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(a) What is the frequency of radiation whose wavelength is \(10.0 \AA\) ? (b) What is the wavelength of radiation that has a frequency of \(7.6 \times 10^{10} \mathrm{~s}^{-1} ?\) (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 \(25.5 \mathrm{fs}\) ?

Short Answer

Expert verified
(a) The frequency of the radiation with a wavelength of \(10.0 \AA\) is \(3.00 \times 10^{18} Hz\). (b) The wavelength of radiation with a frequency of \(7.6 \times 10^{10} s^{-1}\) is \(3.95 \times 10^{-3} m\). (c) The radiation in part (a) would be detected by an X-ray detector, while the radiation in part (b) would not. (d) Electromagnetic radiation travels \(7.65 \times 10^{-6} m\) in \(25.5 fs\).

Step by step solution

01

Calculate the frequency of the radiation with a given wavelength

First, let's find the frequency of radiation with a wavelength of \(10.0 \AA\). The formula we will use is: \(f = \frac{v}{λ}\) where \(v = 3.00 \times 10^8 m/s\), and \(λ = 10.0 \AA = 10.0 \times 10^{-10} m\). So, \(f = \frac{3.00 \times 10^8}{10.0 \times 10^{-10}}\)
02

Calculate the frequency

Now, we can calculate the frequency: \(f = \frac{3.00 \times 10^8}{10.0 \times 10^{-10}} = 3.00 \times 10^{18} Hz\) So, the frequency of the radiation is \(3.00 \times 10^{18} Hz\).
03

Calculate the wavelength of radiation with a given frequency

Next, let's find the wavelength of radiation with a given frequency of \(7.6 \times 10^{10} s^{-1}\). We can rearrange the formula \(v = fλ\) to find the wavelength: \(λ = \frac{v}{f}\) So, \(λ = \frac{3.00 \times 10^8}{7.6 \times 10^{10}}\)
04

Calculate the wavelength

Now we can calculate the wavelength: \(λ = \frac{3.00 \times 10^8}{7.6 \times 10^{10}} = 3.95 \times 10^{-3} m\) So, the wavelength of the radiation is \(3.95 \times 10^{-3} m\).
05

Determine if the radiations can be detected by an X-ray detector

An X-ray detector detects radiation with wavelengths in the range of \(0.01 \AA\) to \(10 \AA\). Since the radiation in part (a) has a wavelength of \(10.0 \AA\), it would be detected by an X-ray detector. The radiation in part (b) has a wavelength of \(3.95 \times 10^{-3}m\), which is outside the range, so it would not be detected by an X-ray detector.
06

Calculate the distance traveled by electromagnetic radiation in a given time

Finally, let's find the distance traveled by electromagnetic radiation in \(25.5 fs\). Since the speed of light is constant, we can use the formula: \(distance = speed \times time\) So, \(distance = (3.00 \times 10^8 m/s)(25.5 \times 10^{-15}s)\)
07

Calculate the distance

Now, we can calculate the distance: \(distance = (3.00 \times 10^8)(25.5 \times 10^{-15}) = 7.65 \times 10^{-6} m\) So, the electromagnetic radiation travels \(7.65 \times 10^{-6} m\) in \(25.5 fs\).

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Key Concepts

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

Wavelength and Frequency
Wavelength and frequency are two key concepts in understanding electromagnetic radiation. They are inversely related to each other. This means that as one increases, the other decreases. The formula that connects them is:\[f = \frac{v}{\lambda}\]where \(f\) is the frequency, \(v\) is the speed of light (approximately \(3.00 \times 10^8\) meters per second), and \(\lambda\) is the wavelength.

Wavelength is the distance between two corresponding points of consecutive waves, usually measured in meters. Frequency tells us how many wave crests pass a given point in one second and is measured in Hertz (Hz).

Take the example of a radiation with a wavelength of \(10.0 \mathrm{~\AA}\) (which is equal to \(10.0 \times 10^{-10}\) meters), we can calculate its frequency by substituting into the formula to find \(f\). So, when using these values, the frequency will be \(3.00 \times 10^{18} \text{ Hz}\). Understanding this relationship allows us to investigate and quantify radiation properties across the electromagnetic spectrum.
X-ray Detection
X-ray detectors are specific devices designed to pick up a range of electromagnetic radiation known as X-rays. These rays have very short wavelengths ranging from \(0.01 \mathrm{~\AA}\) to \(10 \mathrm{~\AA}\).

To understand which waves or radiations can be detected by these devices, we need to look at the wavelengths. For example, the radiation with a wavelength of \(10.0 \mathrm{~\AA}\) would fall neatly within the detection range. Thus, it can be detected by an X-ray detector. Conversely, if the wavelength is significantly longer than this range, as seen in our example with a wavelength of \(3.95 \times 10^{-3}\) meters, X-ray detectors will not pick it up.

The ability to detect such specific ranges is crucial in medical imaging, materials testing, and various scientific research applications. It helps provide clear images or analytical data where high precision is required.
Speed of Light
The speed of light is a fundamental constant and an essential concept in physics, particularly when discussing electromagnetic radiation. This constant is approximately \(3.00 \times 10^8\) meters per second. It represents the fastest speed at which energy or information can travel.

When dealing with electromagnetic waves, their speed in a vacuum is always the speed of light. This consistency allows scientists to use it as a basis for calculations related to wavelength, frequency, and even time. For example, in determining how far electromagnetic radiation will travel over a certain period, we use the formula:\[\text{distance} = \text{speed} \times \text{time}\]Using this formula, if electromagnetic radiation travels for \(25.5 \text{ fs}\) (femtoseconds), the distance covered is \(7.65 \times 10^{-6}\) meters.

This speed underlies many technologies, including GPS, and plays a vital role in theories of relativity. Knowing this constant helps us make precise predictions and measurements in experiments and real-world applications.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, from low-energy radio waves to high-energy gamma rays. It represents the range of wavelengths and frequencies these waves can have.

Each type of radiation in the electromagnetic spectrum exhibits different properties, allowing for various uses and scientific applications:
  • Radio waves: Long wavelengths, used in communications and broadcasting.
  • Microwaves: Used for cooking and satellite transmissions.
  • Infrared: Experienced as heat; used in remote controls and thermal imaging.
  • Visible light: The only part of the spectrum visible to the naked eye.
  • Ultraviolet: Causes sunburn; used in black lights and sterilization.
  • X-rays: Used in medical imaging to view inside the body.
  • Gamma rays: Have very short wavelengths and high energy; used in cancer treatment.
Understanding this spectrum is key to recognizing how different kinds of electromagnetic waves interact with matter and energy. Each segment of the spectrum serves a unique role in technology and scientific exploration. The wavelength and frequency determine where the radiation falls on the spectrum, defining its properties and potential applications.

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Most popular questions from this chapter

The series of emission lines of the hydrogen atom for which \(n_{f}=3\) is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed. (b) Calculate the wavelengths of the first three lines in the Paschen series - those for which \(n_{i}=4,5\), and 6 .

When the spectrum of light from the Sun is examined in high resolution in an experiment similar to that illustrated in Figure 6.11, dark lines are evident. These are called Fraunhofer lines, after the scientist who studied them extensively in the early nineteenth century. Altogether, about 25,000 lines have been identified in the solar spectrum between \(2950 \AA\) and \(10,000 \AA\). The Fraunhofer lines are attributed to absorption of certain wavelengths of the Sun's "white" light by gaseous elements in the Sun's atmosphere. (a) Describe the process that causes absorption of specific wavelengths of light from the solar spectrum. (b) If a scientist wanted to know which Fraunhofer lines belonged to a given element, say neon, what experiments could she conduct here on Earth to provide data?

Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+}\). (a) Why is the Bohr model applicable to \(\mathrm{He}^{+}\) ions but not to neutral He atoms? (b) The ground-state energies of \(\mathrm{H}, \mathrm{He}^{+}\), and \(\mathrm{Li}^{2+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, Z. (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{C}^{5+}\) ion.

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

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, \(\mathrm{ZrO}_{2}\), is reacted 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 \(1: 2 \mathrm{Write}\) 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 Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

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