question 16 (4 points) question 16) repeat this analysis with the light source set to 103 nm. this provides…

Explanation:

Step1: Recall Bohr - model and spectral - line concepts

In the Bohr model, when an electron transitions from a lower energy level ($n_i$) to a higher energy level ($n_f$), it absorbs a photon of a specific energy. When it returns from the higher level to the lower level, it emits photons corresponding to specific spectral lines. For a hydrogen - like atom, the energy of a photon is given by $E = h\nu=\frac{hc}{\lambda}$, and the wavelengths of the spectral lines for hydrogen are described by the Rydberg formula $\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, where $R_H = 1.097\times 10^{7}\ m^{-1}$ is the Rydberg constant for hydrogen.

Step2: Identify the type of transition

The electron is transitioning from $n = 1$ to $n = 3$. When the electron returns to the ground state ($n = 1$), it can make two possible transitions: $3\rightarrow2$ and $3\rightarrow1$.

Step3: Calculate the wavelengths for the transitions

For the $3\rightarrow2$ transition:

Using the Rydberg formula $\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, with $n_1 = 2$ and $n_2 = 3$. [ \begin{align*} \frac{1}{\lambda}&=1.097\times 10^{7}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\ &=1.097\times 10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)\ &=1.097\times 10^{7}\times\frac{9 - 4}{36}\ &=1.097\times 10^{7}\times\frac{5}{36}\ \lambda&=\frac{36}{5\times1.097\times 10^{7}}\ \lambda&\approx 656.3\ nm \end{align*} ] This is a Balmer - series line (visible light).

For the $3\rightarrow1$ transition:

Using the Rydberg formula with $n_1 = 1$ and $n_2 = 3$. [ \begin{align*} \frac{1}{\lambda}&=1.097\times 10^{7}\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\ &=1.097\times 10^{7}\left(1-\frac{1}{9}\right)\ &=1.097\times 10^{7}\times\frac{8}{9}\ \lambda&=\frac{9}{8\times1.097\times 10^{7}}\ \lambda&\approx 102.6\ nm \end{align*} ] This is a Lyman - series line (ultraviolet light).

Answer:

The observed spectral lines are at approximately 102.6 nm (Lyman - series, ultraviolet) and 656.3 nm (Balmer - series, visible).