Assume that the mass of a nucleus is approximately given by $M = A m_{p}$ , where $A$ is the mass number. Estimate the density of matter in $k g m^{- 3}$ inside a nucleus. What is the specific gravity of nuclear matter?

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Solution

$M = A m_{p}, f = M / V, m_{p} = 1.007276 u$
$R = R_{g} A^{1 / 3} = 1.1 \times 10^{- 15} A^{1 / 3}, u = 1.6605402 \times 10^{- 27} k g$
$= \frac{A \times 1.007276 \times 1.6605402 \times 10^{- 27}}{\frac{4}{3} \times 3.14 \times R^{3}} = 0.300159 \times 10^{18} = 3 \times 10^{17} k g / m^{3}$
The specific gravity of the nuclear matter $\frac{D e n s i t y o f m a t t e r}{D e n s i t y o f w a t e r}$
$∴ S p e c i f i c g r a v i t y = \frac{3 \times 10^{17}}{10^{3}} = 3 \times 10^{14} k g / m^{3}$

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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?

M=Amp,f=M/V,mp=1.007276uR=RgA1/3=1.1×10−15A1/3,u=1.6605402×10−27kg=A×1.007276×1.6605402×10−2743×3.14×R3=0.300159×1018=3×1017kg/m3The specific gravity of the nuclear matter Density of matterDensity of water∴Specific gravity=3×1017103=3×1014kg/m3

Assume that the mass of a nucleus is approximately given by $M = A m_{p}$ , where $A$ is the mass number. Estimate the density of matter in $k g m^{- 3}$ inside a nucleus. What is the specific gravity of nuclear matter?