρ=Amp43πR03A=mp43πR03\rho = \dfrac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \dfrac{m_p}{\frac{4}{3}\pi R_0^3}, independent of AA
  2. ρ=1.67×10−2743π(1.3×10−15)3=1.67×10−279.2×10−45≈1.8×1017 kg/m3\rho = \dfrac{1.67 \times 10^{-27}}{\frac{4}{3}\pi (1.3 \times 10^{-15})^3} = \dfrac{1.67 \times 10^{-27}}{9.2 \times 10^{-45}} \approx 1.8 \times 10^{17}\ \text{kg/m}^3
  3. So the answer is (c) 1017 kg m−310^{17}\ \text{kg m}^{-3}.
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