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Chapter 39: Problem 12

Find the wavelengths of a photon and an electron that have the same energy of 25 eV. (Note: The energy of the electron is its kinetic energy.)

Short Answer

Expert verified
Photon wavelength: \(4.97 \times 10^{-8}\) m; Electron wavelength: \(5.48 \times 10^{-10}\) m.

Step by step solution

01

Convert Energy to Joules

First, we need to convert the energy from electronvolts (eV) to joules (J), since the standard unit of energy in the International System of Units (SI) is the joule. The conversion factor is that 1 eV = \(1.602 imes 10^{-19}\) J. Thus, \(25\) eV is equivalent to \(25 imes 1.602 imes 10^{-19} = 4.005 imes 10^{-18}\) J.
02

Calculate Photon Wavelength

The energy of a photon is related to its wavelength by the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant \(6.626 imes 10^{-34}\) J路s, \(c\) is the speed of light \(3 imes 10^{8}\) m/s, and \(\lambda\) is the wavelength. Solving for \(\lambda\), we get \(\lambda = \frac{hc}{E}\). Substitute \(E = 4.005 imes 10^{-18}\) J, we find \(\lambda = \frac{6.626 imes 10^{-34} imes 3 imes 10^{8}}{4.005 imes 10^{-18}} \approx 4.97 \times 10^{-8}\) m.
03

Calculate Electron Wavelength

The wavelength of an electron can be determined using its de Broglie wavelength formula: \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant, and \(p\) is the momentum of the electron. First, find the momentum using \(E = \frac{p^2}{2m}\), rearranging to find \(p = \sqrt{2mE}\). Using the electron mass \(m = 9.11 \times 10^{-31}\) kg and \(E = 4.005 \times 10^{-18}\) J, we have \(p = \sqrt{2 \times 9.11 \times 10^{-31} \times 4.005 \times 10^{-18}}\) J路s/m \(\approx 1.21 \times 10^{-24}\) kg路m/s. Thus, \(\lambda = \frac{6.626 \times 10^{-34}}{1.21 \times 10^{-24}} \approx 5.48 \times 10^{-10}\) m.

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Key Concepts

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

Photon Energy
Photon energy is a fundamental concept in quantum physics. It describes the energy carried by a single photon, which is a packet of electromagnetic radiation. The energy of a photon is directly related to its frequency. The higher the frequency, the greater the energy.

Photon energy is calculated using the formula:
  • \(E = hu\)
where:
  • \(E\) is the energy of the photon,
  • \(h\) is Planck's constant \(6.626 \times 10^{-34}\) J路s,
  • \(u\) (nu) is the frequency of the photon.
Since the speed of light \(c\) is the product of frequency \(u\) and wavelength \(\lambda\), photon energy can also be related to wavelength by using:
  • \(E = \frac{hc}{\lambda}\)
This relationship shows how energy decreases with increasing wavelength. For example, as we calculated, a photon of energy 25 eV corresponds to a wavelength of approximately \(4.97 \times 10^{-8}\) meters.
De Broglie Wavelength
The de Broglie wavelength is a concept that describes the wave-like behavior of particles. It's based on the principle that every moving particle or object has an associated wave.

To find the de Broglie wavelength of a particle like an electron, we use the de Broglie equation:
  • \(\lambda = \frac{h}{p}\)
where:
  • \(\lambda\) is the wavelength,
  • \(h\) is Planck's constant,
  • \(p\) is the momentum of the particle.
The momentum \(p\) is often calculated using the kinetic energy of the particle with the formula:
  • \(p = \sqrt{2mE}\)
For electrons, this provides a way to relate their kinetic energy to wavelength. Calculating with a 25 eV energy, the electron has a de Broglie wavelength of approximately \(5.48 \times 10^{-10}\) meters, which reflects its wave-particle duality.
Electron Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For electrons, which are charged particles, this energy helps to relate their movement with wave properties.

Electron kinetic energy, especially in quantum mechanics, is significant as it affects the electron鈥檚 properties as both particles and waves. It's given by:
  • \(E = \frac{p^2}{2m}\)
where:
  • \(E\) is the kinetic energy,
  • \(p\) is the momentum,
  • \(m\) is the mass of the electron \(9.11 \times 10^{-31}\) kg.
For an electron with kinetic energy of 25 eV, these equations allow us to determine the electron's behavior and its equivalent de Broglie wavelength. Understanding these concepts shows the transition of particles into wave-like behavior at high energies.
Planck's Constant
Planck's constant is a fundamental constant in quantum mechanics that plays a central role in describing the amounts of energy photons carry.

Planck's constant \(h\) has a value of \(6.626 \times 10^{-34}\) J路s, and it appears in many equations within quantum physics, such as the relationship between the energy and frequency of a photon \(E = hu\), or the de Broglie wavelength equation \(\lambda = \frac{h}{p}\).
  • It represents the smallest action and is a bridge between the macroscopic and quantum worlds, setting limits on the accuracy of measurements.
  • It's what allows us to calculate the energy of a photon when we know its frequency or the wavelength when we know the kinetic energy as with electrons.
Using Planck's constant, we can explore phenomena that aren't explained by classical physics, providing insights into the nature of waves and particles alike.

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Most popular questions from this chapter

A beam of electrons is accelerated from rest through a potential difference of 0.100 \(\mathrm{kV}\) and then passes through a thin slit. The diffracted beam shows its first diffraction minima at \(\pm 11.5^{\circ}\) from the original direction of the beam when viewed far from the slit. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

Sirius B. The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is \(24,000 \mathrm{K}\) and that it radiates energy at a total rate of \(1.0 \times 10^{25} \mathrm{W}\) . Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius B? (b) What is the peak-intensity wavelength'? Is this wavelength visible to humans? (c) What is the radius of Sirius \(B ?\) Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more total energy per second, the hot Sirius \(\mathrm{B}\) or the (relatively) cool sun with a surface temperature of 5800 \(\mathrm{K} ?\) To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.

A beam of \(40-\mathrm{eV}\) electrons traveling in the \(+x-\) direction passes through a slit that is parallel to the \(y\) -axis and 5.0\(\mu \mathrm{m}\) wide. The diffraction pattern is recorded on a screen 2.5 \(\mathrm{m}\) from the slit. (a) What is the de Broglie wavelength of the electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the \(y\) -component of momentum of an electron just after it has passed through the slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq. 39.29 for \(y\) ) to estimate the minimum uncertainty in the \(y\) -coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

Electrons go through a single slit 150 \(\mathrm{nm}\) wide and strike a screen 24.0 \(\mathrm{cm}\) away. You find that at angles of \(\pm 20.0^{\circ}\) from the center of the diffraction pattern, no electrons hit the screen but electrons hit at all points closer to the center, (a) How fast were these electrons moving when they went through the slit? (b) What will be the next larger angles at which no electrons hit the screen?

(a) Show that, as \(n\) gets very large, the energy levels of the hydrogen atom get closer and closer together in energy. (b) Do the radii of these energy levels also get closer together?
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