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Chapter 1: Problem 35

If clusters are broken up using lasers in the near-infrared to radio-frequency range of the spectrum, what frequencies could used to break a chemical bond? 1\. Far infrared 2\. Microwave 3\. Ultraviolet 4\. Red light

Short Answer

Expert verified
The frequencies that could be used to break a chemical bond in the near-infrared to radio-frequency range are: 1. Far infrared (with a frequency range of \(10^{12}\) to \(10^{14}\) Hz) 2. Microwave (with a frequency range of \(10^9\) to \(10^{12}\) Hz) Hence, the correct answer consists of options 1 and 2.

Step by step solution

01

Understand the electromagnetic spectrum ranges and frequencies

The first step is to understand the electromagnetic spectrum and the corresponding frequencies of the given options. The electromagnetic spectrum includes various types of waves, and each has a specific frequency range.
02

List the frequency ranges of the given options

Now, let's list down the frequency ranges for the options mentioned in the exercise: 1. Far infrared: \(10^{12}\) to \(10^{14}\) Hz 2. Microwave: \(10^9\) to \(10^{12}\) Hz 3. Ultraviolet: \(7.5 \times 10^{14}\) to \(3 \times 10^{16}\) Hz 4. Red light: \(4 \times 10^{14}\) to \(4.8 \times 10^{14}\) Hz
03

Compare the frequency ranges to the breaking of chemical bonds

In general, breaking of chemical bonds requires a significant amount of energy, which translates to higher frequencies on the electromagnetic spectrum. Hence, we need to identify which of the options has a frequency in the near-infrared to the radio-frequency range that is suitable for breaking chemical bonds. After comparing the given options with their frequency ranges, the most likely frequencies that can break a chemical bond in the near-infrared to radio-frequency range would be: 1. Far infrared 2. Microwave
04

Determine the answer

Based on the analysis, we can conclude that the frequencies that could be used to break a chemical bond in the near-infrared to radio-frequency range are: 1. Far infrared (Option 1) 2. Microwave (Option 2) So, the correct answer consists of options 1 and 2.

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