if the density of a certain spherical atomic nucleus is 1.0x10^14 g cm^-3 and its mass is 2.0x10^-23 g, what is it radius in cm?

Respuesta :

Density as you should know is equal to mass/volume. Solve for the volume.

Now, you know that you are dealing with a sphere. You know the volume of a sphere is V= 4/3pi r ^3.

You know the volume so just solve for r

Answer:

r = [tex]3.63x10^{-13}[/tex] cm

Explanation:

The density (d) is the mass divided by the volume, so:

d = m/V

[tex]1.0x10^{14} = \frac{2.0x10^{-23}}{V}[/tex]

[tex]V = \frac{2.0x10^{-23}}{1.0x10^{14}}[/tex]

V = [tex]2.0x10^{-37} cm^3[/tex]

The volume of a sphere is

[tex]V = \frac{4xpixr^3}{3}[/tex]

For pi = 3.14

[tex]2.0x10^{-37} = \frac{4x3.14xr^3}{3}[/tex]

[tex]12.56r^3 = 6.0x10^{-37}[/tex]

[tex]r^3 = 4.78x10^{-38}[/tex]

[tex]r = \sqrt[3]{4.78x10^{-38}}[/tex]

r = [tex]3.63x10^{-13}[/tex] cm