Assume that the mass of a nucleus is approximately given by M = Am_p where A is the mass number. Estimate the density of matter in kgm⁻³ inside a nucleus. What is the specific gravity of nuclear matter?

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Solution

Given:
Mass of the nucleus, M = Am_p
Volume of the nucleus, V = $\frac{4}{3} π {R_{0}}^{3} A$

Density of the matter, $d = \frac{M}{V} = \frac{A m_{p}}{\frac{4}{3} π {R_{0}}^{3} A}$
$= \frac{3 m_{p}}{4 \times π {R_{0}}^{3}} = \frac{3 \times 1.007276}{4 \times 3.14 (1.1)^{3}} = 3 \times 10^{17} kg / m^{3}$

Specific gravity of the nuclear matter = $\frac{Density of matter}{Density of water}$
$∴$ Specific gravity = $\frac{3 \times 10^{17}}{10^{3}}$ = 3 $\times$ 10¹⁴ kg/m³

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Assume that the mass of a nucleus is approximately given by M = Amp where A is the mass number. Estimate the density of matter in kgm−3 inside a nucleus. What is the specific gravity of nuclear matter?

Given: Mass of the nucleus, M = Amp Volume of the nucleus, V = 43πR03A Density of the matter, d=MV=Amp43πR03A =3mp4×πR03=3×1.0072764×3.14(1.1)3= 3×1017 kg/m3 Specific gravity of the nuclear matter = Density of matter Density of water ∴ Specific gravity = 3×1017103 = 3 × 1014 kg/m3

Assume that the mass of a nucleus is approximately given by M = Am_p where A is the mass number. Estimate the density of matter in kgm⁻³ inside a nucleus. What is the specific gravity of nuclear matter?