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Chapter 6: Problem 20

(a) What is the frequency of radiation whose wavelength is \(0.86 \mathrm{nm} ?(\mathbf{b})\) What is the wavelength of radiation that has a frequency of \(6.4 \times 10^{11} \mathrm{~s}^{-1} ?(\mathbf{c})\) Would the radiations in part (a) or part (b) be detected by an X-ray detector? (d) What distance does electromagnetic radiation travel in \(0.38 \mathrm{ps} ?\)

Short Answer

Expert verified
(a) 3.49 x 10^17 s^-1, (b) 469 碌m, (c) Part (a) can be detected, (d) 114 碌m.

Step by step solution

01

Understanding Radiation Frequency and Wavelength Relationship

To find the frequency of radiation, we can use the relationship between frequency \( \,f\) and wavelength \( \,\lambda\): \( c = \lambda f\), where \( c\) is the speed of light (approximately \( 3.00 \times 10^8 \, m/s\)).
02

Calculating Frequency for Part (a)

Given that the wavelength \( \lambda = 0.86 \, nm = 0.86 \times 10^{-9} \, m\). The formula is rearranged to \( f = \frac{c}{\lambda}\): \[ f = \frac{3.00 \times 10^8}{0.86 \times 10^{-9}} \] Solve this to find \( f = 3.49 \times 10^{17} \, s^{-1}\).
03

Calculating Wavelength for Part (b)

For the radiation with a frequency of \( 6.4 \times 10^{11} \, s^{-1}\), use the formula \( \lambda = \frac{c}{f}\): \[ \lambda = \frac{3.00 \times 10^8}{6.4 \times 10^{11}} \] Solving gives \( \lambda = 4.69 \times 10^{-4} \, m = 469 \, \mu m\).
04

Determining Detectability with an X-ray Detector for Part (c)

X-ray detectors typically detect wavelengths roughly in the range of 0.01 nm to 10 nm. For part (a), the wavelength \( 0.86 \, nm\) falls within this range and can be detected.For part (b), the wavelength \( 469 \, \mu m = 469 \times 10^{-6} \, m = 469,000 \, nm\) is outside the range and cannot be detected.
05

Calculating Distance Traveled for Part (d)

Given \( 0.38 \, ps\) which is \( 0.38 \times 10^{-12} \, s\), use \( \text{distance} = c \times \text{time}\):\[ \text{distance} = 3.00 \times 10^8 \, m/s \times 0.38 \times 10^{-12} \, s \] Calculate the distance to get \( 1.14 \times 10^{-4} \, m = 114 \mu m\).

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Key Concepts

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

Frequency and Wavelength Relationship
In the world of electromagnetic radiation, frequency and wavelength share a unique relationship. These two properties help define the behavior of waves within the electromagnetic spectrum.
The relationship is expressed with the equation \( c = \lambda f \), where \( c \) stands for the speed of light, \( \lambda \) is the wavelength, and \( f \) is the frequency. This equation encapsulates that as one increases, the other decreases, indicating an inverse relationship.
To solve problems like the one given, you rearrange the formula depending on which property you need to find. For instance, finding frequency when wavelength is known involves altering the equation to \( f = \frac{c}{\lambda} \). Similarly, if you know the frequency, you can determine the wavelength via \( \lambda = \frac{c}{f} \).
  • This equation is vital in fields like physics and engineering, making it essential for understanding everything from radio waves to X-rays.
Speed of Light
The speed of light is a constant in a vacuum, symbolized by \( c \). It is approximately \( 3.00 \times 10^8 \) meters per second.
This speed is pivotal because it links frequency and wavelength in the electromagnetic spectrum. All electromagnetic waves travel at this speed in a vacuum.
Light speed is not only fascinating because of its constancy, but it also represents the fastest speed at which information or matter can travel.
  • When dealing with electromagnetic waves, always remember that \( c \) ties wavelength and frequency together consistently.
  • This speed governs the propagation of light and helps in defining the principles of relativity.
X-ray Detection
X-ray detection involves capturing and interpreting electromagnetic radiation in the X-ray spectrum.
X-rays lie on the spectrum at wavelengths from about 0.01 nm to 10 nm. An X-ray detector specializes in identifying these waves, which are shorter and more energetic than visible light.
In the provided exercise, we see that the wavelength of 0.86 nm is suitable for detection as it lies within this range.
  • X-rays are famous for medical imaging because of their ability to pass through soft tissues and offer a view of bones and other dense structures.
  • In technology and security, X-rays are vital, making their detectability a crucial aspect of numerous applications.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from extremely short wavelengths of gamma rays to long wavelengths of radio waves.
Understanding the spectrum is essential. Each type of wave, including radio waves, microwaves, visible light, and X-rays, plays distinct roles in technology and science.
The exercise reveals this spectrum's diversity by addressing X-rays and radio waves within its broad range.
  • X-rays, as discussed, are high-energy waves crucial in medical imaging.
  • Other parts of the spectrum, like radio waves, are vital for communications, showcasing the vast range of purposes these waves serve.
Encompassing all these waves highlights the dynamic nature and utility of the electromagnetic spectrum in our daily lives.

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Most popular questions from this chapter

Certain elements emit light of a specific wavelength when they are burned or heated in a non-luminous flame. Historically, chemists used such emission wavelengths to determine whether specific elements were present in a sample. Some characteristic wavelengths for a few of the elements are given in the following table: $$\begin{array}{llll} \hline \mathrm{Ag} & 328.1 \mathrm{nm} & \mathrm{Fe} & 372.0 \mathrm{nm} \\ \mathrm{Au} & 267.6 \mathrm{nm} & \mathrm{K} & 404.7 \mathrm{nm} \\ \mathrm{Ba} & 455.4 \mathrm{nm} & \mathrm{Mg} & 285.2 \mathrm{nm} \\ \mathrm{Ca} & 422.7 \mathrm{nm} & \mathrm{Na} & 589.6 \mathrm{nm} \\ \mathrm{Cu} & 324.8 \mathrm{nm} & \mathrm{Ni} & 341.5 \mathrm{nm} \\ \hline \end{array}$$ (a) Determine which of these emissions occur in the ultraviolet part of the spectrum. (b) Which emission has the highest frequency and which one has the lowest frequency? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.58 \times 10^{14} \mathrm{~s}^{-1} .\) Which of these elements is probably in the sample?

Write the condensed electron configurations for the following atoms and indicate how many unpaired electrons each has: \((\mathbf{a}) \mathrm{Mg},(\mathbf{b}) \mathrm{Ge},(\mathbf{c}) \mathrm{Br},(\mathbf{d}) \mathrm{V},(\mathbf{e}) \mathrm{Y},(\mathbf{f}) \mathrm{Lu} .\)

A hydrogen atom orbital has \(n=4\) and \(m_{l}=-2 .(\mathbf{a})\) What are the possible values of \(l\) for this orbital? (b) What are the possible values of \(m_{s}\) for the orbital?

Determine whether each of the following sets of quantum numbers for the hydrogen atom are valid. If a set is not valid, indicate which of the quantum numbers has a value that is not valid: (a) \(n=3, l=3, m_{l}=2, m_{\mathrm{s}}=+\frac{1}{2}\) (b) \(n=4, l=3, m_{l}=-3, m_{s}=+\frac{1}{2}\) (c) \(n=3, l=1, m_{l}=2, m_{s}=+\frac{1}{2}\) (d) \(n=5, l=0, m_{l}=0, m_{s}=0\) (e) \(n=2, l=1, m_{l}=1, m_{\mathrm{s}}=-\frac{1}{2}\)

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 -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 \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).
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