The ratio of the radii of hydrogen atom and its nucleus is `~ 10^(5)`.
Assuming the atom and the nucleus to be spherical.
(a) What will be the ratio of their sizes ?
(b) If atom is represented by planet earth `'R_(e)' = 6.4 xx 10^(6)m`, estimate the size of the nucleus.

Explanation

(a) Atomic size is represented in terms of atomic radius, `(r_(H))/(r_(n)) = 10^(5)`
As volume of sphere `= (4)/(3) pi r^(3)`, therefore, `V_(H) = (4)/(3) pi r_(H)^(3)` and `V_(n) = (4)/(3) pi r_(n)^(3)`
Thus, the ratio of volumes `(V_(H))/(V_(n)) = ((4)/(3) pi r_(H)^(3))/((4)/(3) pi r_(n)^(3)) = (r_(H)^(3))/(r_(n)^(3)) = ((r_(H))/(r_(n)))^(3) = (10^(5))^(3) = 10^(15)`
(b) `(V_(n))/(V_(H)) = 10^(-15) " or " V_(n) = 10^(-15) xx V_(H)`
If atom in represented by planet earth with `R_(e) = 6.4 xx 10^(6)m`
Then, volume of atom `(V_(H)) = (4)/(3)piR_(e)^(3) = (4)/(3) xx 3.14 xx (6.4 xx 10^(6)m)^(3)`
`= 1097.5 xx 10^(18)m^(3) = 1.0975 xx 10^(21)m^(3)`
`therefore` Volume of nucleus `= 10^(-15) xx (1.0975 xx 10^(21))m^(3)`
`= 1.0975 xx 10^(6) m^(3)`