Question

The radius of the innermost electron orbit of a hydrogen atom is 5.3 ×10−11 m. What are the radii of the n = 2 and n =3 orbits?

Answer 1

Answer:

Explanation:

The radius of nth orbit is given r = 5.3x10^-11m

So for II orbit n =2 and r= 2x5.3x10^-11 = 10.6x10^-11 m

        III orbit  n=3  and r = 15.9x10^-11 m

Answer 2

Explanation:

Given

• The innermost orbit’s radius of a hydrogen atom, $5.3 \times 10^{-11} m$

Let,

• $r_{2}$ be the radius of the orbit at n = 2

It is related to the radius of the innermost orbit as:

$r_{2} =(n)^{2} r_{1}$

$= 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}m$

For n=3, the respective electron radius can be written as:

$r^{3} = (n)^{2} r_{1}$

$= 9 \times 5.3 \times 10^{-11}= 4.77 \times 10^{-10}m$

Hence,

The electron radii for:-

• $\rm{ n = 2 \: is \: 2.12 \times 10^{-10}m}$

• $\rm{ n = 3\: is \:4.77 \times 10^{-10}m}$