Question

The nucleus of an atom can be modeled as several protons and neutrons closely packed together. Each particle has a mass of 1.67 10-27 kg and radius on the order of 10-15 m. (a) Use this model and the data provided to estimate the density of the nucleus of an atom. 3.9868*10^17kg/m^3 Incorrect: Your answer is incorrect. Check the syntax of your response. (b) Compare your result with the density of a material such as iron (rho = 7874 kg/m3). What do your result and comparison suggest about the structure of matter?

Answer 1

Step-by-step explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle,
`m=1.67* 10^(-27)\ kg`

Radius of the particle,
`R=10^(-15)\ m`

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

`d=(m)/(V)`, V is the volume of the particle

`d=(m)/((4/3)\pi r^3)`

`d=(1.67* 10^(-27))/((4/3)\pi (10^(-15))^3)`

`d=3.98* 10^(17)\ kg/m^3`

So, the density of the nucleus of an atom is
`3.98* 10^(17)\ kg/m^3`.

(b) Density of iron,
`d'=7874\ kg/m^3`

Taking ratio of the density of nucleus of an atom and the density of iron as :

`(d)/(d')=(3.98* 10^(17))/(7874)`

`(d)/(d')=5.05* 10^(13)`

`d=5.05* 10^(13)\ d'`

So, the density of the nucleus of an atom is
`5.05* 10^(13)` times greater than the density of iron. Hence, this is the required solution.