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Chapter 42: Problem 9

What wavelength must electromagnetic radiation have if a photon in the beam is to have the same momentum as an electron moving with a speed of \(2.000 \times 10^{5} \mathrm{~m} / \mathrm{s}\) ? The requirement is that \((m v)_{\text {electron }}=(h / \lambda)_{\text {photon }}\). From this, $$ \lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)\left(2.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\right)}=3.64 \mathrm{~nm} $$ This wavelength is in the X-ray region.

Short Answer

Expert verified
The wavelength is 3.64 nm, which is in the X-ray region.

Step by step solution

01

Understanding the Problem

We need to find the wavelength of electromagnetic radiation whose photon has the same momentum as an electron traveling at a speed of \(2.00 \times 10^5 \text{ m/s}\). The condition given is \((mv)_{\text{electron}} = \left(\frac{h}{\lambda}\right)_{\text{photon}}\), where \(h\) is Planck's constant.
02

Formula for Momentum of Photon

For photons, the momentum \(p\) is given by \(p = \frac{h}{\lambda}\), where \(h\) is Planck's constant and \(\lambda\) is the wavelength of the photon.
03

Setting Electron and Photon Momentum Equal

We equate the momentum of the electron \(mv\) to that of the photon \(\frac{h}{\lambda}\): \[ mv = \frac{h}{\lambda} \]. Here, \(m\) is the mass of the electron, \(v\) is its velocity, and \(\lambda\) is the unknown wavelength of the photon.
04

Rearranging for Wavelength

Rearrange the equation \( mv = \frac{h}{\lambda} \) to solve for \(\lambda\): \[ \lambda = \frac{h}{mv} \].
05

Plugging in Values

Substitute the known values into the equation: \( h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s} \), \( m = 9.11 \times 10^{-31} \text{ kg} \), and \( v = 2.00 \times 10^5 \text{ m/s} \). The calculation becomes: \[ \lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.00 \times 10^5} \].
06

Calculating the Wavelength

Calculate \( \lambda \) using the values plugged in. \[ \lambda = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-25}} = 3.64 \times 10^{-9} \text{ m} = 3.64 \text{ nm} \].

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Key Concepts

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

Wavelength Calculation
The wavelength of an electromagnetic wave is a key property that defines its characteristics and behavior. In the context of photons, wavelength (\(\lambda\)) is inversely related to momentum. When we equate the momentum of a photon to that of an electron, we use the relationship \(p = \frac{h}{\lambda}\), where \(h\)is Planck's constant, a fundamental constant in quantum mechanics. By rearranging the formulas given in the exercise, we can calculate the wavelength as:
  • First, establish momentum equality: \((mv)_{\text{electron}} = \left(\frac{h}{\lambda}\right)_{\text{photon}}\).
  • Then, solve for wavelength: \(\lambda = \frac{h}{mv}\).
  • Insert known values to compute the wavelength, yielding a result consistent with X-ray wavelengths.
This step-by-step approach allows us to find the wavelength necessary for a photon to have the same momentum as a fast-moving electron.
Planck's Constant
Planck's constant (\(h\)) is an essential constant in quantum mechanics. Its value is approximately \(6.63 \times 10^{-34} \text{ J}\cdot\text{s}\). Often seen in equations relating to energy and momentum at quantum scales, it plays a crucial role in our wavelength calculation. In the formula \(p = \frac{h}{\lambda}\), Planck's constant bridges the gap between momentum and wavelength for photons. This enforces the concept that photons, despite having no mass, possess a quantifiable momentum. Planck's constant is employed in various other equations in physics, including the famous E=hf, relating the energy and frequency of a photon. This constant signifies the fundamental limitations of measuring certain properties, like energy and time, with precise accuracy.
Electron Momentum
In classical mechanics, the momentum of an electron is calculated using \(mv\), where \(m\) is mass and \(v\) is velocity. Given the context of the provided problem, we use the electron's mass, \(9.11 \times 10^{-31} \text{ kg}\), and its velocity, \(2.00 \times 10^5 \text{ m/s}\), to determine its momentum. This calculated momentum acts as the benchmark to which the photon's momentum, expressed as \(\frac{h}{\lambda}\), is equated. Understanding electron momentum in quantum contexts highlights the delicate interplay between wave and particle properties. With electrons, we deploy both classical momentum concepts and quantum theories to grasp their behavior fully.
X-ray Wavelength
X-rays are a form of electromagnetic radiation with wavelengths typically ranging from 0.01 to 10 nanometers. The wavelength calculated in the problem, \(3.64 \text{ nm}\), falls squarely within this interval. X-rays are well-known for their ability to penetrate different materials, each to varying depths, which is why they are widely used in imaging and other scientific applications. Knowing that certain wavelengths correspond to specific ranges like X-rays helps in practical applications. When photons have such short wavelengths, they have considerable energy levels, making them valuable for tasks requiring high energy and precision. The problem鈥檚 solution of \(3.64 \text{ nm}\) aligns with expected X-ray wavelengths, confirming that a photon with this much momentum is indeed of the X-ray variety.

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Most popular questions from this chapter

A beam \((\lambda=633 \mathrm{~nm})\) from a typical laser designed for student use has an intensity of \(3.0 \mathrm{~mW}\). How many photons pass a given point in the beam each second? The energy that is carried past the point each second is \(0.0030 \mathrm{~J}\). Because the energy per photon is \(h c / \lambda\), which works out to be \(3.14\) \(\times 10^{-19} \mathrm{~J}\), the number of photons passing the point per second is $$ \text { Number } / \mathrm{s}=\frac{0.0030 \mathrm{~J} / \mathrm{s}}{3.14 \times 10^{-19} \mathrm{~J} / \text { photon }}=9.5 \times 10^{15} \text { photon } / \mathrm{s} $$

What is the de Broglie wavelength for a particle moving with speed \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) if the particle is \((a)\) an electron, \((b)\) a proton, and \((c)\) a \(0.20\) -kg ball? We make use of the definition of the de Broglie wavelength: $$ \lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{m\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}=\frac{3.31 \times 10^{-40} \mathrm{~m} \cdot \mathrm{kg}}{m} $$ Substituting the required values for \(m\), one finds that the wavelength is \(3.6 \times 10^{-10} \mathrm{~m}\) for the electron, \(2.0 \times 10^{-13} \mathrm{~m}\) for the proton, and \(1.7 \times 10^{-39} \mathrm{~m}\) for the \(0.20\) -kg ball.

What must be the wavelength of a photon if it is to have the same momentum as an electron traveling at \(2.2 \mathrm{~km} / \mathrm{s}\) ?

Show that the photons in a 1240 -nm infrared beam have energies of \(1.00 \mathrm{eV}\) $$ \mathrm{E}=h f=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{1240 \times 10^{-9} \mathrm{~m}}=1.602 \times 10^{-19} \mathrm{~J}=1.00 \mathrm{eV} $$

I] A particle of mass \(m\) is confined to a narrow tube of length \(L\). Find \((a)\) the wavelengths of the de Broglie waves which will resonate in the tube, (b) the corresponding particle momenta, and (c) the corresponding energies. (d) Evaluate the energies for an electron in a tube with \(L=0.50 \mathrm{~nm}\). (a) The de Broglie waves will resonate with a node at each end of the tube because the ends are impervious. A few of the possible resonance forms are shown in Fig. \(42-1\). They indicate that, for resonance, \(L=\frac{1}{2} \lambda_{1}, 2\left(\frac{1}{2} \lambda_{2}\right), 3\left(\frac{1}{2} \lambda_{3}\right), \ldots, n\left(\frac{1}{2} \lambda_{n}\right), \ldots\) or (b) Because the de Broglie wavelengths are \(\lambda_{\mathrm{n}}=h / p_{n}\), the resonance momenta are $$ p_{n}=\frac{n h}{2 L} \quad n=1,2,3, \ldots $$ (c) As shown in Problem 42.15, \(p^{2}=(2 m)(\mathrm{KE})\), and so $$ (\mathrm{KE})_{n}=\frac{n^{2} h^{2}}{8 L^{2} m} \quad n=1,2,3, \ldots $$ Notice that the particle can assume only certain discrete energies. The energies are quantized. (d) With \(m=9.1 \times 10^{-31} \mathrm{~kg}\) and \(L=5.0 \times 10^{-10} \mathrm{~m}\), substitution yields $$ (\mathrm{KE})_{n}=2.4 \times 10^{-19} n^{2} \mathrm{~J}=1.5 n^{2} \mathrm{eV} $$
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