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Atoms and Nuclei question

2020 · 3 Sep · Shift 2 · Q54
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Atoms and Nuclei question

2020 · 3 Sep · Shift 2 · Q54

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
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)
  1. A
    1024 kg m–3
  2. B
    1010 kg m–3
  3. C
    1017 kg m–3
  4. D
    103 kg m–3
View written solutionFree

Correct answer: C

  1. Given data

    The nuclear radius is R=1.3×10−15A1/3 mR = 1.3\times 10^{-15} A^{1/3}\,\text{m}R=1.3×10−15A1/3m

    Mass of one nucleon: mn≈1.67×10−27 kgm_n \approx 1.67\times 10^{-27}\,\text{kg}mn​≈1.67×10−27kg

    For a nucleus of mass number AAA, its mass is approximately M≈A (1.67×10−27) kgM \approx A\,(1.67\times 10^{-27})\,\text{kg}M≈A(1.67×10−27)kg

  2. Volume of the nucleus

    Treat the nucleus as a sphere: V=43πR3V = \frac{4}{3}\pi R^3V=34​πR3

    Using R=1.3×10−15A1/3R = 1.3\times 10^{-15} A^{1/3}R=1.3×10−15A1/3, R3=(1.3×10−15)3AR^3 = (1.3\times 10^{-15})^3 AR3=(1.3×10−15)3A

    Hence, V=43π(1.3×10−15)3AV = \frac{4}{3}\pi (1.3\times 10^{-15})^3 AV=34​π(1.3×10−15)3A

  3. Density formula

    Density is ρ=MV\rho = \frac{M}{V}ρ=VM​

    Substituting MMM and VVV: ρ=A(1.67×10−27)43π(1.3×10−15)3A\rho = \frac{A(1.67\times 10^{-27})}{\frac{4}{3}\pi (1.3\times 10^{-15})^3 A}ρ=34​π(1.3×10−15)3AA(1.67×10−27)​

    The factor AAA cancels: ρ=1.67×10−2743π(1.3×10−15)3\rho = \frac{1.67\times 10^{-27}}{\frac{4}{3}\pi (1.3\times 10^{-15})^3}ρ=34​π(1.3×10−15)31.67×10−27​

  4. Numerical calculation

    First, (1.3)3=2.197(1.3)^3 = 2.197(1.3)3=2.197 so (1.3×10−15)3=2.197×10−45(1.3\times 10^{-15})^3 = 2.197\times 10^{-45}(1.3×10−15)3=2.197×10−45

    Now, 43π≈4.19\frac{4}{3}\pi \approx 4.1934​π≈4.19

    Therefore, V factor=4.19×2.197×10−45≈9.2×10−45V \text{ factor} = 4.19\times 2.197\times 10^{-45} \approx 9.2\times 10^{-45}V factor=4.19×2.197×10−45≈9.2×10−45

    Thus, ρ≈1.67×10−279.2×10−45\rho \approx \frac{1.67\times 10^{-27}}{9.2\times 10^{-45}}ρ≈9.2×10−451.67×10−27​

    ρ≈0.181×1018=1.8×1017 kg m−3\rho \approx 0.181\times 10^{18} = 1.8\times 10^{17}\,\text{kg m}^{-3}ρ≈0.181×1018=1.8×1017kg m−3

  5. Order of magnitude

    ρ∼1017 kg m−3\rho \sim 10^{17}\,\text{kg m}^{-3}ρ∼1017kg m−3

  6. Option check

    • A: 1024 kg m−310^{24}\,\text{kg m}^{-3}1024kg m−3 ❌
    • B: 1010 kg m−310^{10}\,\text{kg m}^{-3}1010kg m−3 ❌
    • C: 1017 kg m−310^{17}\,\text{kg m}^{-3}1017kg m−3 ✅
    • D: 103 kg m−310^{3}\,\text{kg m}^{-3}103kg m−3 ❌

Therefore, the correct option is C.

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