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Magazine Chapter 39: Problem 2
For crystal diffraction experiments (discussed in Section 39.1), wavelengths on the order of 0.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle (\(m\) = 6.64 \(\times\) 10\(^{-27}\) kg).
Short Answer
Expert verified
Photon: 6203 eV, Electron: 18800 eV, Alpha particle: 515 eV.
Step by step solution
01
Determine the energy of a photon
To find the energy of a photon, use the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \), \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength \( 0.20 \times 10^{-9} \text{ m} \). Substitute these values: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.20 \times 10^{-9}} \approx 9.939 \times 10^{-16} \text{ J} \]. To convert this to electron volts (eV), divide by the charge of an electron \( 1.602 \times 10^{-19} \text{ C} \): \[ E \approx \frac{9.939 \times 10^{-16}}{1.602 \times 10^{-19}} \approx 6203 \text{ eV} \].
02
Calculate the momentum of the electron and alpha particle
The momentum \( p \) for a particle is given by \( p = \frac{h}{\lambda} \). For \( \lambda = 0.20 \times 10^{-9} \text{ m} \), \[ p = \frac{6.626 \times 10^{-34}}{0.20 \times 10^{-9}} \approx 3.313 \times 10^{-24} \text{ kg}\cdot\text{m/s} \]. This momentum value will be used to find energies for both the electron and the alpha particle.
03
Compute the energy of the electron
For an electron, use its kinetic energy formula related to momentum: \( E = \frac{p^2}{2m} \). The mass of an electron \( m_e = 9.11 \times 10^{-31} \text{ kg} \). Substitute \( p = 3.313 \times 10^{-24} \text{ kg}\cdot\text{m/s} \): \[ E = \frac{(3.313 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} \approx 3.013 \times 10^{-18} \text{ J} \]. Convert \( J \) to \( eV \): \[ E \approx \frac{3.013 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 18,800 \text{ eV} \].
04
Calculate the energy of the alpha particle
Similarly, for an alpha particle, use the same kinetic energy formula: \( E = \frac{p^2}{2m} \) with mass \( m = 6.64 \times 10^{-27} \text{ kg} \): \[ E = \frac{(3.313 \times 10^{-24})^2}{2 \times 6.64 \times 10^{-27}} \approx 8.263 \times 10^{-20} \text{ J} \]. Convert to electron volts: \[ E \approx \frac{8.263 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 515 \text{ eV} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Photon Energy
Photons are crucial in understanding electromagnetic radiation, such as light. The energy of a photon is directly related to its wavelength, which you can calculate using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \) J路s, \( c \) is the speed of light \( 3 \times 10^8 \) m/s, and \( \lambda \) is the wavelength.
- A shorter wavelength results in higher energy.
- In our case, with a wavelength of 0.20 nm, the calculated energy is approximately 6203 eV after conversion from Joules to electron volts.
Electron Energy
Electrons are fundamental particles with defined mass and charge. To find an electron's energy, particularly when dealing with motion such as in a crystal diffraction experiment, you use its kinetic energy. This is given by \( E = \frac{p^2}{2m} \), where \( p \) is momentum and \( m \) is the electron鈥檚 mass.
- For a wavelength of 0.20 nm, the momentum was found to be approximately \( 3.313 \times 10^{-24} \) kg路m/s.
- The mass of an electron is \( 9.11 \times 10^{-31} \) kg.
- Using the formula, the electron energy is about 18,800 eV.
Alpha Particle Energy
Alpha particles are quite different from electrons. They consist of two protons and two neutrons, making them much heavier. To find the kinetic energy of an alpha particle, you apply the same formula used for electrons: \( E = \frac{p^2}{2m} \). However, the mass of an alpha particle \( (m = 6.64 \times 10^{-27} \text{ kg}) \) is much greater.
- With a wavelength of 0.20 nm, the momentum is \( 3.313 \times 10^{-24} \) kg路m/s.
- Calculate the energy using the alpha particle鈥檚 mass, resulting in approximately 515 eV.
Wavelength
Wavelength is a fundamental property of waves, including light, describing the distance between successive peaks. It is inversely related to energy; as wavelength decreases, energy increases.
- This property is essential in techniques like crystal diffraction, where specific wavelengths provide information about crystal structures.
- For example, the 0.20 nm wavelength used in this exercise is typical for X-ray crystallography studies.
- Accurately knowing the wavelength can help determine material properties and lead to advances in fields like chemistry and biology.
Planck's Constant
Planck's constant is a critical value in quantum mechanics, symbolized by \( h \) and equaling \( 6.626 \times 10^{-34} \) J路s. It represents the proportionality between the energy of a photon and the frequency of its electromagnetic wave.
- In equations like \( E = \frac{hc}{\lambda} \), Planck's constant helps relate energy and wavelength.
- Its role is pivotal in understanding phenomena at the atomic and subatomic levels, indicating the quantization of energy.
- Planck's constant is foundational in calculations involving photon energy and wave-particle duality.
