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Chapter 39: Problem 2

For crystal diffraction experiments (discussed in Section \(39.2 ),\) wavelengths on the order of 0.20 \(\mathrm{nm}\) are often appropriate. Find the energy in electron volts for a particle with this wavclength if the particle is \((a)\) a photon; \((b)\) an clectron; \((c)\) an alpha particle \(\left(m=6.64 \times 10^{-27} \mathrm{kg}\right) .\)

Short Answer

Expert verified
6208 eV for a photon, 37.6 eV for an electron, 0.516 eV for an alpha particle.

Step by step solution

01

Photon Energy Calculation

For a photon, the energy is calculated using the equation \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \) is the Planck's constant, \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 0.20 \times 10^{-9} \text{ m} \) is the wavelength. Calculating:\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{0.20 \times 10^{-9}} \]\[ E = 9.939 \times 10^{-16} \text{ J} \]To convert Joules to electron volts, use \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). \[ E = \frac{9.939 \times 10^{-16}}{1.602 \times 10^{-19}} \approx 6208 \text{ eV} \]
02

Electron Energy Calculation

For an electron, we use the de Broglie wavelength formula \( \lambda = \frac{h}{mv} \) to find velocity \( v \), where \( m = 9.11 \times 10^{-31} \text{ kg} \) is the electron mass. First, calculate the momentum \( p = \frac{h}{\lambda} \).\[ p = \frac{6.626 \times 10^{-34}}{0.20 \times 10^{-9}} = 3.313 \times 10^{-24} \text{ kg}\cdot\text{m/s} \]Since \( p = mv \), solve for \( v \):\[ v = \frac{p}{m} = \frac{3.313 \times 10^{-24}}{9.11 \times 10^{-31}} \approx 3.64 \times 10^6 \text{ m/s} \]The kinetic energy \( K = \frac{1}{2}mv^2 \):\[ K = \frac{1}{2}(9.11 \times 10^{-31})(3.64 \times 10^6)^2 \approx 6.02 \times 10^{-18} \text{ J} \]Convert to electron volts:\[ K = \frac{6.02 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 37.6 \text{ eV} \]
03

Alpha Particle Energy Calculation

For an alpha particle, we use the same steps as the electron but with the alpha particle mass \( m = 6.64 \times 10^{-27} \text{ kg} \). Find the momentum \( p = \frac{h}{\lambda} \):\[ p = \frac{6.626 \times 10^{-34}}{0.20 \times 10^{-9}} = 3.313 \times 10^{-24} \text{ kg}\cdot\text{m/s} \]For velocity \( v \):\[ v = \frac{p}{m} = \frac{3.313 \times 10^{-24}}{6.64 \times 10^{-27}} \approx 4.99 \times 10^2 \text{ m/s} \]The kinetic energy \( K = \frac{1}{2}mv^2 \):\[ K = \frac{1}{2}(6.64 \times 10^{-27})(4.99 \times 10^2)^2 \approx 8.27 \times 10^{-23} \text{ J} \]Convert to electron volts:\[ K = \frac{8.27 \times 10^{-23}}{1.602 \times 10^{-19}} \approx 0.516 \text{ eV} \]

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Key Concepts

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

Photon Energy
Photons are the fundamental particles of light and have energy determined by their wavelength. This energy can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon. By plugging these values into the equation, we can determine the energy of a photon at a given wavelength. It is essential to convert this energy into different units like electron volts for various applications.
Understanding the conversion process is vital for experiments like crystal diffraction to determine suitable materials for specific wavelengths.
de Broglie Wavelength
The de Broglie wavelength concept introduces us to the wave-particle duality of matter, a fundamental principle in quantum mechanics. It associates a wavelength with any moving particle based on its momentum. The formula is expressed as \( \lambda = \frac{h}{mv} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, \( m \) is the mass, and \( v \) is the velocity of the particle. This concept is essential for calculating how particles like electrons behave in experiments, such as crystallography.
Using the de Broglie wavelength helps physicists predict and explain interference patterns created during particle diffraction experiments.
Electron Volts
Electron volts (eV) are a unit of energy often used in physics to describe the energy of particles like electrons and photons. One electron volt is the amount of kinetic energy gained by a single electron when it is accelerated through an electric potential difference of one volt. It is a more convenient unit than joules when dealing with atomic-scale particles. For example, converting energy from joules to electron volts simply involves dividing by the conversion factor \( 1.602 \times 10^{-19} \) J/eV. This straightforward conversion is crucial in the context of quantum mechanics and helps make energies comprehensible on a human scale.
Particle Wavelength
Particle wavelength refers to the wavelength associated with particles in motion. According to quantum mechanics, particles like electrons exhibit both wave-like and particle-like behaviors. The de Broglie equation \( \lambda = \frac{h}{mv} \) describes the wavelength of a moving particle. This notion revolutionized physics by demonstrating that matter has wave-like properties.
In practice, understanding particle wavelength is central to designing experiments like electron microscopy and neutron diffraction, which rely on the wave behavior of particles to create detailed images of microscopic structures.
Planck's Constant
Planck's constant is a fundamental quantity in physics that plays a key role in quantum mechanics. Represented by \( h \), it has a value of \( 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \). This constant forms the basis for understanding how energy is quantized, as it relates the energy of a photon to its frequency: \( E = hf \).
Planck's constant is integral to calculations involving photon energy, and it provides the foundation for equations that describe the behavior of particles at the quantum level. This concept is crucial for interpreting the energy levels in atoms and the transmission of energy in quantum systems.

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Most popular questions from this chapter

(a) What is the de Broglie wavelength of an electron accelerated from rest through a potential increase of 125 \(\mathrm{V} ?\) (b) What is the de Broglie wavelength of an alpha particle \((q=+2 e,\) \(m=6.64 \times 10^{-27} \mathrm{kg} )\) accelerated from rest through a potential drop of 125 \(\mathrm{V} ?\)

(a) A particle with mass \(m\) has kinetic energy equal to three times its rest encrgy. What is the de Broglic wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: \(E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}\) and \(K=E-\) \(m c^{2} \cdot(b)\) Determine the numerical value of the kinetic energy (in Mev) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (i) a proton.

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda=h / p\) still holds, but we must use the rclativistic cxpression for momcntum, \(p=m v / \sqrt{1-v^{2} / c^{2}}\) (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is $$ v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}} $$ (b) The quantity \(h / m c\) equals \(2426 \times 10^{-12} \mathrm{m}\) (As we saw in Section 38.7 , this same quantity appears in Eq. (38.23), the expression for Compton scattering of photons by electrons) If \(\lambda\) is small compared to \(h / m c,\) the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to \(c\) . In this case it is convenient to write \(v=(1-\Delta) c\) and express the speed of the electron in terms of \(\Delta\) rather than \(v\) . Find an expression for \(\Delta\) valid when \(\lambda \ll h / m c\) . [Hine. Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots,\) valid for the case \(|z| < 1 .\) (c) How fast must an electron move for its de Broglie wavelength to be \(1.00 \times 10^{-15} \mathrm{m}\) , comparable to the size of a proton? Express your answer in the form \(v=(1-\Delta) c,\) and state the value of \(\Delta .\)

A particle is described by the normalized wave function \(\psi(x, y, z)=A x e^{-\alpha x^{2}} e^{-\beta \beta} e^{-\gamma x^{2}},\) where \(A, \alpha, \beta,\) and \(\gamma\) are all real, positive constants. The probability that the particle will be found in the infinitesimal volume \(d x d y d z\) centered at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is \(\left|\psi\left(x_{0}, y_{0}, z_{0}\right)\right|^{2} d x d y d z\) (a) At what value of \(x_{0}\) is the particle most likely to be found? (b) Are there values of \(x_{0}\) for which the probability of the particle being found is zero? If so, at what \(x_{0} ?\)

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and $(b) v=0.960 c ?(\text { Hint } \text { Use the correct rela- }tivistic expression for linear momentum if necessary.)
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