# Question 90211

Jul 10, 2017

The $n = 3$ orbit is $9 {a}_{0}$ from the nucleus; the energy emitted is 1.936 × 10^"-18"color(white)(l) "J".

#### Explanation:

The radius $r$ of the $n$th orbit in a hydrogen atom is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} r = {n}^{2} {a}_{0} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where ${a}_{0}$ is the Bohr radius.

∴ For $n = 3$,

$r = {3}^{2} {a}_{0} = 9 {a}_{0}$

You use the Rydberg formula to calculate the energy.

Rydberg's original formula was expressed in terms of wavelengths, but we can rewrite the formula to have the units of energy.

The Rydberg formula in terms of energy is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} E = R \left(\frac{1}{n} _ {\textrm{f}}^{2} - \frac{1}{n} _ {\textrm{i}}^{2}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$R = \text{the Rydberg constant", 2.178 × 10^"-18"color(white)(l) "J}$, and
${n}_{\textrm{i}}$ and ${n}_{\textrm{f}}$ are the initial and final energy levels.

In this problem,

${n}_{\textrm{i}} = 3$
${n}_{\textrm{f}} = 1$

E = 2.178 × 10^"-18"color(white)(l) "J" × (1/1 -1/9) = 2.178 × 10^"-18" color(white)(l)"J" × (9-1)/(9×1)

=8/9 ×2.178 × 10^"-18" color(white)(l)"J" = 1.936 × 10^"-18"color(white)(l) "J"#