/*! 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 13 Equation \((3-1)\) expresses Pla... [FREE SOLUTION] | 91Ó°ÊÓ

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Equation \((3-1)\) expresses Planck's spectral energy density as an energy per range \(d f\) of frequencies. Quite of ten, it is more convenient to express it as an energy per range \(d \lambda\) of wavelengths, By differentiating \(f=c / \lambda,\) we find that df \(=-c / \lambda^{2} d \lambda\). Ignoring the minus sign (we are interested only in relating the magnitudes of the ranges \(d f\) and \(d \lambda\) ). show that, in terms of wavelength. Planck's formula is \(\frac{d U}{d \lambda}=\frac{8 \pi V h c}{e^{h c / \lambda k_{B} T}-1} \frac{1}{\lambda^{5}}\)

Short Answer

Expert verified
\(\frac{dU}{d \lambda} = \frac{8 \pi V h c}{e^{hc / \lambda k_{B}T}-1} \frac{1}{\lambda^5}\). The rule of negative signs when switching \(df\) to \(d\lambda\) has been settled by the absolute value, which reflects the physical significance of the magnitudes of the ranges. It took advantage of the differentiation of the reciprocal function to obtain \(\lambda^5\) in the denominator.

Step by step solution

01

Start with the given formula

Planck's spectral energy density is given as \(dU = \frac{8\pi V h f^3}{e^{hf/k_{B}T}-1} df\)
02

Equate and replace

Substitute the given values and write the conversion in terms of frequency \(f\), \(df = -c / \lambda^{2} d \lambda\) to get \(dU = \frac{8 \pi V h (c / \lambda)^3}{e^{hc / \lambda k_{B}T}-1} (-c / \lambda^{2} d \lambda)\)
03

Simplify the Equation

After simplifying the above equation, it is observed that \(dU = \frac{8 \pi V h c^4}{\lambda^5 (e^{hc / \lambda k_{B}T}-1)} d\lambda\)
04

Finalize the Equation

Rearrange the equation into the standard form of Planck's formula, which gives us \(\frac{dU}{d \lambda} = \frac{8 \pi V h c}{e^{hc / \lambda k_{B}T}-1} \frac{1}{\lambda^5}\)

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Key Concepts

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

Spectral Energy Distribution
The concept of spectral energy distribution (SED) is fundamental for understanding how energy is emitted across different wavelengths or frequencies by various objects in physics and astronomy. An object's SED is a representation of the amount of energy it emits at each wavelength or frequency and is crucial for examining the properties of light-emitting sources, such as stars or black bodies.

An SED can be plotted with energy density on the y-axis and frequency or wavelength on the x-axis. This graph provides a visual insight into which part of the electromagnetic spectrum the object is most radiantly efficient. By studying an SED, scientists can infer the temperature, composition, and motion of the object in question. When it comes to blackbody radiation, Planck's spectral energy density formula is often used to calculate the energy emitted per unit frequency or wavelength interval by a perfect blackbody.
Frequency to Wavelength Conversion
Frequency and wavelength are two essential characteristics of waves, including electromagnetic radiation. The frequency (\f) of a wave refers to the number of wave cycles that pass a point per unit of time, typically measured in hertz (Hz), whereas the wavelength (\f\begin{small}\(\backslash\)lambda\f\begin{small}) is the distance between successive wave peaks, measured in meters (m) for electromagnetic waves.

The relationship between frequency and wavelength is inversely proportional and is dictated by the equation \f\begin{small}f = c / \begin{small}\(\backslash\)lambda\begin{small}\f\begin{small}, where \begin{small}c\f\begin{small} is the speed of light. This relationship is crucial in converting Planck's spectral energy density from a frequency basis to a wavelength basis. Understanding this conversion allows for the comparison of measurements taken in different units and aids in computational simplification across various physics fields, including optics and quantum mechanics.
Blackbody Radiation
Blackbody radiation is a cornerstone concept in thermodynamics and quantum mechanics. A blackbody is an idealized physical body that perfectly absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Consequently, it also emits radiation with maximum efficiency at every frequency.

The spectral energy density of a blackbody is described by Planck's law, which provides the amount of electromagnetic radiation emitted by a blackbody in thermal equilibrium at a given temperature, per unit area, per unit solid angle, and within a given frequency. This emission is characterized by the blackbody's temperature, and the SED peaks at a wavelength inversely related to the temperature (Wien's Displacement Law). The quantization of energy and the concept of photons are integral to the explanation of blackbody radiation, leading to the development of quantum mechanics.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the behaviors and interactions of particles at microscopic scales, where the classical laws of physics are no longer sufficient. Max Planck's introduction of the concept of quantization of energy is seen as the birth of quantum mechanics. He proposed that energy is emitted or absorbed in discrete units called quanta.

Planck's spectral energy density formula is a groundbreaking application of quantum mechanics. It incorporates the idea that electromagnetic waves can exhibit particle-like properties, leading to the conclusion that light can be absorbed or emitted in quantized packets (photons). This quantization resolved the classical ultraviolet catastrophe problem by correctly describing the emitted radiation from black bodies. Quantum mechanics radically changed our understanding of fundamental processes at the smallest scales, profoundly affecting fields such as chemistry, material science, and electronics.

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Most popular questions from this chapter

Suppose we produce X-rays not by smashing elections into targets but by smashing protons, which are far more massive. If the same accelerating potential difference were used for both, how would the cutoff wavelengths of the two X-ray spectra compare? Explain.

An electron moving to the lefi at \(0.8 c\) collides with an incoming photon moving to the right. Afuer the collision, the elactron is moving to the right at \(0.6 c\) and an outgoing photon moves to the lefi. What was the wavelength of the incoming photon?

A bedrock topic in quantum mechanics is the uncertainty principle. It is discussed mostly for massive objects in Chapter \(4,\) but the idea also applies to light: Increasing certainty in knowledge of photon position implies increasing uncertainty in knowledge of its momentum, and vice versa. A single-slit pattern that is developed (like the double-slit pattern of Section 3.6 ) one photon at a time provides a good example. Depicted in the accompanying figure, the pattern shows that photons emerging from a narrow slit are spread allover; a photon's \(x\) -component of momentum can be any value over a broad range and is thus uncertain. On the other hand, the \(x\) -coordinate of position of an emerging photon covers a fairly small range, for \(w\) is small. Using the single-slit diffraction formula \(n \lambda=w \sin \theta,\) show that the range of likely values of \(p_{x}\), which is roughly \(p \sin \theta\), is inversely proportional to the range \(w\) of likely position values. Thus, an inherent wave nature implies that the precisions with which the particle properties of position and momentum can be known are inversely proportional.

Show that the laws of momentum and enargy corservadion forbid the complete absorprion of a photon by a free electron. (Note: This is not the photoelectric effect In the pholoelectric effect, the electron is not free: the metal participates in momentum and eoergy onservation.)

Light of \(300 \mathrm{nm}\) wavelength strikes a metal plate, and photoclectrons are produced moving as \(f\) ast as \(0.002 c\). (a) What is the work function of the metal? (b) What is the threshold wavelength for this metal?

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