# Question 5da42

Nov 25, 2017

Here's what I got.

#### Explanation:

Right from the start, you should know that you can use the Rydberg equation to find the wavelength of the photon emitted when an electron in a hydrogen atom falls from ${n}_{i} = 4$ to ${n}_{f} = 2$.

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

Here

• $l a m \mathrm{da}$ si the wavelength of the emittted photon
• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$

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

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

and plug in your values to find

$l a m \mathrm{da} = \frac{1}{1.097 \cdot {10}^{7} \textcolor{w h i t e}{.} {\text{m}}^{- 1}} \cdot \frac{{4}^{2} \cdot {2}^{2}}{{4}^{2} - {2}^{2}}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = 4.86 \cdot {10}^{- 7} \textcolor{w h i t e}{.} \text{m}}}}$

I'll leave the answer rounded to three sig figs.

To find the frequency of the photon, use the fact that the wavelength of the photon and its frequency have an inverse relationship described by the equation

$l a m \mathrm{da} \cdot \nu = c$

Here

• $\nu$ is the frequency of the photon
• $c$ is the speed of light in a vacuum, usually given as $3 \cdot {10}^{8}$ ${\text{m s}}^{- 1}$

You will end up with

$\nu = \frac{c}{l a m \mathrm{da}}$

$\nu = \left(3 \cdot {10}^{8} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{m"))) "s"^(-1))/(4.86 * 10^(-7)color(red)(cancel(color(black)("m}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\nu = 6.17 \cdot {10}^{14} \textcolor{w h i t e}{.} {\text{s}}^{- 1}}}}$

Finally, to find the energy of the photon, use the Planck - Einstein relation

$E = h \cdot \nu$

Here

• $E$ is the energy of the photon
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34}$ $\text{J s}$

Plug in your values to find

E = 6.626 * 10^(-34)color(white)(.)"J" color(red)(cancel(color(black)("s"))) * 6.17 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))#

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{E = 4.09 \cdot {10}^{- 19} \textcolor{w h i t e}{.} \text{J}}}}$

As a final note, the ${n}_{i} = 4 \to {n}_{f} = 2$ transition is part of the Balmer series.