Question

What is the approximate radius of the n = 1orbit of gold ( Z
=19 )?

Answer 1

`r=6.72* 10^(-13)\ m`

Step-by-step explanation:

Let r is the radius of the n = 1 orbit of the gold. According to Bohr's model, the radius of orbit is given by :

`r=(n^2h^2\epsilon_o)/(Z\pi me^2)`

Where

n = number of orbit

h = Planck's constant

Z = atomic number (for gold, Z = 79)

m = mass of electron

e = charge on electron

`r=((6.63* 10^(-34))^2 * 8.85* 10^(-12))/(79\pi * 9.1* 10^(-31)* (1.6* 10^(-19))^2)`

`r=6.72* 10^(-13)\ m`

So, the radius of the n = 1 orbit of gold is
`6.72* 10^(-13)\ m`. Hence, this is the required solution.