The radius R of a nucleus of mass number A can be estimated by the formula $R=\left(1.3 \times 10^{-15}\right) A^{1 / 3} \mathrm{~m}$. It follows that the mass density of a nucleus is of the order of: $\left(M_{\text {prot }}=M_{\text {neut. }} \simeq 1.67 \times 10^{-27} \mathrm{~kg}\right)$

(B) $10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$

(A) $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$

(C) $10^{10} \mathrm{~kg} \mathrm{~m}^{-3}$

(D) $10^{24} \mathrm{~kg} \mathrm{~m}^{-3}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Modern Physics

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

R = (1.3 \times 1 0^{- 15}) A^{1/3} m

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

(M_{prot} = M_{neut.} ≃ 1.67 \times 1 0^{- 27} kg)

Answer: (B)

Nuclear density is mass over volume,

ρ = \frac{A M _{nucleon}}{\frac{4}{3} π R ^{3}}

. Writing

R = 1.3 \times 1 0^{- 15} A^{1/3}

makes

R^{3} \propto A

, so the mass number cancels and the density is the same for every nucleus:

ρ = \frac{1.67 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.3 ) ^{3} \times 1 0 ^{- 45}} \approx \frac{1.67 \times 1 0 ^{- 27}}{9.2 \times 1 0 ^{- 45}} \approx 1.8 \times 1 0^{17}

, which is of the order of

1 0^{17} kg m^{- 3}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option