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

Answer:

r = $3.63x10^{-13}$ cm

Explanation:

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

d = m/V

$1.0x10^{14} = \frac{2.0x10^{-23}}{V}$

$V = \frac{2.0x10^{-23}}{1.0x10^{14}}$

V = $2.0x10^{-37} cm^3$

The volume of a sphere is

$V = \frac{4xpixr^3}{3}$

For pi = 3.14

$2.0x10^{-37} = \frac{4x3.14xr^3}{3}$

$12.56r^3 = 6.0x10^{-37}$

$r^3 = 4.78x10^{-38}$

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

r = $3.63x10^{-13}$ cm