Question

If you wanted to make an accurate scale model of the hydrogen atom and decided that the nucleus would have a diameter of 1mm, what would be the diameter of the entire model?

Answer 1

The diameter of the entire model is 41,680 mm

Step-by-step explanation:

Let the radius of the model be = x

Nuclear radius of the model = y = 0.5 mm *(radius = 2 × diameter)

`(y)/(x)=(0.5 mm)/(x)`...(1)

`R=r_(o)* A^(1/3)`

`R` = radius of the nucleus

`r_o` = r = 1.25 fm =
`1.25* 10^(-15) m`

(Constant value for all nuclei)

A = Number of nucleons or atomic mass number

Radius of hydrogen nucleus =

`R=1.25* 10^(-15) m* 1^(1/3)=1.25* 10^(-15) m`

Radius of the nth shell:
`r_n`

`r_n=(52.9* n^2)/(Z) pm`

`r_n` = radius of the nth shell

Z = Atomic number

Radius of hydrogen atom :
`r_h`

Z =1, n = 1 (1 electron in first shell)

`r_h=r_1=(52.9* 1^2)/(1) pm=52.1 pm`

1 pm =
`10^(-12) m`

`r_h=5.21* 10^(-11) m`

Ratio of R to
`r_h` :

`(R)/(r_h)=(1.25* 10^(-15) m)/(5.21* 10^(-11) m)`..(2)

(1)=(2)

`(0.5 mm)/(x)=(1.25* 10^(-15) m)/(5.21* 10^(-11) m)`

`x=20,840 mm`

Diameter of the model = 2 × 20,840 mm = 41,680 mm

Answer 2

The diameter of the model would be 63.1 m.

The diameter of a hydrogen atom is 1.06 × 10⁻¹⁰ m.
The diameter of a proton is 1.68 × 10⁻¹⁵ m.

The ratio of their diameters is
`(1.06* 10^(-10) m)/(1.68 * 10^(-15) m)` = 6.31 × 10⁴

So the diameter of your model would be 6.31 × 10⁴ mm
.
6.31 × 10⁴ mm ×
`(1 m)/(1000 mm)` = 63.1 m