Respuesta :

Answer:

[tex]6.70\times 10^{-13}\ m[/tex]

Explanation:

Given:

  • [tex]n = n^{th}[/tex] orbit of gold = 1
  • [tex]Z[/tex] = atomic number of gold = 79

Assumptions:

  • [tex]h[/tex] = Planck's constant = [tex]6.62\times 10^{-34}\ m^2kg/s[/tex]
  • [tex]k[/tex] = Boltzmann constant = [tex]9\times 10^{9}\ Nm^2/C^2[/tex]
  • [tex]e[/tex] = magnitude of charge on an electron = [tex]1.6\times 10^{-19}\ C[/tex]
  • [tex]m[/tex] = mass of an electron = [tex]9.1\times 10^{-31}\ kg[/tex]
  • [tex]r[/tex] = radius of the [tex]n^{th}[/tex] orbit of the atom

WE know that the radius of the [tex]n^{th}[/tex] orbit of an atom is given by:

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\[/tex]

Let us find out the radius of the 1st orbit of the gold atom for which n = 1 and Z = 79.

[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\\Rightarrow r = \dfrac{(1)^2(6.62\times 10^{-34})^2}{4\pi^2\times 9\times 10^9\times 79\times (1.6\times 10^{-19})^2\times 9.1\times 10^{-31}}\\\Rightarrow r =6.70\times 10^{-13}\ m[/tex]

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