Radius of nucleus may be given as $R_{N} = R_{0} . A^{1 / 3}$ , where $A = m a s s n u m b e r$ and $R_{0} = C o n s t a n t$ . Calculate the density of nucleus of an atom if $R_{0} = 1.2 \times 10^{- 15}$
$[N_{A} = 6 t i m e s 10^{23} \times 8 π \times (1.2)^{3}$ = $1000$

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Solution

$Density of nucleus = \frac{mass of nucleus}{volume of nucleus} = \frac{A}{\frac{4}{3} π R_{N}^{3}} = \frac{A}{\frac{4}{3} π {(R_{\circ} A^{\frac{1}{3}})}^{3}} = \frac{3}{4 π R_{\circ}^{3}} = \frac{3}{4 π (1.2 \times 10^{- 15})^{3}} = 1.4 \times 10^{44} m^{- 3}$

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