Question

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

Answer 1

Answer:

$r=6.72\times 10^{-13}\ m$

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 :

$r=\dfrac{n^2h^2\epsilon_o}{Z\pi me^2}$

Where

n = number of orbit

h = Planck's constant

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

m = mass of electron

e = charge on electron

$r=\dfrac{(6.63\times 10^{-34})^2 \times 8.85\times 10^{-12}}{79\pi \times 9.1\times 10^{-31}\times (1.6\times 10^{-19})^2}$

$r=6.72\times 10^{-13}\ m$

So, the radius of the n = 1 orbit of gold is $6.72\times 10^{-13}\ m$. Hence, this is the required solution.