Question

What is the approximate radius of the n = 1orbit of gold ( Z
=79 )?

Answer 1

`6.70* 10^(-13)\ m`

Step-by-step explanation:

Given:

• 
  `n = n^(th)` orbit of gold = 1
• 
  `Z` = atomic number of gold = 79

Assumptions:

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

WE know that the radius of the
`n^(th)` orbit of an atom is given by:

`r = (n^2h^2)/(4\pi^2kZe^2m)\\`

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

`r = (n^2h^2)/(4\pi^2kZe^2m)\\\Rightarrow r = ((1)^2(6.62* 10^(-34))^2)/(4\pi^2* 9* 10^9* 79* (1.6* 10^(-19))^2* 9.1* 10^(-31))\\\Rightarrow r =6.70* 10^(-13)\ m`