Question

An atom has a diameter of 3.00 Å and the nucleus of that atom has a diameter of 8.50×10−5 Å . Determine the fraction of the volume of the atom that is taken up by the nucleus. Assume the atom and the nucleus are a sphere. fraction of atomic volume: Calculate the density of a proton, given that the mass of a proton is 1.0073 amu and the diameter of a proton is 1.72×10−15 m .

Answer 1

Step-by-step explanation:

Given the following :

Diameter of atom = 3Å

Diameter of Nucleus = 8.50×10^-5 Å

Since Nucleus is assumed to be a sphere ;

Volume of a sphere (V) = (4/3)πr³

Radius of nucleus = diameter of nucleus /2 = 8.50×10^-5Å / 2 = 4.25 × 10^-5Å

Radius of atom = diameter of atom /2 = 3Å / 2 = 1.5Å

Volume of atom = (4/3)×π×(1.5)³

V = 4/3 × π × 3.375 = 14.137166

Volume of nucleus = (4/3)×π×(4.25 × 10^-5)³

V = 4/3 × π × 76.765625 × 10^-15= 14.137166

V = 321.55509 × 10^-15

Fraction of volume of atom taken up by the nucleus :

=[321.55509 × 10^-15]÷ [14.137166]

= 227.453 × 10^-16

Density = mass / volume