Respuesta :
Answer:
1.5 x 10¹⁴
Explanation:
For the nucleus :
[tex]r_{p}[/tex] = radius of proton = 1.0 x 10⁻¹⁵ m
[tex]m_{p}[/tex] = mass of proton = 1.67 x 10⁻²⁷ kg
density of nucleus is given as
[tex]\rho _{n} = \frac{m_{p}}{\frac{4}{3}\pi r_{p}^{3}}[/tex]
[tex]\rho _{n} = \frac{1.67\times 10^{-27}}{\frac{4}{3}(3.14) (1.0\times 10^{-15})^{3}}[/tex]
[tex]\rho _{n} [/tex] = 3.98 x 10¹⁷
For the atom :
[tex]r_{a}[/tex] = radius of atom = radius of proton + radius of orbit = (1.0 x 10⁻¹⁵ + 5.3 x 10⁻¹¹) m
[tex]m_{a}[/tex] = mass of atom = mass of proton + mass of electron = 1.67 x 10⁻²⁷ + 9.31 x 10⁻³¹ kg
density of atom is given as
[tex]\rho _{a} = \frac{m_{a}}{\frac{4}{3}\pi r_{a}^{3}}[/tex]
[tex]\rho _{a} = \frac{(1.67\times 10^{-27} + 9.1\times 10^{-31})}{\frac{4}{3}(3.14) (1.0\times 10^{-15}+ 5.3\times 10^{-11})^{3}}[/tex]
[tex]\rho _{a} [/tex] = 2680.5
Ratio is given as
[tex]\frac{\rho _{n}}{\rho _{a}} = \frac{4\times 10^{17}}{2680.5}[/tex]
[tex]\frac{\rho _{n}}{\rho _{a}} [/tex] = 1.5 x 10¹⁴
