# Question 64a98

Mar 26, 2017

I get a value of the order of ${10}^{14}$ for Hydrogen
and a value of the order of ${10}^{13}$ for Uranium"^238

#### Explanation:

A. Let us take the example of hydrogen atom.
Atomic radius, Bohr's radius ${a}_{0}$: $5.29 \times {10}^{-} 11 m$
Atomic mass: $1.00794 u$
Nuclear diameter: 1.75×10^-15 m
Mass of a proton: $1.00728 u$

Average atomic density ${\rho}_{a} = \text{mass"/"volume}$
$\implies {\rho}_{a} = \frac{1.00794}{\frac{4}{3} \pi {\left(5.29 \times {10}^{-} 11\right)}^{3}}$
Similarly Nuclear density rho_n=1.00728/(4/3pi(1.75/2×10^-15)^3)
Ratio of two densities rho_n/rho_a=1.00728/1.00794xx((5.29xx10^-11)^3)/((0.875×10^-15)^3)
${\rho}_{n} / {\rho}_{a}$~${10}^{14}$
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
B. Let us now take the example of Uranium"^238 atom.
Atomic radius, empirical: $1.56 \times {10}^{-} 10 m$
Average Atomic mass: $238.029 u$
Nuclear diameter: 15×10^-15 m
Mass of nucleas: $238.0508 u$

Average atomic density ${\rho}_{a} = \text{mass"/"volume}$
$\implies {\rho}_{a} = \frac{238.029}{\frac{4}{3} \pi {\left(1.56 \times {10}^{-} 10\right)}^{3}}$
Similarly Nuclear density rho_n=238.0508/(4/3pi(15/2×10^-15)^3)
Ratio of two densities rho_n/rho_a=238.0508/238.029xx((1.56 xx10^-10)^3)/((7.5×10^-15)^3)#
${\rho}_{n} / {\rho}_{a}$~${10}^{13}$