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Chapter 37: Problem 67

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is \(\lambda=656.3 \mathrm{nm},\) in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler- shifted to \(\lambda=953.4 \mathrm{nm},\) in the infrared portion of the spectrum. How fast are the emitting atoms moving relative to the earth? Are they approaching the earth or receding from it?

Short Answer

Expert verified
Velocity is \(1.36 \times 10^8\) m/s away from Earth.

Step by step solution

01

Understanding the Doppler Effect

The Doppler effect is the change in the frequency or wavelength of a wave in relation to an observer moving relative to the wave source. When light is redshifted, it indicates that the source is moving away from the observer.
02

Identify Known Values

We are given the initial wavelength of hydrogen, \( \lambda_0 = 656.3 \text{ nm} \), and the observed wavelength as \( \lambda = 953.4 \text{ nm} \).
03

Calculate the Change in Wavelength

Calculate the change in wavelength \( \Delta \lambda = \lambda - \lambda_0 \). This gives us \( \Delta \lambda = 953.4 \text{ nm} - 656.3 \text{ nm} = 297.1 \text{ nm} \).
04

Use the Redshift Formula

The redshift \( z \) is defined as \( z = \frac{\Delta \lambda}{\lambda_0} \). Substitute the values to get \( z = \frac{297.1}{656.3} \approx 0.4526 \).
05

Relate Redshift to Velocity

For small values of \( z \), the velocity \( v \) can be approximated by \( v = c \cdot z \), where \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \). Thus, \( v = 3 \times 10^8 \times 0.4526 \approx 1.36 \times 10^8 \text{ m/s} \).
06

Determine Direction of Movement

Since the wavelength is longer in the infrared region, the light is redshifted. Thus, the galaxy is receding from the Earth.

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Key Concepts

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

Wavelength
Wavelength is a fundamental concept in understanding waves and how they propagate through different media. It refers to the distance between two consecutive peaks or troughs in a wave. In the context of light, it is the distance between successive crests of a light wave. Wavelength is usually denoted by the Greek letter lambda (\(\lambda\)).
- In this exercise, we deal with the wavelengths of light emitted by hydrogen atoms.
- The initial wavelength given is \(\lambda_0 = 656.3 \, \text{nm}\) while under the influence of the Doppler effect, it changes to \(\lambda = 953.4 \, \text{nm}\).

Wavelength plays a crucial role in determining the color of light in the visible spectrum. In this scenario, the change in wavelength takes the light from the red part of the spectrum to the infrared part, which helps in understanding the motion of celestial bodies like galaxies.
Redshift
Redshift is a key concept when discussing distant celestial objects and the universe's dynamic nature. It refers to the increase in the wavelength of light emitted by an object as it moves away from an observer.
- This increase in wavelength causes the light to shift towards the red end of the spectrum.
- Redshift is denoted by \(z\) and is calculated using the formula \(z = \frac{\Delta \lambda}{\lambda_0}\).

In the given problem, the redshift value \(z\) tells us how fast the galaxy is receding from us. The larger the redshift, the faster the object is moving away. With a calculated redshift of \(0.4526\), it indicates a significant departure speed. Redshift not only helps in astronomical measurements but also in understanding the expansion of the universe.
Velocity Calculation
Calculating the velocity of a distant galaxy or any celestial object involves understanding its movement relative to Earth. The Doppler effect allows us to determine this velocity using the redshift of light.
- For this exercise, we use the approximation for small redshift values, \(v = c \times z\), where \(c\) is the speed of light (approximately \(3 \times 10^8 \, \text{m/s}\)).

In our exercise:
  • We substitute \(z = 0.4526\)
  • The calculated velocity \(v\) is approximately \(1.36 \times 10^8 \, \text{m/s}\)
This calculation shows how fast the galaxy is moving away, providing insight into both individual celestial movements and larger cosmic phenomena like the expansion of the universe.
Hydrogen Spectral Lines
Hydrogen spectral lines are specific wavelengths of light emitted by hydrogen atoms when electrons transition between energy levels. These spectral lines are fundamental to astronomy because they appear as bright lines in a spectrum.
- Each line corresponds to a specific wavelength, characteristic of hydrogen.
- The most famous of these lines is the Balmer series in the visible region.

In the context of this exercise, we focus on the hydrogen line at \(656.3 \, \text{nm}\). Observing a shift in this line allows scientists to determine the motion of astronomical objects like galaxies. Any change from the expected spectral line indicates movement, such as redshift indicating the object is moving away. Thus, hydrogen spectral lines are invaluable for studying both nearby stars and distant galaxies.

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Most popular questions from this chapter

As you pilot your space utility vehicle at a constant speed toward the moon, a race pilot fies past you in her spaceracer at a constant speed of 0.800\(c\) relative to you. At the instant the spaceracer passes you, both of you start timers at zero. (a) At the instant when you measure that the spaceracer has traveled \(1.20 \times 10^{8} \mathrm{m}\) past you, what does the race pilot read on her timer? (b) When the race pilot reads the value calculated in part (a) on her timer, what does she measure to be your distance from her? (c) At the instant when the race pilot reads the value calculated in part (a) on her timer, what do you read on yours?

The French physicist Armand Fizeau was the first to measure the speed of light accurately. He also found experimentally that the speed, relative to the lab frame, of light traveling in a tank of water that is itself moving at a speed \(V\) relative to the lab frame is $$v=\frac{c}{n}+k V$$ where \(n=1.333\) is the index of refraction of water. Fizeau called \(k\) the draging coefficient and obtained an experimental value of \(k=0.44 .\) What value of \(k\) do you calculate from relativistic transformations?

An extraterrestrial spaceship is moving away from the earth after an unpleasant encounter with its inhabitants. As it departs, the spaceship fires a missile toward the earth. An observer on earth measures that the spaceship is moving away with a speed of 0.600\(c .\) An observer in the spaceship measures that the missile is moving away from him at a speed of 0.800\(c .\) As measured by an observer on earth, how fast is the missile approaching the earth?

As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 \(\mathrm{m} .\) (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of \(4.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

In high-energy physics, new particles can be created by collisions of fast- moving projectile par- ticles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive kaon \(\left(\mathrm{K}^{+}\right)\) $$p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 \(\mathrm{MeV}\) . (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)
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