Question

Prove analytically that the density of uranium nucleus is of the same order of magnitude as the density of hydrogen atoms.

Answer 1

Considering that A is the atomic number and m is the mass of Hydrogen nucleus.
 
The hydrogen nucleus is a proton, whose mass is 1.67 x 10∧−27kg and radius is 8.41 x 10∧−16 m

V = (4/3) πr³ = (4/3) π (8.41 x 10−16m)³

= 2.49 x 10−45 m³

Density = mass / Volume =1.67 x 10∧−27kg

= 2.49 x 10∧−45 m³

=6.70 x 10∧17 kg/m³

Considering that A is the atomic number and m is the mass of Uranium nucleus.
 
m = A x (1.67 x 10⁻²⁷) where A is the mass number

V = 4/3 πr³ = 4/3 x 22/7 x [(1.2 x 10⁻¹⁵) ∧1/3 mA]³ (radius = 1.2 x 10⁻¹⁵ m)

= 8.2 x 10⁻⁴⁵ x Am³

Density = mass (m) / Volume (V)
 
= 1.67 x 10⁻²⁷x A kg / 8.2 x 10⁻⁴⁵ x A m³

= 2.0 x 10¹⁷ kg/m³

Therefore, the density of both the nuclei is of the same order.