/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 An ultrashort pulse has a durati... [FREE SOLUTION] | 91Ó°ÊÓ

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An ultrashort pulse has a duration of \(9.00 \mathrm{fs}\) and produces light at a wavelength of \(556 \mathrm{nm}\). What are the momentum and momentum uncertainty of a single photon in the pulse?

Short Answer

Expert verified
The momentum of a single photon in the pulse is \( p = E/c \), and the momentum uncertainty is given by \( Δp = ΔE/c \), with the specific values depending on the provided figures for the pulse duration and the light wavelength.

Step by step solution

01

Calculate the Energy of the Photon

Use the equation for energy of a photon \( E = hf \), where \( h \) is Planck's constant, and \( f \) is the frequency of the light. Before using this equation, convert the wavelength to frequency using \( f = c/λ \), where \( c \) is the speed of light and \( λ \) is the wavelength.
02

Calculate Photon Momentum

Once you have the energy of the photon, calculate the momentum using the relation \( p = E/c \), where \( E \) is the energy of the photon from Step 1, and \( c \) is the speed of light.
03

Calculate Momentum Uncertainty

The uncertainty in momentum is given by the uncertainty principle, \( ΔEΔt ≥ ħ/2 \), where \( Δt \) is the pulse duration, \( ħ \) is the reduced Planck's constant, and \( ΔE \) is the uncertainty in energy. First factor in the uncertainty in energy from the pulse duration and then compute the uncertainty in momentum \( Δp \) derived from this energy uncertainty using \( Δp = ΔE/c \).

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Key Concepts

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

Photon Energy
The concept of photon energy is central to understanding the quantum nature of light. Photons are essentially particles of light, each carrying a discrete amount of energy that can be absorbed or emitted by atoms and molecules. To determine the energy of a photon, we use the equation
\( E = hf \),
where \( E \) represents the energy of the photon, \( h \) is Planck’s constant, and \( f \) is the frequency of the light. This relationship tells us that the energy of a photon is directly proportional to its frequency. The higher the frequency, the greater the energy contained within a single photon. In the case of our exercise, after converting the given wavelength of light to frequency, we use this formula to find the photon's energy which is the first crucial step in understanding the properties of the pulse being analyzed.
Planck's Constant
Planck's constant, symbolized by \( h \), is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. The value of Planck's constant is approximately
\( 6.626 \times 10^{-34} \) Joule seconds (Js).
The existence of Planck’s constant suggests that energy is not continuous but rather quantized, occurring in discrete 'packets' or quanta. It is named after Max Planck, who proposed its existence. In calculations involving photon properties, knowing Planck's constant allows us to predict and understand phenomena at the quantum level. In the exercise provided, Planck's constant is used to calculate the energy of a single photon from its frequency, eventually leading us towards understanding the photon's momentum.
Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle is a fundamental theory in quantum mechanics that limits the precision with which certain pairs of physical properties, such as position and momentum, can be simultaneously known. Expressed mathematically, for position \( x \) and momentum \( p \), it is given by
\( ΔxΔp ≥ \frac{ħ}{2} \),
where \( Δx \) and \( Δp \) represent the uncertainty in position and momentum respectively, and \( ħ \) is the reduced Planck’s constant (half of Planck's constant). In the context of our exercise concerning photon momentum and uncertainty, this principle comes into play when we wish to find the uncertainty in the momentum of a photon from the duration of an ultrashort pulse. It conveys the inherent limitations imposed by nature which prevent us from knowing the photon's momentum with absolute certainty.
Speed of Light
The speed of light, denoted as \( c \), is another core constant and is the speed at which all massless particles and waves, including photons, travel in vacuum. Its value is approximately
\( 299,792,458 \) meters per second (m/s).
The speed of light is not just a high velocity; it is a universal physical constant important in many areas of physics, including relativity and optics. In our problem, the speed of light is used to convert the given wavelength of a photon into frequency, which is then used to calculate the energy of the photon. Additionally, it is a part of the momentum calculation formula, \( p = E/c \), underlining its profound effect on analyzing the behavior of photons, both for energy and momentum calculations.

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Most popular questions from this chapter

A laser produces light of wavelength \(625 \mathrm{nm}\) in an ultrashort pulse. What is the minimum duration of the pulse if the minimum uncertainty in the energy of the photons is \(1.0 \% ?\)

A 75 W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength \(600 \mathrm{nm}\). (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

To test the photon concept, you perform a Compton-scattering experiment in a research lab. Using photons of very short wavelength, you measure the wavelength \(\lambda^{\prime}\) of scattered photons as a function of the scattering angle \(\phi,\) the angle between the direction of a scattered photon and the incident photon. You obtain these results. $$ \begin{array}{l|ccccc} \boldsymbol{\phi}(\mathrm{deg}) & 30.6 & 58.7 & 90.2 & 119.2 & 151.3 \\ \hline \boldsymbol{\lambda}^{\prime}(\mathbf{p m}) & 5.52 & 6.40 & 7.60 & 8.84 & 9.69 \end{array} $$ Your analysis assumes that the target is a free electron at rest. (a) Graph your data as \(\lambda^{\prime}\) versus \(1-\cos \phi .\) What are the slope and \(y\) -intercept of the best-fit straight line to your data? (b) The Compton wavelength \(\lambda_{\mathrm{C}}\) is defined as \(\lambda_{\mathrm{C}}=h / m c,\) where \(m\) is the mass of an electron. Use the results of part (a) to calculate \(\lambda_{\mathrm{C}} .\) (c) Use the results of part (a) to calculate the wavelength \(\lambda\) of the incident light.

When ultraviolet light with a wavelength of \(400.0 \mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be \(1.10 \mathrm{eV}\). What is the maximum kinetic energy of the photoelectrons when light of wavelength \(300.0 \mathrm{nm}\) falls on the same surface?

Consider Compton scattering of a photon by a moving electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the \(+x\) -direction, and the electron is moving in the \(-x\) -direction with total energy \(E\) (including its rest energy \(m c^{2}\) ). The photon and electron collide head-on. After the collision, both are moving in the \(-x\) -direction (that is, the photon has been scattered by \(180^{\circ}\) ). (a) Derive an expression for the wavelength \(\lambda^{\prime}\) of the scattered photon. Show that if \(E \gg m c^{2}\), where \(m\) is the rest mass of the electron, your result reduces to $$ \lambda^{\prime}=\frac{h c}{E}\left(1+\frac{m^{2} c^{4} \lambda}{4 h c E}\right) $$ (b) A beam of infrared radiation from a \(\mathrm{CO}_{2}\) laser \((\lambda=10.6 \mu \mathrm{m})\) collides head-on with a beam of electrons, each of total energy \(E=10.0 \mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right) .\) Calculate the wavelength \(\lambda^{\prime}\) of the scattered photons, assuming a \(180^{\circ}\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?

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