Q125CP - Chemistry: Molecular Nature Of Matter And Change

Question

Helium is the lightest noble gas and the second most abundant element (after hydrogen) in the universe.
(a) The radius of a helium atom is 3.1×10^-11 m; the radius of its nucleus is 2.5×10^-15 m. What fraction of the spherical atomic volume is occupied by the nucleus (V of a sphere =4/3 πr³)?
(b) The mass of a helium-4 atom is 6.64648X10^-24 g, and each of its two electrons has a mass of 9.10939×10^-28 g. What fraction of this atom's mass is contributed by its nucleus?

Step-by-Step Solution

Answer

1. The fraction of the spherical atomic volume is $V=5.{2410}^{-13} \mathrm{or} 5.{2410}^{-10} \mathrm{percent}$
2. The fraction of the atom mass is $\mathrm{m}=0.999726 \mathrm{or} 99.9726 \mathrm{percent}$

a. Step 1: Finding the fraction of the spherical atomic volume

The radius of the helium and nucleus can be given as,

$$\begin{gathered} {\mathrm{r}}_1=2.5\times {10}^{-15} \\ \mathrm{r} =3.1\times {10}^{-11} \\ {\mathrm{v}}_1\left(\mathrm{nucleus}\right)=\frac{4}{3}{{\mathrm{r}}_1}^3 \mathrm{II} \\ {\mathrm{v}}_1\left(\mathrm{nucleus}\right)=\frac{4}{3}{\left(2.5\times {10}^{-15}\right)}^3\parallel \\ {\mathrm{v}}_1\left(\mathrm{nucleus}\right)=6.54\times {10}^{-44} \\ \mathrm{v}\left(\mathrm{atom}\right)\frac{4}{3}{{\mathrm{r}}_{ }}^3\mathrm{II} \\ \mathrm{v}\left(\mathrm{atom}\right)\frac{4}{3}{\left(3.1\times {10}^{-11}\right)}^3\mathrm{II} \\ {\mathrm{v}}_1\left(\mathrm{atom}\right)=1.25\times {10}^{-31} \\ \mathrm{fraction}\left(\mathrm{v}\right)=\frac{{\mathrm{v}}_1}{\mathrm{v}} \\ \mathrm{fraction}\left(\mathrm{v}\right)=\frac{6.54\times {10}^{-44}}{1.25\times {10}^{-31}} \\ \mathrm{fraction}\left(\mathrm{v}\right)=5.24\times {10}^{-13} \mathrm{or} 5.24\times {10}^{-11} \mathrm{percent} \end{gathered}$$

b. Step 2: Finding the fraction of this atom's mass is contributed by its nucleus

The mass of the atom and electron can be denoted as,

$$\begin{gathered} \mathrm{m}\left(\mathrm{atom}\right)=6.64648\times {10}^{-24} \mathrm{g} \\ {\mathrm{m}}_1\left(\mathrm{electron}\right)=9.{1093910}^{-28} \mathrm{g} \end{gathered}$$

The electron contains 2 electrons and masses can be sum up.

$$\begin{gathered} {\mathrm{m}}_1\left(\mathrm{electrons}\right)=1.8219\times {10}^{-27} \mathrm{g} \\ {\mathrm{m}}_2\left(\mathrm{nucleus}\right)=\mathrm{m}-{\mathrm{m}}_1 \\ {\mathrm{m}}_2\left(\mathrm{nucleus}\right)=6.64648\times {10}^{-24}-1.821910\times {10}^{-27} \\ {\mathrm{m}}_2\left(\mathrm{nucleus}\right)=6.64466\times {10}^{-24} \mathrm{g} \\ \mathrm{fraction}=\frac{{\mathrm{m}}_2}{\mathrm{m}} \\ \mathrm{fraction}=\frac{6.64466\times {10}^{-24}}{6.64648\times {10}^{-24}} \\ \mathrm{fraction}\left(\mathrm{m}\right)=0.999726 \mathrm{or} 99.9726\% \end{gathered}$$