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Answer 1

Given that:

A = 56

Iron nuclei mass = 55.85 u

Answer:

$\large\rm { = 2.30 × 10^{17} kg/m^{3}}$

Formula to be used:

$\large\rm { r = r_{0} × A^{⅓} }$

and

$\large\rm { ( \rho_{nucleus} ) }$ = $\large\rm { \frac{m}{\frac{4}{3} \pi r^{2}} }$

Step by step explanation:-

Radius ( by Rutherford) is given as :

$\large\rm { r = r_{0} × A^{⅓} }$

So, $\large\rm { r = 1.2 × 10^{-15} × 56^{⅓}}$

$\large\rm { = 4.59 × 10^{-15} m }$

mass of iron nucleus = $\large\rm { 9.327 × 10^{-26} kg}$

so, nuclear density $\large\rm { ( \rho_{nucleus} ) }$ = $\large\rm { \frac{m}{\frac{4}{3} \pi r^{2}} }$

$\large\rm { = \frac{9.327×10^{-26}}{4.05×10^{-43}} }$

$\large\rm { = 2.30 × 10^{17} kg/m^{3}}$

Answer 2

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