Question

The ratio of the radii of the hydrogen atom and its nucleus is$~{10}^5$. Assuming the atom and the nucleus to be spherical,

(i) What will be the ratio of their sizes?

(ii) If the atom is represented by planet earth $‘{\mathrm{R}}_{\mathrm{e}}’=6.4\times {10}^6\mathrm{m}$, estimate the size of the nucleus.

Answer 1

(i)

Step 1: Volume of a sphere

• The volume of a sphere is given by $\mathrm{V}=\frac{3}{4}{\mathrm{\pi r}}^3$
• Let $"\mathrm{R}"$ be the radius of the atom and $"\mathrm{r}"$be that of the nucleus.
• For the radii of the hydrogen atom and its nucleus, its volume is

$$\begin{gathered} {\mathrm{V}}_{\mathrm{H}}=\frac{3}{4}{{\mathrm{\pi r}}_{\mathrm{H}}}^3\cdots 1 \\ {\mathrm{V}}_{\mathrm{n}}=\frac{3}{4}{{\mathrm{\pi r}}_{\mathrm{n}}}^3\cdots 2 \end{gathered}$$

Step 2: Volume of the atom and its nucleus:

• The ratio of the radii of the hydrogen atom and its nucleus is $~{10}^5$
  $\frac{{\mathrm{r}}_{\mathrm{H}}}{{\mathrm{r}}_{\mathrm{n}}}={10}^5$

Step 3: Ratio of their volumes:

• The ratio of their volumes,

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{V}}_{\mathrm{n}}}=\frac{\frac{4}{3}{{\mathrm{\pi r}}_{\mathrm{H}}}^3}{\frac{4}{3}{{\mathrm{\pi r}}_{\mathrm{n}}}^3} \\ \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{V}}_{\mathrm{n}}}=\frac{{{\mathrm{r}}_{\mathrm{H}}}^3}{{{\mathrm{r}}_{\mathrm{n}}}^3}(\frac{{\mathrm{r}}_{\mathrm{H}}}{{\mathrm{r}}_{\mathrm{n}}}={10}^5) \\ \frac{{\mathrm{V}}_{\mathrm{H}}}{{\mathrm{V}}_{\mathrm{n}}}={\left({10}^5\right)}^3 \end{gathered}$$
(ii)

Step 1: Ratio of the radius of earth and nucleus:

• If the atom is represented by the planet Earth, $‘{\mathrm{R}}_{\mathrm{e}}’=6.4\times {10}^6\mathrm{m}$ then the radius of the nucleus would be given by ${\mathrm{r}}_{\mathrm{n}}$

$‘{\mathrm{R}}_{\mathrm{e}}’=6.4\times {10}^6\mathrm{m}$

• We know that the ratio of the radius of the earth and the radius of the nucleus,

$\frac{{\mathrm{R}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{n}}}={10}^5$

Step 2: Radius of the nucleus:

$$\begin{gathered} \frac{{\mathrm{R}}_{\mathrm{e}}}{{\mathrm{R}}_{\mathrm{n}}}={10}^5 \\ \frac{6.4\mathrm{x}{10}^6}{{10}^5}={\mathrm{R}}_{\mathrm{n}} \\ {\mathrm{R}}_{\mathrm{n}}=6.4\mathrm{x}10 \\ {\mathrm{R}}_{\mathrm{n}}=64\mathrm{m} \end{gathered}$$