Oct 17, 2017

$n = 2$

#### Explanation:

The idea here is that the difference in energy between the initial energy level of the electron and the final energy level of the electron will give you the wavelength of the photon emitted during this transition.

In your case, you know that the electron falls from the sixth energy level, $n = 6$, to a lower energy level. When this transition takes place, a photon of wavelength

$\text{410 nm" = 410 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 4.10 * 10^(-7)color(white)(.)"m}$

is emitted. As you know, the Rydberg equation allows you to use the wavelength of the emitted photon, let's say $l a m \mathrm{da}$, to figure out the final energy level of the electron.

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

Here

• $R$ is the Rydberg constant, equal to $1.097 \cdot {10}^{7}$ ${\text{m}}^{- 1}$
• ${n}_{i}$ is the initial energy level of the electron
• ${n}_{f}$ is the final energy level of the electron

Now, your goal here is to figure out the value of the final energy level, so rearrange the Rydberg equation to solve for ${n}_{f}$.

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

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

This is equivalent to

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

which gets you

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

Plug in your values to find

${n}_{f} = \sqrt{\left(4.10 \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{10}^{- 7}}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))) * 6^2)/(4.10 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m}}^{- 1}}}} + {6}^{2}\right)}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{n}_{f} = 1.9995 \approx 2}}}$

Therefore, you can say that the electron will fall to the second energy level, $n = 2$, which implies that this transition, i.e. ${n}_{i} = 6 \to {n}_{f} = 2$, is part of the Balmer series.

To double-check the result, you can use the Balmer formula to confirm that the electron is initially located on the sixth energy level.

$l a m \mathrm{da} = B \cdot {n}_{i}^{2} / \left({n}_{i}^{2} - {\textcolor{b l u e}{2}}^{2}\right)$

Here

• $B$ is a constant that is $\approx 3.6451 \cdot {10}^{- 7} \textcolor{w h i t e}{.} \text{m}$
• ${n}_{i}$ is the initial energy level of the electron
• $\textcolor{b l u e}{2}$ is the final energy level for the Balmer series, ${n}_{f} = 2$

This time, you will have

${n}^{2} \cdot \left(l a m \mathrm{da} - B\right) = 4 \cdot l a m \mathrm{da}$

$n = \sqrt{\frac{4 \cdot l a m \mathrm{da}}{l a m \mathrm{da} - B}}$

Plug in your values to find

${n}_{i} = 6.0043 \approx 6$

which confirms that you're dealing with the ${n}_{i} = 6 \to {n}_{f} = 2$ transition of the Balmer series.