# How much energy is released when "1 mol" of hydrogen atoms transition from n=3 to n=2?

Sep 30, 2017

Here's what I got.

#### Explanation:

The first thing that you need to do here is to figure out the wavelength of the photon emitted by the electron as it's making the $n = 3 \to n = 2$ transition.

To do that, you need to use the Rydberg equation, which looks like this

$\frac{1}{l a m \mathrm{da}} = R \cdot \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$

Here

• $l a m \mathrm{da}$ is the wavelength of the photon
• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• ${n}_{I}$ is the initial energy level of the transition
• ${n}_{f}$ is the final energy level of the transition

Plug in your values to find

$\frac{1}{l a m \mathrm{da}} = 1.097 \cdot {10}^{7} \textcolor{w h i t e}{.} {\text{m}}^{- 1} \cdot \left(\frac{1}{2} ^ 2 - \frac{1}{3} ^ 2\right)$

$\frac{1}{l a m \mathrm{da}} = 1.097 \cdot {10}^{7} \textcolor{w h i t e}{.} {\text{m}}^{- 1} \cdot \left(\frac{9 - 4}{9 \cdot 4}\right)$

Rearrange to solve for $l a m \mathrm{da}$

$l a m \mathrm{da} = \frac{36}{1.097 \cdot {10}^{7} \cdot 5} \textcolor{w h i t e}{.} \text{m}$

$l a m \mathrm{da} = 6.563 \cdot {10}^{- 7} \textcolor{w h i t e}{.} \text{m}$

The wavelength of the emitted photon confirms that you're dealing with a transition that's part of the Balmer series. Now, the energy of a photon is directly proportional to its frequency, which implies that it is inversely proportional to its wavelength.

This relationship is described by the Planck - Einstein relation

$E = h \cdot \frac{c}{l a m \mathrm{da}}$

Here

• $E$ is the energy of the photon
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34}$ $\text{J s}$
• $c$ is the speed of light in a vacuum, usually given as $3 \cdot {10}^{8}$ ${\text{m s}}^{- 1}$

Plug in your value to find the energy of a single photon that is emitted when an electron makes this transition.

E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(6.563 * 10^(-7) color(red)(cancel(color(black)("m"))))

$E = 3.029 \cdot {10}^{- 19}$ $\text{J}$

So, you need the energy of the photon because it is equal to the difference between the two energy levels of the transition.

In other words, the difference between the energy of the $n = 3$ level and that of the $n = 2$ energy level, $\Delta {E}_{3 \to 2}$, is equal to

$\Delta {E}_{3 \to 2} = 3.029 \cdot {10}^{- 19}$ $\text{J}$

This value corresponds to the energy of $1$ photon emitted when $1$ electron makes that transition, i.e. for $1$ atom of hydrogen.

To find the energy difference for $1.00$ moles of hydrogen atoms, you need to find the total energy of $1.00$ moles of photons. Your tool of choice here will be Avogadro's constant, which tells you that $1$ mole of photons will contain $6.022 \cdot {10}^{23}$ photons.

$6.022 \cdot {10}^{23} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{photons"))) * (3.029 * 10^(-19)color(white)(.)"J")/(1color(red)(cancel(color(black)("photon")))) = color(darkgreen)(ul(color(black)(1.82 * 10^(5)color(white)(.)"J}}}}$

The answer is rounded to three sig figs. If you want, you can express the answer in kilojoules to get $\text{182 kJ}$.