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 \(n_{1}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11

Short answer

The best explanation for why the lines with \(n_{f}=3\) are not observed in the visible portion of the spectrum is (ii) transitions to \(n_{f} = 3\) emit photons in the infrared portion of the spectrum. The wavelengths of the first three lines in the Balmer series are: \(\lambda_1 \approx 656.28 ~nm\) (red, Hα), \(\lambda_2 \approx 486.13 ~nm\) (green-blue, Hβ), and \(\lambda_3 \approx 434.05 ~nm\) (violet, Hγ).

Step-by-step solution

Understand Balmer series and the formula for wavelengths

The Balmer series represents a set of spectral emission lines due to electronic transitions in a hydrogen atom where the electron reaches the second energy level (n=2). To calculate the wavelengths of these lines, we use the Balmer formula: \[ \frac{1}{\lambda} = R_H\left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right) \] where \(\lambda\) is the wavelength of the line, \(R_H\) is the Rydberg constant \((1.097\times10^7 m^{-1})\), \(n_{f}\) is the final energy level (in our case, \(n_{f} = 2\)), and \(n_{i}\) is the initial energy level.

Determine the best explanation

Among the given options, the second option (ii) states that transitions to \(n_{f} = 3\) emit photons in the infrared portion of the spectrum. This is accurate because, in a hydrogen atom, electronic transitions to levels higher than \(n_{f} = 2\) have lower energy differences between the levels leading to the release of light in the infrared region. Therefore, (ii) is the best explanation.

Calculate the wavelengths of the first three lines in the Balmer series

Now we will use the Balmer formula to calculate the wavelengths of the first three lines in the Balmer series with \(n_i = 3,~ 4,~ 5\). 1. For \(n_i = 3\): \[ \frac{1}{\lambda_1} = R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \Rightarrow \lambda_1 = \frac{1}{R_H\left(\frac{1}{4} - \frac{1}{9}\right)} \approx 656.28~ nm\] 2. For \(n_i = 4\): \[ \frac{1}{\lambda_2} = R_H\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \Rightarrow \lambda_2 = \frac{1}{R_H\left(\frac{1}{4} - \frac{1}{16}\right)} \approx 486.13~ nm\] 3. For \(n_i = 5\): \[ \frac{1}{\lambda_3} = R_H\left(\frac{1}{2^2} - \frac{1}{5^2}\right) \Rightarrow \lambda_3 = \frac{1}{R_H\left(\frac{1}{4} - \frac{1}{25}\right)} \approx 434.05~ nm\]

Identify the lines in the emission spectrum

According to the calculated wavelengths, the first three lines in the Balmer series are: 1. \(\lambda_1 = 656.28~ nm\): This line falls in the red part of the visible spectrum and is also known as the H-alpha (Hα) line. 2. \(\lambda_2 = 486.13~ nm\): This line falls in the green-blue part of the visible spectrum and is also known as the H-beta (Hβ) line. 3. \(\lambda_3 = 434.05~ nm\): This line falls in the violet part of the visible spectrum and is also known as the H-gamma (Hγ) line.

Key concepts

Spectral Lines

Spectral lines are the distinct colored lines in an emission or absorption spectrum that denote specific transitions of electrons within atoms. When an electron in an atom transitions between different energy levels, it either absorbs or emits a photon of light of a specific wavelength.

In the case of hydrogen atoms, these transitions give rise to well-known spectral series, like the Balmer series, which are observed in the visible spectrum. Each line in the series corresponds to a different electronic transition. These transitions happen when an electron jumps from a higher energy level to a lower one.

The color and position of these spectral lines help scientists identify elements and analyze various physical characteristics of the emitter, such as temperature and density. The uniqueness of the spectral lines is akin to an atomic fingerprint for each element.

Rydberg Formula

The Rydberg formula is a fundamental tool in atomic physics, used to predict the wavelengths of photons emitted or absorbed during electron transitions between energy levels in hydrogen-like atoms. It is given by the formula:

\[ \frac{1}{\lambda} = R_H\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

where \(\lambda\) is the wavelength of the light, \(R_H\) is the Rydberg constant, approximately \(1.097 \times 10^7 \text{ m}^{-1}\), \(n_f\) is the final energy level, and \(n_i\) is the initial energy level.

This formula is particularly significant in calculating the wavelengths in the Balmer series, where the final level \(n_f\) is 2. It is a precise equation that helps in understanding the energy transition between quantized atomic energy levels and is critical for predicting the lines in an atom's emission spectrum.

By inserting the values for \(n_i\) and \(n_f\) into the formula, one can compute the specific wavelengths for different spectral lines.

Hydrogen Atom Transitions

In the hydrogen atom, electron transitions between energy levels result in the emission or absorption of light. These transitions are the basis for the spectral series, such as the Balmer series, which occurs when electrons drop to the second energy level \(n=2\).

When an electron falls from a higher energy level \(n_i\) to a lower one like \(n_f\), the energy difference between these levels is released as a photon. This energy difference also determines the wavelength of the emitted light, as described by the Rydberg formula.

For transitions to \(n=2\), the light is visible, yielding the characteristic emission lines like H-alpha and H-beta. However, for transitions to levels like \(n_f = 3\), the emitted radiation falls into the infrared region, which is not visible to the naked eye.

• H-alpha: Electron transition from \(n=3\) to \(n=2\)
• H-beta: Electron transition from \(n=4\) to \(n=2\)
• H-gamma: Electron transition from \(n=5\) to \(n=2\)

Understanding these transitions is essential for interpreting the hydrogen spectrum and provides insights into the quantum nature of atoms.