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Medical X rays. Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10\(\mathrm { nm }\) . What are the frequency, period, and wave number of such waves?

Short Answer

Expert verified
Frequency: 3.00 x 10^18 Hz, Period: 3.33 x 10^{-19} s, Wave number: 6.28 x 10^10 m^{-1}.

Step by step solution

01

Convert Wavelength to Meters

First, we need to convert the given wavelength from nanometers to meters because the standard units for wavelength in physics are meters. Given that 1 nanometer is equal to \(10^{-9}\) meters, we have:\[0.10 \text{ nm} = 0.10 \times 10^{-9} \text{ m} = 1.0 \times 10^{-10} \text{ m}\]
02

Calculate Frequency using the Speed of Light

The speed of light \(c\) is approximately \(3.00 \times 10^8\) m/s. The relationship between speed \(v\), frequency \(f\), and wavelength \(\lambda\) is given by the equation:\[c = f \times \lambda\]We can rearrange this to solve for frequency:\[f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{1.0 \times 10^{-10} \text{ m}} = 3.00 \times 10^{18} \text{ Hz}\]
03

Determine the Period of the Wave

The period \(T\) is the reciprocal of the frequency \(f\). We use the formula:\[T = \frac{1}{f}\]Substituting in the value we found for frequency:\[T = \frac{1}{3.00 \times 10^{18} \text{ Hz}} \approx 3.33 \times 10^{-19} \text{ s}\]
04

Calculate the Wave Number

The wave number \(k\) is defined as the number of wavelengths per unit distance, calculated as \(k = \frac{2\pi}{\lambda}\). With our given wavelength:\[k = \frac{2\pi}{1.0 \times 10^{-10} \text{ m}} = 2\pi \times 10^{10} \text{ m}^{-1} \approx 6.28 \times 10^{10} \text{ m}^{-1}\]

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Key Concepts

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

Electromagnetic waves
Electromagnetic waves are fascinating waves that carry energy through space. Unlike sound waves, they do not need a medium to travel. This means they can move through a vacuum, like the emptiness of space. These waves are composed of oscillating electric and magnetic fields. You can think of the electric field and the magnetic field as two invisible sheets, one riding on top of the other, and they are always perpendicular to each other.
These waves include a range of different types, each with different properties, from radio waves to gamma rays. The family of electromagnetic waves can be described by their varying wavelengths and frequencies. Medical X-rays fall into the part of the spectrum with very short wavelengths and high frequencies, allowing them to penetrate body tissues and provide vital medical imaging.
Wavelength conversion
Understanding wavelength conversion is crucial in studying electromagnetic waves. Wavelength is the distance between two consecutive peaks of a wave, usually measured in meters. However, many electromagnetic waves, like X-rays, have very short wavelengths often measured in nanometers (nm) or even smaller units.
To work with these waves in physics, we often need to convert wavelengths into standard units of meters. The conversion is straightforward: 1 nanometer equals \(10^{-9}\) meters. For example, if an X-ray has a wavelength of 0.10 nm, converting it to meters gives \(0.10 \times 10^{-9}\) meters, which simplifies to \(1.0 \times 10^{-10}\) meters.
Speed of light
The speed of light is a central concept in physics, especially when studying electromagnetic waves. Denoted as \(c\), the speed of light in a vacuum is approximately \(3.00 \times 10^8\) meters per second (m/s). It's the universal speed limit for how fast information or matter can travel.
This speed connects with the wave's frequency \(f\) and wavelength \(\lambda\) through the equation \(c = f \times \lambda\). When you know two of these values, you can always calculate the third. This relationship allows us to determine how frequently the wave oscillates—or its frequency—based on its known wavelength and the speed of light.
Wave frequency
Wave frequency tells us how many wave peaks pass a fixed point every second. It is measured in Hertz (Hz), where 1 Hz is equal to 1 cycle per second. For waves like medical X-rays, which have very short wavelengths, the frequency is extremely high.
Using the formula \(f = \frac{c}{\lambda}\), where \(c\) is the speed of light and \(\lambda\) is the wavelength, we can calculate frequency. For example, given a wavelength of \(1.0 \times 10^{-10}\) meters, the frequency of these X-rays would be \(3.00 \times 10^{18}\) Hz, meaning that many oscillations occur every second.
Wave period
The wave period \(T\) is the time it takes for one complete wavelength to pass a point. Since it's the inverse of frequency, if you know the frequency of a wave, you can easily find its period using \(T = \frac{1}{f}\).
For high-frequency waves, like medical X-rays, the period is very short. If the frequency is \(3.00 \times 10^{18}\) Hz, the period comes out to approximately \(3.33 \times 10^{-19}\) seconds. This tiny period reflects the rapid oscillation of high-frequency waves.
Wave number
Wave number \(k\) provides another way to describe waves by focusing on the spatial aspect rather than temporal. It quantifies how many waves exist per unit of distance and is given by \(k = \frac{2\pi}{\lambda}\). The wave number is often expressed in units of \(m^{-1}\), meaning "per meter."
For example, with a wavelength of \(1.0 \times 10^{-10}\) meters, the wave number becomes \(2\pi \times 10^{10} \space m^{-1}\), approximately \(6.28 \times 10^{10} \space m^{-1}\). This indicates a very high density of waves in a given space, which aligns with the characteristics of high-energy waves such as X-rays.

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Most popular questions from this chapter

A standing electromagnetic wave in a certain material has frequency \(2.20 \times 10 ^ { 10 } \mathrm { Hz }\) . The nodal planes of \(\vec { \boldsymbol { B } }\) are 3.55\(\mathrm { mm }\) apart. Find (a) the wavelength of the wave in this material; (b) the distance between adjacent nodal planes of the \(\vec { \boldsymbol { E } }\) field; (c) the speed of propagation of the wave.

Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 316\(\mathrm { kW }\) . Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00\(\mathrm { km }\) away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

In the 25 -ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead are lamps can produce light of intensity 2500\(\mathrm { W } / \mathrm { m } ^ { 2 }\) at the floor of the facility. This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflection section of the floor. (c) Find the average momentum densit (momentum per unit volume) in the light at the floor.

Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is 2.5\(\mathrm { km }\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090\(\mathrm { V } / \mathrm { m }\) and that the frequency of the waves is 244\(\mathrm { MH } z\) . Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) Verify this by substituting \(E _ { y } ( x , t )\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) m; permeability \(\mu = \mu _ { 0 } )\) . since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

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