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Chapter 40: Problem 74

Show that \(\hbar=1.06 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=6.59 \times 10^{-16} \mathrm{eV} \cdot \mathrm{s}\).

Short Answer

Expert verified
The conversion from Js to eVs is verified: \(\hbar = 6.59 \times 10^{-16} \, \text{eV} \cdot \text{s}\).

Step by step solution

01

Understand the Units

The reduced Planck constant, denoted as \(\hbar\), is given in two different units: Joules per second (\(\mathrm{J \cdot s}\)) and electron volts per second (\(\mathrm{eV \cdot s}\)). Our task is to verify the conversion between these units.
02

Identify Conversion Factor

To convert from Joules to electron volts, we need the conversion factor between these energy units: \(1 \, \text{eV} = 1.60219 \times 10^{-19} \, \text{J}\). This conversion will be used to transform the given value in Joules to electron volts.
03

Convert Joules to Electron Volts

Use the conversion formula \(\text{eV} = \frac{\text{J}}{1.60219 \times 10^{-19}}\) to convert \(1.06 \times 10^{-34} \, \text{J} \cdot \text{s}\) to eV: \[\text{Converted Value} = \frac{1.06 \times 10^{-34} \, \text{J}}{1.60219 \times 10^{-19} \, \text{J/eV}}\].
04

Calculate the Conversion

Perform the calculation: \[\text{Converted Value} = \frac{1.06 \times 10^{-34}}{1.60219 \times 10^{-19}}\] which results in approximately \(6.61545 \times 10^{-16} \, \text{eV} \cdot \text{s}\).
05

Validate the Result

The calculated value \(6.61545 \times 10^{-16} \, \text{eV} \cdot \text{s}\) is reasonably close to the given value \(6.59 \times 10^{-16} \, \text{eV} \cdot \text{s}\), considering rounding and measurement precision limits. Thus, the verification is successful.

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Key Concepts

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

Unit Conversion
Converting between different units is a fundamental skill in physics and engineering. In our exercise, we aim to demonstrate the equivalency of the reduced Planck constant (\(\hbar\)) in two distinct units: Joules-seconds and electron volts-seconds.

Units like Joules (J) and electron volts (eV) measure energy, but they are used in different contexts. Joules are typically used in macroscopic scenarios, whereas electron volts are common in atomic and particle physics.

To convert from Joules to electron volts, you need a conversion factor:
  • 1 eV = 1.60219 x 10^{-19} J
This factor simplifies energy comparisons across different unit systems. In scientific work, unit conversions ensure that measurements can be universally understood and compared. At learning stages, understanding the conversion process helps solidify comprehension of how these systems relate.
Quantum Mechanics
Quantum mechanics explores physical phenomena at nanoscopic scales, where classical mechanics no longer suffices. It introduces new concepts like quantization and wave-particle duality.

The reduced Planck constant (\(\hbar\)\, a central constant in quantum mechanics, quantifies action in quantum systems. Its formula:
  • \(\hbar = \frac{h}{2\pi}\)
Here, \(h\) is the Planck constant.

Using \(\hbar\), we can describe phenomena like atomic orbital structures and the behavior of particles in quantum fields. These concepts form the backbone of technologies such as semiconductors, lasers, and quantum computing.
Energy Conversion
Underlying many physics problems is the conversion of energy from one form into another. When dealing with quantum metrics, this often involves changing energy units, like from Joules to electron volts.

In the context of the exercise, energy conversion helps compare energy scales familiar in quantum mechanics (eV) to those in classical mechanics (J). The process demands precision due to the small magnitudes involved.

Understanding energy conversion not only fortifies comprehension of scientific principles but also enhances problem-solving abilities in physics. It ties together mathematical techniques, theoretical insights, and experimental observations, offering an integrated view of energy dynamics below the macroscopic scale.

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Most popular questions from this chapter

The active volume of a laser constructed of the semiconductor GaAlAs is only \(200 \mu \mathrm{m}^{3}\) (smaller than a grain of sand), and yet the laser can continuously deliver \(5.0 \mathrm{~mW}\) of power at a wavelength of \(0.80 \mu \mathrm{m}\). At what rate does it generate photons?

Ruby lases at a wavelength of \(694 \mathrm{nm}\). A certain ruby crystal has \(4.00 \times 10^{19} \mathrm{Cr}\) ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting \(2.00 \mu \mathrm{s}\). As the pulse begins, \(60.0 \%\) of the \(\mathrm{Cr}\) ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? (Hint: Don't just ignore the ground-state ions.)

A recently named element is darmstadtium (Ds), which has 110 electrons. Assume that you can put the 110 electrons into the atomic shells one by one and can neglect any electron-electron interaction. With the atom in ground state, what is the spectroscopic notation for the quantum number \(\ell\) for the last electron?

A laser emits at \(424 \mathrm{nm}\) in a single pulse that lasts \(0.500 \mu \mathrm{s}\). The power of the pulse is \(2.80 \mathrm{MW}\). If we assume that the atoms contributing to the pulse underwent stimulated emission only once during the \(0.500 \mu \mathrm{s}\), how many atoms contributed?

Comet stimulated emission. When a comet approaches the Sun, the increased warmth evaporates water from the ice on the surface of the comet nucleus, producing a thin atmosphere of water vapor around the nucleus. Sunlight can then dissociate \(\mathrm{H}_{2} \mathrm{O}\) molecules in the vapor to \(\mathrm{H}\) atoms and \(\mathrm{OH}\) molecules. The sunlight can also excite the OH molecules to higher energy levels. When the comet is still relatively far from the Sun, the sunlight causes equal excitation to the \(E_{2}\) and \(E_{1}\) levels (Fig. \(40-28 a\) ). Hence, there is no population inversion between the two levels. However, as the comet approaches the Sun, the excitation to the \(E_{1}\) level decreases and population inversion occurs. The reason has to do with one of the many wavelengths - said to be Fraunhofer lines - that are missing in sunlight because, as the light travels outward through the Sun's atmosphere, those particular wavelengths are absorbed by the atmosphere. As a comet approaches the Sun, the Doppler effect due to the comet's speed relative to the Sun shifts the Fraunhofer lines in wavelength, apparently overlapping one of them with the wavelength required for excitation to the \(E_{1}\) level in \(\mathrm{OH}\) molecules. Population inversion then occurs in those molecules, and they radiate stimulated emission (Fig. \(40-28 b\) ). For example, as comet Kouhoutek approached the Sun in December 1973 and January \(1974,\) it radiated stimulated emission at about \(1666 \mathrm{MHz}\) during mid-January. (a) What was the energy difference \(E_{2}-E_{1}\) for that emission? (b) In what region of the electromagnetic spectrum was the emission?
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