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For reasons having to do with quantum mechanics, a given kind of atom can emit only certain wavelengths of light. These spectral lines serve as a "fingerprint." For instance, hydrogen's only visible spectral lines are \(656,486.434,\) and \(410 \mathrm{nm} .\) If spectral lines were of absolutely precise wavelength, they would be very difficult to discern. Fortunately, two factors broaden them: the uncertainty principle (discussed in Chapter 4 ) and Doppler broadening. Atoms in a gas are in motion, so some light will arrive that was emitted by atoms moving toward the observer and some from atoms moving away. Thus, the light reaching the observer will cover 8 range of wavelengths. (a) Making the assumption that atoms move no faster than their rms speed-given by \(v_{\mathrm{nns}}=\sqrt{2 k_{\mathrm{B}} T / m},\) where \(k_{\mathrm{B}}\) is the Boltanann constant obtain a formula for the range of wavelengths in terms of the wavelength \(\lambda\) of the spectral line, the atomic mass \(m,\) and the temperature \(T\). (Note: \(\left.v_{\text {rms }} \ll c .\right)\) (b) Evaluate this range for the 656 nm hydrogen spectral line, assuming a temperature of \(5 \times 10^{4} \mathrm{~K}\).

Short Answer

Expert verified
The range of wavelengths is calculated by using the derived formula \(\Delta \lambda = (\lambda v_{rms}) / c\), where \(v_{rms} = \sqrt{2k_BT/m}\), and substituting given values into this equation.

Step by step solution

01

Understanding the Concept

From the physics principle of Doppler effect, the change in frequency (or wave length) due to the relative motion between the source and observer is given by \(\Delta f = f' - f\), where \(f'\) and \(f\) are the observed and original frequency. This difference can be expressed in terms of wavelength as \(\Delta f = c(1/\lambda' - 1/\lambda)\), where \(\lambda'\) and \(\lambda\) are the observed and original wavelength, and \(c\) is the speed of light. Hence, the change in wavelength \(\Delta \lambda = \lambda' - \lambda\). Since \(v_{rms} << c\), we have \(\lambda' = \lambda + \Delta \lambda\).
02

Deriving the Formula

Rearranging for \(\Delta \lambda\), we get \(\Delta \lambda = \lambda' - \lambda = (c/\Delta f) - \lambda\). Substituting \(\Delta f = v_{rms}/\lambda\) and rearranging, we get \(\Delta \lambda = (\lambda^2 v_{rms}) / c - \lambda\). Simplifying this, we get \( \Delta \lambda =\lambda ((\lambda v_{rms}/ c) - 1) = \lambda ((v_{rms}/c) - 1) = (\lambda v_{rms}/c) \), where \(v_{rms} = \sqrt{2k_BT/m}\). Hence the formula for the range of wavelengths is \(\Delta \lambda = (\lambda v_{rms}) / c\).
03

Evaluating the Range

Substitute the given values \(\lambda = 656 nm = 656 x 10^{-9} m, T = 5 x 10^4 K, k_B = 1.38 x 10^{-23} JK^{-1}, m = 1.67 x 10^{-27} kg\) (mass of a hydrogen atom), and \(c = 2.998 x 10^8 ms^{-1}\) into the formula \(\Delta \lambda = (\lambda v_{rms}) / c\) to obtain the range.

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behavior of particles on an atomic and subatomic level. It revolutionizes our understanding of how nature operates at these small scales, contrasting with classical physics.
One of the distinctive features of quantum mechanics is the idea that energy levels in an atom are quantized. This means electrons can only occupy certain energy levels. As a result, when an electron transitions between two levels, it releases or absorbs photons with specific energies corresponding to those levels.
This energy translates to specific wavelengths of light, which is why atoms emit spectral lines. These lines are unique to each element, making them akin to a fingerprint, assisting in the identification of substances in various fields, including astrophysics and chemistry.
Uncertainty Principle
One of the most intriguing concepts in quantum mechanics is the Heisenberg Uncertainty Principle. This principle states that it is impossible to simultaneously know both the exact position and momentum of a particle with absolute certainty. In simpler terms, if we pinpoint a particle's position very precisely, its momentum will be uncertain, and vice versa.
This principle significantly impacts the behavior of atomic systems. Because of this inherent uncertainty, spectral lines emitted by atoms are not infinitely sharp. Instead, they have a slight broadening. This broadening arises due to the uncertainties in energy levels, contributing to the practical observation of spectral lines.
Remember, this broadening allows these lines to remain discernible. Without it, overlapping lines would be difficult to separate, making spectroscopic observations impractical.
Spectral Lines
Spectral lines are a vital aspect of the study of light and matter. When atoms absorb or emit light, they do so at specific wavelengths, producing a series of lines known as spectral lines.
Each chemical element has its own set of spectral lines, including distinctive patterns and characteristics. These patterns act like fingerprints for elements, aiding in their detection in stars and other celestial bodies.
The Doppler effect and the uncertainty principle influence these lines. Broadening due to these effects can provide information about the temperature, density, and motion of the atoms in a system, making spectral lines essential in both laboratory analysis and astrophysical observations.
Atomic Mass
Atomic mass refers to the mass of an atom, typically expressed in atomic mass units (amu). The atomic mass considers the masses of all the protons, neutrons, and electrons in an atom, with the former two contributing most significantly.
The mass of atoms, particularly in gases, influences their motion, and consequently, their diffusion properties and the spectrum of light they emit or absorb. The movement and momentum of atoms depend on their masses, which ties into the Doppler effect and spectral line broadening.
Understanding atomic mass is crucial when applying concepts like the root-mean-square speed, which determines the thermal velocity distribution of gas particles, linking mass to properties like temperature and pressure.
Wavelength Range
The wavelength range describes the interval between the shortest and longest wavelengths of light that a system can emit or absorb. This is particularly significant when discussing light from atoms, as it provides insight into the characteristics of the emitting or absorbing materials.
The range of wavelengths is influenced by factors like temperature, pressure, and the atomic mass of the particles involved. In the context of the Doppler effect, atoms in motion due to temperature changes will shift the wavelengths of emitted or absorbed light.
This range plays a vital role in the precision of spectroscopic measurements, contributing to our understanding of physical conditions in various environments, such as stars or laboratory conditions, and assisting in the exploration and identification of foreign bodies through spectroscopic techniques.

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Most popular questions from this chapter

A famous experiment detected 527 muons per hour at the top of Mt. Washington, New Hampshire, elevation \(1910 \mathrm{~m}\). At sea level. the same equipment detected 395 muons per hour. A discriminator selected for muons whose speed was between \(0.9950 c\) and \(0.9954 c\). Given that the mean lifetime \(\tau\) of a muons in a frame in which it is at rest is \(2.2 \mu \mathrm{s}\) and that in this frame the number of moons decays exponentially with time accounting \(\operatorname{to} N=N_{0} e^{-d r},\) show that the results obtained in the experiment are sensible.

Both classically and relativistically, the force on an object is what causes a time rate of change of its momentum: \(F=d p / d t\). (a) Using the relativistically cotrect expression for momentum, show that $$ F=\gamma_{u}^{3} m \frac{d u}{d l} $$ (b) Under what condition does the classical equation \(F=m a\) hold? \(?\) (c) Assuming a constant force and that the speed is zero at \(t=0\), separate \(t\) and \(u\), then integrate to show that $$ u=\frac{1}{\sqrt{1+(F t / m c)^{2}}} \frac{F}{m} t $$ (d) Plot \(u\) versus \(t\). What happens to the velocity of an object when a constant force is applied for an indefinite length of time?

A \(3.000\) u ob ject moving to the right through a laboratory at \(0.8 c\) collides with a \(4.000\) u ob ject moving to the left through the laboratory at \(0.6 c\). Afterward, there are two objects. one of which is a \(6.000\) u mass at rest. (a) What are the mass and speed of the other object? (b) Determine the change in kinetic energy in this collision.

From a standstill, you begin jogging at \(5 \mathrm{~m} / \mathrm{s}\) directly toward the galaxy Centaurus A. which is on the horizon \(2 \times 10^{23} \mathrm{~m}\) away, (a) There is a clock in Centaurus A. According to you, how will readings on this clock differ before and after you begin jogging? (Remember: You change frames.) (b) The planet Neptune is between Earth and Centaurus A. \(4.5 \times 10^{9} \mathrm{~m}\) from Earth. How much would readings on a clock there differ? (c) What would be the time differences if you had instead begun jogging in the opposite direction? (d) What do these results tell you about the observations of a traveling twin who accelerates toward his Earth-bound twin? How do these observations depend on the distance between the twins?

Particle 1, of mass \(m_{1}\), moving at \(0.8 c\) relative to the lab. collides head-on with particle 2 , of mass \(m_{2}\), mov ing at \(0.6 c\) relative to the lab. Afterward, there is a single stationary object. Find, in terms of \(m_{1},(a) m_{2}:\) (b) the mass of the final stationary ob ject: and (c) the change in kinetic energy in this collision.

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