Problem 55 - Chemistry: An Atoms-Focused Approach

Question

Balmer observed a hydrogen emission line for the transition from \(n=6\) to \(n=2,\) but not for the transition from \(n=7\) to \(n=2 .\) Why?

Step-by-Step Solution

Answer

Balmer observed the hydrogen emission line for the transition from \(n=6\) to \(n=2\) because it produced a wavelength of 410.2 nm, which falls within the visible light range. On the other hand, the transition from \(n=7\) to \(n=2\) had a wavelength of 397.0 nm, which is also within the visible light range but at the extreme violet end of the spectrum. Due to the limited sensitivity of the human eye and possible limitations of the experimental setup, the hydrogen emission line for the \(n=7\) to \(n=2\) transition was not observed.

Step 1: Understanding the Balmer series

The Balmer series is a specific subset of hydrogen spectral lines that correspond to transitions in which the electron falls down to the second energy level (n=2) from higher energy levels. These specific hydrogen emission lines are in the visible light region of the electromagnetic spectrum. The Balmer series for hydrogen can be represented using the Rydberg formula: $$\frac{1}{\lambda} = R_{\text{H}} \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$$ Here, \(\lambda\) is the wavelength of the emitted light, \(R_{\text{H}}\) is the Rydberg constant for hydrogen \((R_{\text{H}} \approx 1.097\times10^7 \text{m}^{-1})\), and \(n\) is the principal quantum number of the initial energy level.

Step 2: Analyzing the two transitions

To understand why the hydrogen emission line was observed for the transition from \(n=6\) to \(n=2\) but not for the transition from \(n=7\) to \(n=2\), let's find the wavelength of the emitted light for both transitions using the Balmer formula: 1) For the transition from \(n=6\) to \(n=2\): $$\frac{1}{\lambda_{6 \rightarrow 2}} = R_{\text{H}} \left( \frac{1}{2^2} - \frac{1}{6^2} \right)$$ 2) For the transition from \(n=7\) to \(n=2\): $$\frac{1}{\lambda_{7 \rightarrow 2}} = R_{\text{H}} \left( \frac{1}{2^2} - \frac{1}{7^2} \right)$$ Now, let's calculate the wavelength for both transitions.

Step 3: Calculate the wavelength of the emitted light

Using the Rydberg constant and the formula, we calculate the values for the wavelengths of the emitted light in both transitions: 1) For the transition from \(n=6\) to \(n=2\): $$\lambda_{6 \rightarrow 2} = \frac{1}{R_{\text{H}} \left( \frac{1}{2^2} - \frac{1}{6^2} \right)} \approx 410.2 \text{ nm}$$ 2) For the transition from \(n=7\) to \(n=2\): $$\lambda_{7 \rightarrow 2} = \frac{1}{R_{\text{H}} \left( \frac{1}{2^2} - \frac{1}{7^2} \right)} \approx 397.0 \text{ nm}$$

Step 4: Explain the Balmer observation

The visible light region of the electromagnetic spectrum ranges from about 380 nm to 750 nm. Both calculated wavelengths fall within this range, but due to the limited sensitivity of the human eye, 397 nm lies at the extreme violet end of the spectrum and is difficult to observe. Additionally, detection of the spectral lines also depends on the sensitivity and resolution of the instruments used by Balmer. Thus, the hydrogen emission line for the transition from \(n=6\) to \(n=2\) was observed as it lies comfortably within the visible light range, while the transition from \(n=7\) to \(n=2\) might not have been observed due to its location at the extreme end of the visible light spectrum or the limitations of the experimental setup.