Problem 151 - Chemistry An Atoms First Approach

Question

Calculate, to four significant figures, the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=5\) state and then fall to states with smaller values of \(n\).

Step-by-Step Solution

Answer

The longest wavelength of light emitted by electrons in the hydrogen atom when transitioning from the \(n=5\) state to a smaller value of \(n\) is approximately \(6.564 \times 10^{-7}\ m\), and the shortest wavelength is approximately \(4.342 \times 10^{-7}\ m\).

Step 1: Recall the Rydberg Formula

The Rydberg formula is used to calculate the wavelengths of light emitted during electron transitions in hydrogen. The formula is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_j^2} - \frac{1}{n_i^2}\right)\] Here, \(\lambda\) is the wavelength of light emitted, \(R_H\) is the Rydberg constant for hydrogen (\(R_H \approx 1.097\times10^7 m^{-1}\)), \(n_i\) is the initial quantum state, and \(n_j\) is the final quantum state.

Step 2: Calculate the longest wavelength

To calculate the longest wavelength, we'll use the values \(n_i = 5\) and \(n_j = 4\) in the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right)\] Now, solve for \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right)}\] Plug in the value of \(R_H\) and solve for \(\lambda\): \[\lambda \approx \frac{1}{1.097\times10^7 \left(\frac{1}{16} - \frac{1}{25}\right)} \approx 6.564\times10^{-7} m\] The longest wavelength the electron can emit, when falling from \(n=5\) to \(n=4\), is approximately \(6.564\times10^{-7} m\).

Step 3: Calculate the shortest wavelength

To calculate the shortest wavelength, we'll use the values \(n_i = 5\) and \(n_j = 1\) in the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{5^2}\right)\] Now, solve for \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{1^2} - \frac{1}{5^2}\right)}\] Plug in the value of \(R_H\) and solve for \(\lambda\): \[\lambda \approx \frac{1}{1.097\times10^7 \left(\frac{1}{1} - \frac{1}{25}\right)} \approx 4.342\times10^{-7} m\] The shortest wavelength the electron can emit, when falling from \(n=5\) to \(n=1\), is approximately \(4.342\times10^{-7} m\).

Step 4: Express the results to four significant figures

Express the results of the longest and shortest wavelengths to four significant figures: \[\lambda_{longest} \approx 6.564\times10^{-7} m \approx 6.564 \times 10^{-7}\ m\] \[\lambda_{shortest} \approx 4.342\times10^{-7} m \approx 4.342 \times 10^{-7}\ m\] Thus, the longest wavelength of light emitted is approximately \(6.564 \times 10^{-7}\ m\) and the shortest wavelength is approximately \(4.342 \times 10^{-7}\ m\).