Answer to Question #163570 in Physics for Sam

Question

A carbon atom has a mass of 2.00x 10-26kg, an atomic radius of 7.0 x 10-11m, and a nuclear radius of 3.2 x 10-15m , (a) Calculate its mean atomic density   . (b) Calculate its mean nuclear density

Answer 1

a) The volume of the atom is given as the volume of the ball with the radius $r_a = 7\times 10^{-11}m$:

$$V_a = \dfrac43\pi r_a^3$$

By definition, the mean atomic density is given as the ratio of the mass $m = 2\times 10^{-26}kg$ and the volume of the atom:

$$\rho_a = \dfrac{m}{V_a} = \dfrac{3m}{\pi r_a^3}\\ \rho_a=\dfrac{3\cdot 2\times 10^{-26}}{4\pi\cdot (7\times 10^{-11})^3} \approx 1.39\times 10^{4}\space\dfrac{kg}{m^3}$$

b) The volume of the nucleus is:

$$V_n = \dfrac43\pi r_n^3$$

where $r_n = 3.2\times 10^{-15}m$ is the nuclear radius.

The mean nuclear density is then:

$$\rho_n = \dfrac{m}{V_n} = \dfrac{3m}{\pi r_n^3}\\$$

where $m$ is again the mass of the atom. Since protons and neutrons are much havier then electrons, we can consider the whole mass of the atom to be concentrated in its nucleus. Thus, obtain;

$$\rho_n=\dfrac{3\cdot 2\times 10^{-26}}{4\pi\cdot (3.2\times 10^{-15})^3} \approx 1.46\times 10^{17}\space\dfrac{kg}{m^3}$$

Answer. a) $1.39\times 10^{4}\space\dfrac{kg}{m^3}$, b) $1.46\times 10^{17}\space\dfrac{kg}{m^3}$.