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(a) What is the energy of a photon that has wavelength 0.10\(\mu \mathrm{m} ?\) (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10\(\mu \mathrm{m}\) in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

Short Answer

Expert verified
a) The photon energy is approximately \(1.986 \times 10^{-18}\) J. b) Electrons need \(\approx 150.4\) V and have speed \(\approx 5.78 \times 10^6\) m/s. c) Protons need \(\approx 0.081\) V and have speed \(\approx 1.24 \times 10^4\) m/s.

Step by step solution

01

Calculate Photon Energy

The energy of a photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \) Js (Planck's constant), \( c = 3 \times 10^8 \) m/s (speed of light), and \( \lambda = 0.10 \times 10^{-6} \) m (wavelength in meters). Substitute these values to find the energy. \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.10 \times 10^{-6}} \]. After the calculations, \( E \approx 1.986 \times 10^{-18} \) J.
02

Relate Electron Wavelength to Potential Difference

For an electron exhibiting wave nature, its wavelength \( \lambda \) is governed by \( \lambda = \frac{h}{p} \), where \( p \) is the momentum. Using the de Broglie relation and kinetic energy \( K = \frac{1}{2}mv^2 = eV \) (where \( e \) is the electron charge and \( V \) is the potential difference), we have \( p = \sqrt{2meV} \). Substitute into de Broglie: \( \lambda = \frac{h}{\sqrt{2meV}} \). Set \( \lambda = 0.10 \times 10^{-6} \) m to find \( V \): \[ V = \frac{h^2}{2me\lambda^2} \]. Replace with known values: \( m = 9.11 \times 10^{-31} \) kg (electron mass), and \( e = 1.6 \times 10^{-19} \) C. Calculating yields \( V \approx 150.4 \) volts.
03

Calculate Electron Speed

Using the kinetic energy principle, \( eV = \frac{1}{2}mv^2 \), solve for \( v \). Rearrange to find \( v \): \[ v = \sqrt{\frac{2eV}{m}} \]. Substitute the known values: \( e = 1.6 \times 10^{-19} \) C, \( V \approx 150.4 \) volts, \( m = 9.11 \times 10^{-31} \) kg. This gives \( v \approx 5.78 \times 10^6 \) m/s.
04

Relate Proton Wavelength to Potential Difference

Apply identical principles as in Step 2 but replace electron mass with proton mass \( m_p = 1.67 \times 10^{-27} \) kg. Recalculate \( V \) using \[ V = \frac{h^2}{2m_pe\lambda^2} \], with the same \( \lambda = 0.10 \) µm. Substitute known values to solve for \( V \), which results in \( V \approx 0.081 \) volts for protons.
05

Calculate Proton Speed

Using the kinetic energy relation \( eV = \frac{1}{2}m_pv^2 \), find \( v \) for protons. Rearrange to \( v = \sqrt{\frac{2eV}{m_p}} \), and plug values \( e = 1.6 \times 10^{-19} \) C, \( V \approx 0.081 \) volts, \( m_p = 1.67 \times 10^{-27} \) kg, yielding \( v \approx 1.24 \times 10^4 \) m/s.

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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 mechanics. It quantifies the energy carried by a photon, which is a particle of light. This energy depends on the photon's wavelength, as described by the formula:
  • \( E = \frac{hc}{\lambda} \)
Here, \( E \) is the energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \; \text{Js}) \), \( c \) is the speed of light \( (3 \times 10^8 \; \text{m/s}) \), and \( \lambda \) is the wavelength. The shorter the wavelength, the higher the energy of the photon. In this calculation, for a 0.10 µm wavelength, the energy approximates to \( 1.986 \times 10^{-18} \; \text{J} \).
Photon energy plays a crucial role in various applications such as solar power, where photons are transformed into electrical energy. Understanding this energy is also key in fields like photonics and quantum computing, where light's discrete nature is explored.
de Broglie Wavelength
The de Broglie wavelength reflects the wave-particle duality of matter, a centerpiece of quantum mechanics. Louis de Broglie postulated that not just light has wave properties, but all matter does, too. This wavelength is defined as:
  • \( \lambda = \frac{h}{p} \)
Where \( \lambda \) is the de Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. For a particle with velocity \( v \), momentum \( p = mv \).
In applications like electron microscopy and quantum physics, the wave properties of particles allow for phenomena like interference and diffraction. For an electron in this exercise, expressing wave nature requires acceleration through a voltage so its de Broglie wavelength corresponds to a certain physical size (like a 0.10 µm pinhole). Calculating this yields an acceleration potential difference of around 150.4 volts.
Electron Acceleration
Electron acceleration involves increasing the speed of electrons using an electric field, accomplished by applying a potential difference, \( V \). This involves converting electric potential energy to kinetic energy
  • \( eV = \frac{1}{2}mv^2 \)
where \( e \) is the charge of an electron, \( m \) is its mass, and \( v \) is its velocity. Rearranging for \( v \) allows us to find how fast an electron moves after being accelerated through a potential \( V \).
In this example, applying a potential of 150.4 volts accelerates the electrons to speeds of approximately \( 5.78 \times 10^6 \; \text{m/s} \). Electron acceleration is pivotal in devices like cathode ray tubes and particle accelerators, showing how energy transformations manifest across various technologies.
Wave-Particle Duality
Wave-particle duality is a core concept of quantum mechanics, suggesting that all particles exhibit both wave and particle characteristics. This paradigm shift from classical mechanics revolutionizes our understanding of fundamental processes.
  • Light behaves as particles (photons) in phenomena like the photoelectric effect, but as waves in interference and diffraction.
  • Matter, such as electrons, shows wave properties under certain conditions, evidenced when they pass through small apertures.
Understanding this duality helps explain experimental results that one could not rationalize using classical physics alone. For example, electrons demonstrating interference patterns after passing through a pinhole highlight their wave nature, contributing to technologies like quantum computing and tunneling in semiconductors. Explaining everyday world observations goes beyond tangible perceptions, accounting for quantum oddities described by duality.

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Most popular questions from this chapter

Blue Supergiants. A typical blue supergiant star (the type that explodes and leaves behind a black hole) has a surface temperature of \(30,000 \mathrm{K}\) and a visual luminosity \(100,000\) times that of our sun. Our sun radiates at the rate of \(3.86 \times 10^{26} \mathrm{W}\) .(Visual luminosity is the total power radiated at visible wavelengths.) (a) Assuming that this star behaves like an ideal blackbody, what is the principal wavelength it radiates? Is this light visible? Use your answer to explain why these stars are blue. (b) If we assume that the power radiated by the star is also \(100,000\) times that of our sun, what is the radius of this star? Compare its size to that of our sun, which has a radius of \(6.96 \times 10^{5} \mathrm{km}\) . (c) Is it really correct to say that the visual luminosity is proportional to the total power radiated? Explain.

An atom with mass \(m\) emits a photon of wavelength \(\lambda\) . (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K / E,\) where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K / E\) for a hydrogen atom (mass \(1.67 \times\) \(10^{-27} \mathrm{kg} )\) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n=1,2,\) and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is \(1.0 \times 10^{-8}\) s. In the Bohr model, how many orbits does an electron in the \(n=2\) level complete before returning to the ground level?

Find the longest and shortest wavelengths in the Lyman and Paschen series for hydrogen. In what region of the electromagnetic spectrum does each series lie?

Bohr Orbits of a Satellite. A 20.0 -kg satellite circles the earth once every 2.00 \(\mathrm{h}\) in an orbit having a radius of 8060 \(\mathrm{km} .\) (a) Assuming that Bohr's angular-momentum result \((L=n h / 2 \pi)\) applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number \(n\) of the orbit of the satellite. (b) Show from Bohr's angular momentum result and Newton's law of gravitation that the radius of an earth-satellite orbit is directly proportional to the square of the quantum number, \(r=k n^{2},\) where \(k\) is the constant of proportionality. (c) Using the result from part \((\mathrm{b}),\) find the distance between the orbit of the satellite in this problem and its next "allowed" orbit. (Calculate a numerical value.) (d) Comment on the possibility of observing the separation of the two adjacent orbits. (e) Do quantized and classical orbits correspond for this satellite? Which is the "correct" method for calculating the orbits?

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