# Question 5dc82

Oct 6, 2017

Here's what I got.

#### Explanation:

The first thing that you need to do here is to calculate the wavelength of the photon emitted when your electron undergoes a ${n}_{i} = 5 \to {n}_{f} = 2$ transition in a hydrogen atom.

The wavelength of the photon will then help you determine its energy.

So, you know that when an electron in a hydrogen atom makes a transition from an initial energy level ${n}_{i}$ to a final energy level ${n}_{f}$, it emits a photon whose wavelength is equal to

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

Here

• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$

You can rearrange the above equation, which is called the Rydberg equation, as

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

which gets you

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

In your case, you have ${n}_{i} = 5$ and ${n}_{f} = 2$, which means that the emitted photon will have a wavelength of

$l a m \mathrm{da} = \frac{1}{1.097 \cdot {10}^{7} \textcolor{w h i t e}{.} {\text{m}}^{- 1}} \cdot \frac{{5}^{2} \cdot {2}^{2}}{{5}^{2} - {2}^{2}}$

$l a m \mathrm{da} = 4.34 \cdot {10}^{- 7} \textcolor{w h i t e}{.} \text{m}$

Expressed in nanometers--recall that $\text{1 m} = {10}^{9}$ $\text{nm}$--the wavelength of the emitted photon will be

$l a m \mathrm{da} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{434 nm}}}}$

Now, in order to find the energy of the photon, you need to use a variation of 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

E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * (3 * 10^8 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))/(4.34 * 10^(-7) color(red)(cancel(color(black)("m"))))#

$E = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{4.58 \cdot {10}^{- 19} \textcolor{w h i t e}{.} \text{J}}}}$

I'll leave both answers rounded to three sig figs.

It's worth mentioning that the ${n}_{i} = 5 \to {n}_{f} = 2$ transition, which is part of the Balmer series, is located in the visible portion of the EM spectrum. 