Question

The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.0 × 10-15 m. The single electron in a hydrogen atom orbits the nucleus at a distance of 5.3 × 10-11 m. What is the ratio of the density of the hydrogen nucleus to the density of the complete hydrogen atom?

Answer 1

1.5 x 10¹⁴

Step-by-step explanation:

For the nucleus :

`r_(p)` = radius of proton = 1.0 x 10⁻¹⁵ m

`m_(p)` = mass of proton = 1.67 x 10⁻²⁷ kg

density of nucleus is given as

`\rho _(n) = (m_(p))/((4)/(3)\pi r_(p)^(3))`

`\rho _(n) = (1.67* 10^(-27))/((4)/(3)(3.14) (1.0* 10^(-15))^(3))`

`\rho _(n)` = 3.98 x 10¹⁷

For the atom :

`r_(a)` = radius of atom = radius of proton + radius of orbit = (1.0 x 10⁻¹⁵ + 5.3 x 10⁻¹¹) m

`m_(a)` = mass of atom = mass of proton + mass of electron = 1.67 x 10⁻²⁷ + 9.31 x 10⁻³¹ kg

density of atom is given as

`\rho _(a) = (m_(a))/((4)/(3)\pi r_(a)^(3))`

`\rho _(a) = ((1.67* 10^(-27) + 9.1* 10^(-31)))/((4)/(3)(3.14) (1.0* 10^(-15)+ 5.3* 10^(-11))^(3))`

`\rho _(a)` = 2680.5

Ratio is given as

`(\rho _(n))/(\rho _(a)) = (4* 10^(17))/(2680.5)`

`(\rho _(n))/(\rho _(a))` = 1.5 x 10¹⁴