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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

Expert verified
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

01

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.
02

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.
03

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 \(n_i = 5\): \[ \frac{1}{\lambda_3} = R_H\left(\frac{1}{2^2} - \frac{1}{5^2}\right) \Rightarrow \lambda_3 = \frac{1}{R_H\left(\frac{1}{4} - \frac{1}{25}\right)} \approx 434.05~ nm\]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

If a sample of calcium chloride is introduced into a nonluminous flame, the color of the flame turns to orange ("flame test"). The light is emitted because calcium atoms become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(422.7 \mathrm{nm} .\) Calculate its frequency. (b) What is the energy of \(1.00 \mathrm{~mol}\) of these photons (a mole of photons is called an Einstein)? (c) Calculate the energy gap between the excited and ground states for the calcium atom.

Molybdenum metal must absorb radiation with an energy higher than \(7.22 \times 10^{-19} \mathrm{~J}\) ( "energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

Give the numerical values of \(n\) and \(l\) corresponding to each of the following orbital designations: (a) \(3 p\), (b) \(2 s,(\mathbf{c}) 4 f,\) (d) \(5 d\).

Does the hydrogen atom "expand" or "contract" when an electron is excited from the \(n=1\) state to the \(n=3\) state?

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

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