Question

From the information in this chapter on the mass of the proton, the mass of the electron, and thesizes of the nucleus and the atom, calculate the densities of a hydrogen nucleus and a hydrogenatom.

Short answer

The density of hydrogen nucleus and atom are,

$\begin{array}{l}{\rho }_{\mathrm{nucleus}}=3.2\times {10}^{15}\frac{g}{c{m}^3}\\ {\rho }_{\mathrm{Hatom}} =0.39\frac{g}{c{m}^3}\end{array}$

Step-by-step solution

Calculating the density using the equation

On calculating the density,

$\rho =\frac{m}{v}$

Here, m is mass; v is volume and $\rho$is density.

Calculating the density of H nucleus

Diameter of H nucleus is $5\times {10}^{-14}cm=r$. On calculating the nucleus volume, the equation can be denoted as,

$\begin{array}{l}V=\frac{4}{3}{r}^3\pi \\ V_{\mathrm{nucleus}}=\frac{4}{3}\times 3.14\times (5\times {10}^{-14}{)}^3\\ V_{\mathrm{nucleus}} =523.3\times {10}^{24}g\\ \\ 1 \mathrm{proton}=1.67\times {10}^{24}g\\ {\rho }_{\mathrm{nucleus}}=\frac{m_{\mathrm{nuclues}}}{V_{\mathrm{nucleus}}}\\ =\frac{1.67\times {10}^{-24}g}{523.3\times {10}^{-42}c{m}^3}\\ {\rho }_{\mathrm{nucleus}}=3.2\times {10}^{15}\frac{g}{c{m}^3}\end{array}$

Calculating the density of H atom

Density of H atom is$1\times {10}^{-8}cm.$It can be calculated as,

$\begin{array}{l}{V}_{\mathrm{Hatom}}=V\\ =\frac{4}{3}{r}^3\pi \\ =\frac{4}{3}.3.14.(1\times {10}^{-8}{)}^3\\ V_{\mathrm{Hatom}}=4.186\times {10}^{-24}c{m}^3\\ \\ m_{\mathrm{Hatom}}=m\left(proton\right).m\left(electron\right)\\ m_{\mathrm{Hatom}} =9.11\times {10}^{-28}g+1.67\times {10}^{-244}g\\ \\ {\rho }_{\mathrm{Hatom}}=\frac{m_{\mathrm{Hatom}}}{V_{\mathrm{Hatom}}}\\ =\frac{1.67\times {10}^{-24}g+9.11\times {10}^{-28}g}{4.186\times {10}^{-24}c{m}^3}\\ {\rho }_{\mathrm{Hatom}}=0.39\frac{g}{c{m}^3}\end{array}$

Calculating the densities of Hydrogen nucleus and atom

It can be denoted as,

$\begin{array}{l}{\rho }_{\mathrm{nucleus}}=\frac{m_{\mathrm{nucleus}}}{V_{\mathrm{nucleus}}}\\ =\frac{1.67\times {10}^{-24}g}{523.3\times {10}^{-42}c{m}^3}\\ {\rho }_{\mathrm{nucleus}}=3.2\times {10}^{15}\frac{g}{c{m}^3}\\ \\ {\rho }_{\mathrm{Hatom}}=\frac{m_{\mathrm{Hatom}}}{V_{\mathrm{Hatom}}}\\ =\frac{1.67\times {10}^{-24}+9.11\times {10}^{-28}g}{4.186\times {10}^{-24}c{m}^3}\\ {\rho }_{\mathrm{Hatom}}=0.39\frac{g}{c{m}^3}\end{array}$