Answer:
[tex]r=6.72\times 10^{-13}\ m[/tex]
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 :
[tex]r=\dfrac{n^2h^2\epsilon_o}{Z\pi me^2}[/tex]
Where
n = number of orbit
h = Planck's constant
Z = atomic number (for gold, Z = 79)
m = mass of electron
e = charge on electron
[tex]r=\dfrac{(6.63\times 10^{-34})^2 \times 8.85\times 10^{-12}}{79\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19})^2}[/tex]
[tex]r=6.72\times 10^{-13}\ m[/tex]
So, the radius of the n = 1 orbit of gold is [tex]6.72\times 10^{-13}\ m[/tex]. Hence, this is the required solution.
