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
