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The hydrogen atom can absorb light of wavelength \(1094 \mathrm{~nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

Short Answer

Expert verified
The absorption of light with a wavelength of 1094 nm occurs in the infrared region of the electromagnetic spectrum. Using the Rydberg formula and testing integer values for the initial and final principal quantum numbers \(n_1\) and \(n_2\), we find that the initial value is \(n_1 = 1\) and the final value is \(n_2 = 3\).

Step by step solution

01

Identify the region in the electromagnetic spectrum

Look up the wavelength intervals for each region of the spectrum to see where 1094 nm falls. The electromagnetic spectrum can be divided into these regions: 1. Radio waves: greater than 1 mm 2. Microwaves: 1 mm - 100 µm 3. Infrared: 100 µm - 750 nm 4. Visible light: 750 nm - 400 nm 5. Ultraviolet: 400 nm - 10 nm 6. X-rays: 10 nm - 0.1 nm 7. Gamma rays: less than 0.1 nm 1094 nm falls within the range of infrared light, so this absorption occurs in the infrared region of the spectrum.
02

Determine the initial and final values of n using the Rydberg formula

The Rydberg formula is given by: \[\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (\(1.097373 \times 10^7 \mathrm{m^{-1}}\)), and \(n_1\) and \(n_2\) are the initial and final principal quantum numbers, respectively. Since the hydrogen atom absorbs light, we know that an electron transitions to a higher energy level, so \(n_2 > n_1\). First, let's convert the wavelength to meters and calculate the reciprocal: \[\lambda = 1094 \mathrm{~nm} = 1.094 \times 10^{-6} \mathrm{m}\] \[\frac{1}{\lambda} = \frac{1}{1.094 \times 10^{-6}} = 9.137 \times 10^6 \mathrm{m^{-1}}\] Now, plug in the known values into the Rydberg formula: \[(9.137 \times 10^6) = (1.097373 \times 10^7) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] Divide by the Rydberg constant so that we can solve for the difference in \(n_1^2\) and \(n_2^2\): \[\frac{9.137 \times 10^6}{1.097373 \times 10^7} = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] \[0.833 = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] Now we need to test different integer values of \(n_1\) and \(n_2\) to find a pair of numbers that satisfies this equation. After testing, we find that: \[0.833 \approx \frac{1}{1^2} - \frac{1}{3^2} = \frac{8}{9}\] So, the initial value, \(n_1\), is 1, and the final value, \(n_2\), is 3. This electron transition corresponds to the absorption of light with a wavelength of 1094 nm.

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Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum is a range that includes all types of electromagnetic radiation. Each type of radiation differs based on wavelength and energy.
The range is vast, covering energy forms from very low-energy radio waves to high-energy gamma rays. Here’s a quick breakdown of the spectrum:
  • Radio waves: These have the longest wavelengths, typically greater than 1 mm.
  • Microwaves: Shorter than radio waves, ranging from 1 mm to 100 µm.
  • Infrared: These have wavelengths between 100 µm and 750 nm, which make them invisible to the naked eye.
  • Visible light: The only portion visible to humans, with wavelengths between 750 nm and 400 nm.
  • Ultraviolet: Invisible rays between 400 nm and 10 nm, capable of causing sunburn.
  • X-rays and Gamma rays: These have very short wavelengths, less than 10 nm, and carry high energy, useful in medical imaging.
Understanding where electromagnetic radiation like light fits into this spectrum helps us determine its properties and potential applications.
Rydberg Formula
The Rydberg formula is a critical tool in understanding the atomic spectra, especially for hydrogen. It helps to calculate the wavelength of light emitted or absorbed when an electron transitions between different energy levels of an atom.
The formula is expressed as: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
  • \(\lambda\) is the wavelength of the absorbed or emitted light.
  • \(R_H\) is the Rydberg constant for hydrogen, \(1.097373 \times 10^7\, \mathrm{m}^{-1}\).
  • \(n_1\) and \(n_2\) are the principal quantum numbers, representing the initial and final energy levels of the electron.
For instance, when a hydrogen electron absorbs a photon, it jumps from a lower to a higher energy level. This concept is vital in understanding why hydrogen emits or absorbs light at specific wavelengths. By using observed wavelengths, we can predict electron transitions and gain insights into atomic structure.
Infrared Region
The infrared region of the electromagnetic spectrum ranges from wavelengths of 750 nm to 1 mm. It is a part of the spectrum that is invisible to the naked eye but can be felt as heat.
This region extends itself more into the light spectrum than visible light does and consists of several subranges based on wavelength:
  • Near Infrared: Closest to visible light, with wavelengths from 750 nm to 1400 nm.
  • Mid Infrared: Wavelengths range from 1400 nm to 3000 nm, where many thermal sensors operate.
  • Far Infrared: Goes from 3000 nm to 1 mm, mostly associated with thermal radiation.
In our specific example, the hydrogen atom absorbs light with a wavelength of 1094 nm, placing it in the near-infrared subrange. Infrared radiation finds extensive use in various technologies such as night-vision equipment, remote control devices, and fiber optic communications. Understanding this region enhances our grasp of how atoms and molecules interact with heat and light.

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Most popular questions from this chapter

In the experiment shown schematically below, a beam of neutral atoms is passed through a magnetic field. Atoms that have unpaired electrons are deflected in different directions in the magnetic field depending on the value of the electron spin quantum number. In the experiment illustrated, we envision that a beam of hydrogen atoms splits into two beams. (a) What is the significance of the observation that the single beam splits into two beams? (b) What do you think would happen if the strength of the magnet were increased? (c) What do you think would happen if the beam of hydrogen atoms were replaced with a beam of helium atoms? Why? (d) The relevant experiment was first performed by Otto Stern and Walter Gerlach in 1921 . They used a beam of Ag atoms in the experiment. By considering the electron configuration of a silver atom, explain why the single beam splits into two beams.

It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

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.

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m}\), about the size of a bacterium? (b) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1}\) ? (c) Would the radiations in part (a) or part (b) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu \mathrm{s}\) ?

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitted because sodium ions in the pickle become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(589 \mathrm{~nm}\). Calculate its frequency. (b) What is the energy of \(0.10 \mathrm{~mol}\) of these photons? (c) Calculate the energy gap between the excited and ground states for the sodium ion. (d) If you soaked the pickle for a long time in a different salt solution, such as strontium chloride, would you still observe \(589-\mathrm{nm}\) light emission?

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