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 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 $\frac{1.06\times 10^{-10} m}{1.68 \times 10^{-15} m}$ = 6.31 × 10⁴

So the diameter of your model would be 6.31 × 10⁴ mm
.
6.31 × 10⁴ mm × $\frac{1 m}{1000 mm}$ = 63.1 m

Answer 2

Answer:

The diameter of the entire model is 41,680 mm

Explanation:

Let the radius of the model be = x

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

$\frac{y}{x}=\frac{0.5 mm}{x}$...(1)

$R=r_{o}\times A^{1/3}$

$R$ = radius of the nucleus

$r_o$ = r = 1.25 fm = $1.25\times 10^{-15} m$

(Constant value for all nuclei)

A = Number of nucleons or atomic mass number

Radius of hydrogen nucleus =

$R=1.25\times 10^{-15} m\times 1^{1/3}=1.25\times 10^{-15} m$

Radius of the nth shell: $r_n$

$r_n=\frac{52.9\times n^2}{Z} pm$

$r_n$ = radius of the nth shell

Z = Atomic number

Radius of hydrogen atom :$r_h$

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

$r_h=r_1=\frac{52.9\times 1^2}{1} pm=52.1 pm$

1 pm = $10^{-12} m$

$r_h=5.21\times 10^{-11} m$

Ratio of R to $r_h$ :

$\frac{R}{r_h}=\frac{1.25\times 10^{-15} m}{5.21\times 10^{-11} m}$..(2)

(1)=(2)

$\frac{0.5 mm}{x}=\frac{1.25\times 10^{-15} m}{5.21\times 10^{-11} m}$

$x=20,840 mm$

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