Question

The visible emission lines observed by Balmer all involved \(n_{\mathrm{f}}=2\). (a) Which of the following is the best explanation of why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum: (i) Transitions to \(n_{\mathrm{f}}=3\) are not allowed to happen, (ii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum, (iii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to \(n_{\mathrm{f}}=3\) emit photons that are at exactly the same wavelengths as those to \(n_{\mathrm{f}}=2\). (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which \(n_{\mathrm{i}}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11.

Short answer

(a) The best explanation for why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum is (ii): transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum. (b) The wavelengths of the first three lines in the Balmer series are: \(\lambda_1 = 656.3 \, nm\) \(\lambda_2 = 486.1 \, nm\) \(\lambda_3 = 434.1 \, nm\) These lines can be identified in the emission spectrum shown in Figure 6.11 by their corresponding wavelengths.

Step-by-step solution

Part (a): Explanation for non-observed transitions to \(n_{\mathrm{f}}=3\)

To answer this question, we can first find the wavelengths of the emitted photons for transitions to \(n_{\mathrm{f}}=3\) and compare them to the wavelengths of visible light. The formula to find the wavelength of emitted photons (\(\lambda\)) in a hydrogen atom when an electron transitions from one energy level (ni) to another (nf) is: \(\frac{1}{\lambda}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\) where R is the Rydberg constant for hydrogen, \(R=1.097\times10^7\,m^{-1}\), and nf and ni are the final and initial energy levels, respectively. For \(n_{\mathrm{f}}=3\) and \(n_{i}=4,5,6\) we get the following wavelengths: \(\frac{1}{\lambda_1}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)\) \(\frac{1}{\lambda_2}=R\left(\frac{1}{3^2}-\frac{1}{5^2}\right)\) \(\frac{1}{\lambda_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)\) Wavelengths for visible light range from 380 nm to 750 nm. We can find the values of λ1, λ2, and λ3 to see if they belong to the range of visible light. We find that the transitions involving \(n_{\mathrm{f}}=3\) emit photons with wavelengths longer than the visible range, which corresponds to infrared light. Hence, the best answer to part (a) is option (ii): transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum.

Part (b): Wavelengths of the first three lines in the Balmer series

To find the wavelengths for the first three lines in the Balmer series, we should use the same wavelength formula but with \(n_{\mathrm{f}}=2\). This corresponds to the Balmer series, which is observed in the visible portion of the spectrum. The initial energy levels (\(n_{i}\)) are 3, 4, and 5. Therefore, we can write the following equations to find the wavelengths: \(\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\) \(\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\) \(\frac{1}{\lambda_3}=R\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\) Now, we can plug the value of the Rydberg constant for hydrogen and find the corresponding wavelengths for each transition: \(\lambda_1 = 656.3 \, nm\) \(\lambda_2 = 486.1 \, nm\) \(\lambda_3 = 434.1 \, nm\) These wavelengths correspond to the first three lines in the Balmer series. In the emission spectrum shown in Figure 6.11, the lines will be marked with their corresponding wavelengths.

Key concepts

Hydrogen Emission Spectrum

The hydrogen emission spectrum is a set of lines that represent the wavelengths of light emitted by electrons in hydrogen atoms as they transition from higher energy levels to lower ones. Each line corresponds to a specific color in the spectrum, and the entire set forms what is known as the hydrogen spectral series. The Balmer series, in particular, includes the spectral lines visible to the human eye.

When an electron in a hydrogen atom drops from a higher energy level to energy level 2 (designated as \( n_{\mathrm{f}}=2 \)), it emits a photon that falls within the visible light spectrum, leading to the bright lines that we can observe. This is why the Balmer series is so valuable in studying the properties of hydrogen and plays a crucial role in understanding atomic structure. The presence of these lines and their specific wavelengths provide us with invaluable information about the energy levels within a hydrogen atom and the differences in energy between these levels.

Rydberg Constant

The Rydberg constant (R) is a fundamental constant of physics that plays a vital role in atomic spectroscopy. It is the basis of the Rydberg formula, which predicts the wavelengths of spectral lines of many chemical elements, most notably hydrogen. With a value of approximately \( 1.097 \times 10^7 \, m^{-1} \), it reflects the fundamental structure and transitions of an electron in hydrogen.

The constant is named after Johannes Rydberg, who first proposed a formula to predict the wavelengths of spectral lines. His work led to the formulation of the Rydberg formula we use today: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right) \], where \( \lambda \) is the wavelength of the emitted or absorbed light, \( n_{f} \) is the principal quantum number of the final state, and \( n_{i} \) is the initial state's principal quantum number. This formula is a monumental achievement in the field of quantum mechanics and is extensively used for calculating the wavelengths of emissions for any given transition.

Visible Light Spectrum

The visible light spectrum represents the range of electromagnetic radiation that our eyes can detect. This spectrum spans wavelengths from approximately 380 nanometers (nm), which correspond to violet light, to around 750 nm, associated with red light. The colors of the rainbow—red, orange, yellow, green, blue, indigo, and violet—are often used to remember the sequential order of colors within this spectrum.

The wavelengths of the Balmer series that fall within this range, specifically between 380 nm and 750 nm, are the ones observed as visible light. That's why the transitions to \( n_{\mathrm{f}}=2 \) produce colors that are part of the Balmer series in the hydrogen emission spectrum. This series includes the wavelengths calculated in the textbook exercise that can be seen with the naked eye, such as the first three lines with wavelengths 656.3 nm (red), 486.1 nm (blue-green), and 434.1 nm (blue). Understanding the visible light spectrum allows us to comprehend why some emissions are visible while others, like those transitioning to \( n_{\mathrm{f}}=3 \), result in infrared emissions and are not visible to us.