Question

Q.27. From the information in this chapter on the mass of the proton, the mass of the electron, and the sizes of the nucleus and the atom, calculate the densities of a hydrogen nucleus and a hydrogen atom.

Short answer

Answer

The density of hydrogen nucleus and H-atom is:

$$\begin{gathered} {\rho }_{nucleus}=3.2\times {10}^{15}\frac{g}{c{m}^3} \\ {\rho }_{H-atom}=0.39\frac{g}{c{m}^3} \end{gathered}$$

Step-by-step solution

Step-by-Step SolutionStep 1: Density

The density of an atom or the density of an object can be calculated as:

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

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

Calculating the density of H nucleus

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

role="math" localid="1648230197186" $$\begin{gathered} V=\frac{4}{3}{r}^3\pi \\ {V}_{nucleus}=\frac{4}{3}\times 3.14\times {(5\times {10}^{-14})}^3 \\ {V}_{nuceus}=523.3\times {10}^{24}g \end{gathered}$$

The mass of one proton is

1 proton = 1.67 $\times$ 10^-24g

$$\begin{gathered} {\rho }_{nucleus}=\frac{m_{nucleus}}{V_{nucleus}} \\ =\frac{1.67\times {10}^{-24}g}{523.3\times {10}^{-42}c{m}^3} \\ {\rho }_{nucleus}=3.2\times {10}^{15}\frac{g}{c{m}^3} \end{gathered}$$

Calculation of the density of H-atom

The density of H-atom is 1 x 10^-8 cmIt can be calculated as:

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

Calculation of the densities of Hydrogen nucleus and H- atom

It can be denoted as:

$$\begin{gathered} {\rho }_{nucleus}=\frac{m_{nucleus}}{V_{nucleus}} \\ =\frac{1.67\times {10}^{-24}g}{523.3\times {10}^{-42}c{m}^3} \\ {\rho }_{nucleus}=3.2\times {10}^{15}\frac{g}{c{m}^3} \end{gathered}$$