Question

How can I calculate the densities of a hydrogen nucleus and a hydrogen atom?

Answer 1

Hint: As a first step, you could recall the known values of certain quantities, such as, the mass of hydrogen nuclei and hydrogen atom, the radius of the two. Then you could find the volume by using this radius. Then, you can divide their mass by their respective volume and thus get their respective densities.

Formula used:
Density,
$\rho =\dfrac{M}{V}$

Complete Step by step solution:
In the question, we are asked as to how the calculation of densities of hydrogen nucleus and hydrogen atom is done.
Calculation of density is pretty simple. All you have to do is divide the mass by volume. That is,
$\rho =\dfrac{M}{V}$

We have to find the densities of hydrogen nuclei as well as hydrogen atoms. In order to calculate this we have to use certain standard values. For example,

Radius of the proton,
${{r}_{p}}=8.41\times {{10}^{-16}}m$

The volume of the hydrogen nuclei will be,
$V=\dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi {{\left( 8.41\times {{10}^{-16}} \right)}^{3}}$
$\therefore V=2.49\times {{10}^{-45}}{{m}^{3}}$

The mass of the proton,
${{m}_{p}}=1.67\times {{10}^{-27}}kg$

Now the density of hydrogen nuclei will be,
${{\rho }_{n}}=\dfrac{{{M}_{n}}}{{{V}_{n}}}$
${{\rho }_{n}}=\dfrac{1.67\times {{10}^{-27}}}{2.49\times {{10}^{-45}}}$
$\therefore {{\rho }_{n}}=6.70\times {{10}^{17}}kg{{m}^{-3}}$

Therefore, we found the density of the hydrogen nuclei to be,
${{\rho }_{n}}=6.70\times {{10}^{17}}kg{{m}^{-3}}$

Now we know that a hydrogen atom’s mass is contributed by the proton that is present in the nucleus. So the mass of the nucleus and the mass of the atom will be the same.

The radius of a hydrogen atom is known to have the value given by,
${{r}_{a}}=53\times {{10}^{-12}}m$

The volume of the hydrogen atom will be,
${{V}_{a}}=\dfrac{4}{3}\pi {{r}_{a}}^{3}$
$\Rightarrow {{V}_{a}}=\dfrac{4}{3}\pi {{\left( 53\times {{10}^{-12}} \right)}^{3}}$
$\therefore {{V}_{a}}=6.24\times {{10}^{-31}}{{m}^{3}}$

Now we will get the value of density as,
${{\rho }_{a}}=\dfrac{1.67\times {{10}^{-27}}}{6.24\times {{10}^{-31}}}$
$\therefore {{\rho }_{a}}=2.7\times {{10}^{3}}kg{{m}^{-3}}$

Therefore, we found the densities of hydrogen nuclei and hydrogen atom respectively to be,
${{\rho }_{n}}=6.70\times {{10}^{17}}kg{{m}^{-3}}$
${{\rho }_{a}}=2.7\times {{10}^{3}}kg{{m}^{-3}}$

Note:
From the value obtained we could conclude that the mass density of hydrogen nucleus is way greater than the mass density of hydrogen atom. This is because, though both these quantities weigh the same, the volume in which this mass occupies, varies from hydrogen nuclei to hydrogen atoms to a great extent.