Question #163570

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


1
Expert's answer
2021-02-15T00:55:12-0500

a) The volume of the atom is given as the volume of the ball with the radius ra=7×1011mr_a = 7\times 10^{-11}m:


Va=43πra3V_a = \dfrac43\pi r_a^3

By definition, the mean atomic density is given as the ratio of the mass m=2×1026kgm = 2\times 10^{-26}kg and the volume of the atom:


ρa=mVa=3mπra3ρa=32×10264π(7×1011)31.39×104 kgm3\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:


Vn=43πrn3V_n = \dfrac43\pi r_n^3

where rn=3.2×1015mr_n = 3.2\times 10^{-15}m is the nuclear radius.

The mean nuclear density is then:


ρn=mVn=3mπrn3\rho_n = \dfrac{m}{V_n} = \dfrac{3m}{\pi r_n^3}\\

where mm 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;


ρn=32×10264π(3.2×1015)31.46×1017 kgm3\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) 1.39×104 kgm31.39\times 10^{4}\space\dfrac{kg}{m^3}, b) 1.46×1017 kgm31.46\times 10^{17}\space\dfrac{kg}{m^3}.


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