Question

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?

Answer 1

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 2

r =
`3.63x10^(-13)` cm

Step-by-step explanation:

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

d = m/V

`1.0x10^(14) = (2.0x10^(-23))/(V)`

`V = (2.0x10^(-23))/(1.0x10^(14))`

V =
`2.0x10^(-37) cm^3`

The volume of a sphere is

`V = (4xpixr^3)/(3)`

For pi = 3.14

`2.0x10^(-37) = (4x3.14xr^3)/(3)`

`12.56r^3 = 6.0x10^(-37)`

`r^3 = 4.78x10^(-38)`

`r = \sqrt[3]{4.78x10^(-38)}`

r =
`3.63x10^(-13)` cm