Skip to main content
Ch 39: Wave Functions and Uncertainty
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 39: Wave Functions and UncertaintyProblem 40d
Chapter 39, Problem 40d

A pulse of light is created by the superposition of many waves that span the frequency range f₀ − (1/2) Δf ≤ f ≤ f₀ + (1/2) Δf, where f₀ = c/λ is called the center frequency of the pulse. Laser technology can generate a pulse of light that has a wavelength of 600 nm and lasts a mere 6.0 fs (1 fs = 1 femtosecond =10−15 s). What is the spatial length of the laser pulse as it travels through space?

Verified step by step guidance
1
Step 1: Understand the problem. The spatial length of the laser pulse is the distance the pulse travels in space during its duration. This can be calculated using the relationship between speed, time, and distance: distance = speed × time. Here, the speed of the pulse is the speed of light in a vacuum, c = 3.0 × 10⁸ m/s, and the duration of the pulse is given as 6.0 fs (femtoseconds).
Step 2: Convert the duration of the pulse from femtoseconds to seconds. Since 1 femtosecond (fs) = 10⁻¹⁵ seconds, multiply the given duration (6.0 fs) by 10⁻¹⁵ to express it in seconds.
Step 3: Use the formula for distance: distance = speed × time. Substitute the speed of light (c = 3.0 × 10⁸ m/s) and the duration of the pulse (in seconds, calculated in Step 2) into the formula.
Step 4: Perform the multiplication to find the spatial length of the laser pulse. This will give the distance the pulse travels in meters during its duration.
Step 5: If needed, convert the result into a more convenient unit (e.g., nanometers or micrometers) for better interpretation, depending on the context of the problem.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Pulse of Light

A pulse of light is a short burst of electromagnetic radiation characterized by its duration and frequency range. It can be described as a superposition of multiple waves, each with different frequencies, which together create a localized wave packet. The center frequency, f₀, represents the average frequency of the pulse, while the bandwidth, Δf, indicates the range of frequencies present.
Recommended video:
10:30
Ray Nature of Light

Wavelength and Frequency Relationship

The relationship between wavelength (λ) and frequency (f) of light is given by the equation f = c/λ, where c is the speed of light in a vacuum. This relationship implies that as the wavelength decreases, the frequency increases, and vice versa. Understanding this relationship is crucial for determining the properties of light pulses, including their spatial characteristics.
Recommended video:
Guided course
05:08
Circumference, Period, and Frequency in UCM

Spatial Length of a Pulse

The spatial length of a light pulse can be calculated using the product of its speed and duration. Since light travels at a constant speed (c), the length of the pulse as it propagates through space can be found by multiplying the speed of light by the pulse duration. For a pulse lasting 6.0 femtoseconds, this calculation provides insight into how far the pulse travels during its brief existence.
Recommended video:
Guided course
10:54
Spinning on a string of variable length
Related Practice
Textbook Question

Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it’s reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted. A 238U nucleus, which decays by alpha emission, is 15 fm in diameter. Model an alpha particle within a 238U nucleus as being in a one-dimensional box. What is the maximum speed an alpha particle is likely to have?

Textbook Question

A particle is described by the wave function ψ(x)={cex/Lx≤0 mmce−x/Lx≥0 mm\(\psi\) (x)=\(\begin{cases}\) ce^{x/L} & x\(\leq\) 0\(\text{ mm}\) \\ ce^{-x/L} & x\(\geq\) 0\(\text{ mm}\) \(\end{cases}\) where L = 2.0 mm. Interpret your answer to part b by shading the region representing this probability on the appropriate graph in part a.

1
views
Textbook Question

Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it’s reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted. The probability that a nucleus will undergo alpha decay is proportional to the frequency with which the alpha particle reflects from the walls of the nucleus. What is that frequency (reflections/s) for a maximum-speed alpha particle within a 238U nucleus?

Textbook Question

Consider the electron wave function ψ(x)={c1−x2∣x∣≤1 cm0∣x∣≥1 cm\(\psi\) (x)=\(\begin{cases}\) c\(\sqrt{1-x^{2}\)} & \(\left\)|x\(\right\)|\(\leq\) 1\(\text{ cm}\) \\ 0 & \(\left\)|x\(\right\)|\(\geq\) 1\(\text{ cm}\) \(\end{cases}\) where x is in cm. If 104 electrons are detected, how many will be in the interval 0.00 cm ≤ x ≤ 0.50 cm?

Textbook Question

A particle is described by the wave function ψ(x)={cex/Lx≤0 mmce−x/Lx≥0 mm\(\psi\) (x)=\(\begin{cases}\) ce^{x/L} & x\(\leq\) 0\(\text{ mm}\) \\ ce^{-x/L} & x\(\geq\) 0\(\text{ mm}\) \(\end{cases}\) where L = 2.0 mm. Calculate the probability of finding the particle within 1.0 mm of the origin.

1
views