# Question #2e787

Oct 6, 2017

${n}_{i} = 8$

#### Explanation:

Your tool of choice here will be 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

Now, you know that the electron starts on an initial energy level ${n}_{i}$ and ends up on the third energy level, so

${n}_{f} = 3$

Moreover, you know that this transition is accompanied by the emission of a photon of wavelength

$l a m \mathrm{da} = 9.54 \cdot {10}^{- 7}$ $\text{m}$

Rearrange the Rydberg equation to solve for ${n}_{i}$

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

This is equivalent to

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

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

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

Finally, you should end up with

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

Plug in your values to find

${n}_{i} = \sqrt{\left(9.54 \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{10}^{- 7}}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))) * 3^2)/(9.54 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m}}^{- 1}}}} - {3}^{2}\right)}$

${n}_{i} = 8.017 \approx \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{8}}}$

Therefore, you can say that this electron underwent a ${n}_{i} = 8 \to {n}_{f} = 3$ transition.