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Chapter 38: Problem 23

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 \(1.0 \% ?\)

Short Answer

Expert verified
The minimum duration of the laser pulse is derived from the Heisenberg's Uncertainty Principle. With the given uncertainty in energy as 1% of the energy expressed in terms of wavelength, the minimum duration (uncertainty in time) can be derived by solving the Uncertainty Principle equation. The exact result depends on the values of the Planck's constant and the speed of light used.

Step by step solution

01

Express Energy in terms of Wavelength

The energy \(E\) of a photon can be expressed in terms of its wavelength \(\lambda\) by using Planck's relation: \(E = \frac{hc}{\lambda} \), where \(h\) is the Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. With the given wavelength \(625nm = 625 \times 10^{-9} m\), solve for \(E\).
02

Find the Uncertainty in Energy

The minimum uncertainty in the energy of the photons is given to be 1%. That means the uncertainty \( \Delta E \) is \(1\%\) of the calculated energy \(E\) from Step 1.
03

Solve for Minimum Duration

Substitute the calculated energy uncertainty \( \Delta E \) and Planck's constant \(h\) into the Heisenberg's Uncertainty Principle equation \( \Delta E \Delta t \geq \frac{h}{4\pi} \) to solve for \(\Delta t\). Make sure to rearrange the formula to \( \Delta t = \frac{h}{4\pi \Delta E}\) before substituting the values. This will give the minimum duration of the pulse.

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Key Concepts

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

Photon Energy
Photon energy describes the amount of energy carried by a photon, the fundamental particle of light. The energy of a single photon is directly related to its electromagnetic wave properties, particularly its frequency and wavelength. Understanding photon energy is crucial when studying the behavior of light in both classical and quantum mechanics.

In the context of lasers and light sources, determining the photon energy is essential for a range of applications, such as spectroscopy, medical treatments, and communications technologies. By knowing the energy of the photons produced by a laser, one can infer various characteristics of the laser light, like its potential to cause transitions in atoms or molecules, and its ability to convey information.
Planck's Relation
Planck's relation is a fundamental equation in quantum physics, connecting the energy of a photon to its frequency. The equation is given by \( E = h u \), where \( E \) is the energy of the photon, \( h \) is Planck's constant (approximately \( 6.626 \times 10^{-34} \) Js), and \( u \) is the frequency of the photon. An alternative form of Planck's relation involves the wavelength \( \lambda \) of the photon: \( E = \frac{hc}{\lambda} \), with \( c \) being the speed of light in a vacuum.

This relation implies that the energy of a photon is inversely proportional to its wavelength: the shorter the wavelength, the higher the energy. This concept is key to understanding the emission and absorption spectra of atoms, the functioning of various electronic devices, and the principles underlying different types of spectroscopy.
Laser Pulse Duration
Laser pulse duration refers to the length of time over which the laser emits light during a single pulse. In the case of ultrashort laser pulses, durations can range from femtoseconds (\(10^{-15}\) seconds) to picoseconds (\(10^{-12}\) seconds) and are a key characteristic in precision applications such as in cutting materials, in medical procedures like LASIK eye surgery, and in scientific research where the interaction of light and matter at very short timescales is studied.

The pulse duration is not only important for how much energy is delivered in each pulse but also relates to how accurately energy can be targeted. This is vital in avoiding damage to surrounding material or tissues in delicate procedures. Additionally, the duration of the pulse has implications for the minimum uncertainty in the energy and position of photons, as described by Heisenberg's Uncertainty Principle.

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Most popular questions from this chapter

In developing night-vision equipment, you need to measure the work function for a metal surface, so you perform a photoelectric-effect experiment. You measure the stopping potential \(V_{0}\) as a function of the wavelength \(\lambda\) of the light that is incident on the surface. You get the results in the table. $$ \begin{array}{l|llllll} \boldsymbol{\lambda}(\mathbf{n m}) & 100 & 120 & 140 & 160 & 180 & 200 \\ \hline \boldsymbol{V}_{0}(\mathbf{V}) & 7.53 & 5.59 & 3.98 & 2.92 & 2.06 & 1.43 \end{array} $$ In your analysis, you use \(c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and \(e=1.602 \times 10^{-19} \mathrm{C}\) which are values obtained in other experiments. (a) Select a way to plot your results so that the data points fall close to a straight line. Using that plot, find the slope and \(y\) -intercept of the best-fit straight line to the data. (b) Use the results of part (a) to calculate Planck's constant \(h\) (as a test of your data) and the work function (in eV) of the surface. (c) What is the longest wavelength of light that will produce photoelectrons from this surface? (d) What wavelength of light is required to produce photoelectrons with kinetic energy \(10.0 \mathrm{eV} ?\)

A beam of photons with just barely enough individual photon energy to create electron-positron pairs undergoes Compton scattering from free electrons before any pair production occurs. (a) Can the Compton-scattered photons create electron-positron pairs? (b) Calculate the momentum magnitude of the photons found at a \(20.0^{\circ}\) angle from the direction of the original beam.

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.

A \(2.50 \mathrm{~W}\) beam of light of wavelength \(124 \mathrm{nm}\) falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is \(4.16 \mathrm{eV}\). Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

A photon with wavelength \(\lambda=0.0980 \mathrm{nm}\) is incident on an electron that is initially at rest. If the photon scatters in the backward direction, what is the magnitude of the linear momentum of the electron just after the collision with the photon?
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