Question

In the Pfund series, \(\mathbf{n}_{\mathrm{lo}}=5 .\) Calculate the longest wavelength (in nanometers) possible for a transition in this series.

Short answer

In the Pfund series with an initial low-energy level of 5, the longest wavelength possible for a transition is 284 nm. This is obtained by finding the smallest energy difference between the initial energy level (5) and the nearest higher energy level (6) using the Rydberg formula and then converting the wavelength to nanometers.

Step-by-step solution

Calculate the energy difference: Initial Energy Level and Final Energy Level (Using Rydberg Formula)

The Rydberg formula for the Pfund series is given by: \( \frac{1}{\lambda} = R_H \left(\frac{1}{n_{lo}^2} - \frac{1}{n_{hi}^2}\right) \) Where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant (\(1.097 \times 10^7 \, m^{-1}\)), \(n_{lo}\) is the initial low-energy level, and \(n_{hi}\) is the final high-energy level. We have been given \(n_{lo}=5\) and need to find the longest wavelength, so we need the smallest energy difference, which occurs when \(n_{hi}=6\).

Apply the Rydberg Formula

Plug in the values \(n_{lo}=5\) and \(n_{hi}=6\) into the Rydberg formula: \( \frac{1}{\lambda} = R_H \left(\frac{1}{5^2} - \frac{1}{6^2}\right) \)

Solve for Wavelength

Now, solve for the wavelength (\(\lambda\)): \( \frac{1}{\lambda} = (1.097 \times 10^7 \, m^{-1})\left(\frac{1}{25}-\frac{1}{36}\right) \) \( \frac{1}{\lambda} = (1.097 \times 10^7 \, m^{-1})\left(0.032\right) \) \( \lambda = \frac{1}{0.035104 \times 10^7 \, m^{-1}} = 2.84 \times 10^{-7} \, m \)

Convert Wavelength to Nanometers

Finally, convert the wavelength from meters to nanometers: \( \lambda = 2.84 \times 10^{-7} \, m \times \frac{10^9 \, nm}{1 \, m} = 284 \, nm \) Thus, the longest wavelength possible for a transition in the Pfund series with \(n_{lo}=5\) is 284 nm.