/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 The silicon-silicon single bond ... [FREE SOLUTION] | 91影视

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Chapter 39: Problem 24

The silicon-silicon single bond that forms the basis of the mythical silicon- based creature the Horta has a bond strength of 3.80 eV. What wavelength of photon would you need in a (mythical) phasor disintegration gun to destroy the Horta?

Short Answer

Expert verified
The required wavelength is approximately 326.3 nm.

Step by step solution

01

Understand the Energy and Wavelength Relationship

To determine the wavelength needed to break a bond, we need to use the relationship between energy and wavelength, given by the formula: \[E = \frac{hc}{\lambda}\]where \(E\) is the energy of the photon, \(h\) is Planck's constant (6.626 x 10^{-34} \, \text{J} \cdot \text{s}), \(c\) is the speed of light in a vacuum (3.00 x 10^8 \, \text{m/s}), and \(\lambda\) is the wavelength of the photon.
02

Convert Energy to Joules

The given bond strength is 3.80 eV. Since 1 eV = 1.60218 x 10^{-19} Joules, we convert 3.80 eV to Joules:\[3.80 \, \text{eV} \times 1.60218 \times 10^{-19} \, \text{J/eV} = 6.088284 \times 10^{-19} \, \text{J}\]
03

Rearrange the Formula to Solve for Wavelength

Using the formula \(E = \frac{hc}{\lambda}\), rearrange it to solve for the wavelength \(\lambda\):\[\lambda = \frac{hc}{E}\]Substitute the values of \(h\), \(c\), and \(E\) into this equation to find \(\lambda\).
04

Plug Values into the Wavelength Formula

Insert the known values into the equation:\[\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{6.088284 \times 10^{-19} \, \text{J}}\]Calculate \(\lambda\):
05

Calculate the Wavelength

Perform the arithmetic calculation to get:\[\lambda = \frac{1.9878 \times 10^{-25} \, \text{J} \cdot \text{m}}{6.088284 \times 10^{-19} \, \text{J}} = 3.263 \times 10^{-7} \, \text{m}\]Thus, the required wavelength is approximately 326.3 nm.

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Key Concepts

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

Planck's Constant
Planck's constant is a fundamental constant in physics, represented by the symbol \( h \). It plays a crucial role in quantum mechanics and was introduced by Max Planck in the early 1900s. Planck's constant has a value of \( 6.626 \times 10^{-34} \, \text{J s} \). This constant is pivotal because it relates the energy \( E \) of a photon to its frequency \( f \). The formula that links these quantities is given by:
\[ E = h \cdot f \]
This means that energy is directly proportional to frequency, with Planck's constant being the proportionality factor. In the context of photons and light, Planck's constant helps us understand how energy changes as light changes either in color or intensity, especially in quantum phenomena. Because of its smallness, Planck's constant indicates the quantization of energy on a minuscule scale, making it a cornerstone in explaining the behavior of particles at atomic and subatomic levels.
Wavelength Calculation
Calculating the wavelength of a photon is crucial when analyzing how light interacts with matter. The key formula to find the wavelength of a photon is:
\[ \lambda = \frac{hc}{E} \]
Where:
  • \( \lambda \) is the wavelength of the photon.
  • \( h \) is Planck's constant.
  • \( c \) is the speed of light in a vacuum, \( 3.00 \times 10^8 \, \text{m/s} \).
  • \( E \) is the energy of the photon.
In practice, to determine the wavelength when energy is given, first ensure that the energy is in Joules since that aligns with the units of Planck's constant and the speed of light. Then, simply substitute the values into the equation. This calculation provides the wavelength in meters. At times, converting it to nanometers, microscopes, or other units might be useful for practical applications.For instance, in the original problem, converting the energy from electron volts to Joules aligns all units correctly to solve for the wavelength effectively.
Energy Conversion
Energy conversion is a fundamental process in physics, especially when dealing with photon energy and particles. The energy of photons is often given in electron volts (eV) and needs conversion to Joules (J) for computations involving other physical constants. This is due to the standard units of physical constants like Planck's constant being based on the metric system. The conversion factor is:
1 eV = \( 1.60218 \times 10^{-19} \) Joules
This conversion is straightforward: multiply the electron volts by the conversion factor to attain energy in Joules.Understanding energy conversion allows us to use consistent units and apply them effectively within formulas, like the one that relates photon energy, Planck's constant, and wavelength. Taking care to convert energy correctly ensures accurate and meaningful calculations, making it essential in scientific equations and problem-solving scenarios where energy transitions from one form or unit to another.

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Most popular questions from this chapter

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?

In a TV picture tube the accelerating voltage is 15.0 \(\mathrm{kV}\) , and the electron beam passes through an aperture 0.50 \(\mathrm{mm}\) in diameter to a screen 0.300 \(\mathrm{m}\) away. (a) Calculate the uncertainty in the component of the electron's velocity perpendicular to the line between aperture and screen. (b) What is the uncertainty in position of the point where the electrons strike the screen? (c) Does this uncertainty affect the clarity of the picture significantly? (Use nonrelativistic expressions for the motion of the electrons. This is fairly accurate and is certainly adequate for obtaining an estimate of uncertainty effects.)

Zero-Point Energy. Consider a particle with mass \(m\) moving in a potential \(U=\frac{1}{2} k x^{2},\) as in a mass-spring system. The total energy of the particle is \(E=p^{2} / 2 m+\frac{1}{2} k x^{2}\) Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(p x \approx h .\) (a) Calculate the minimum possible value of the energy \(E,\) and the value of \(x\) that gives this minimum \(E .\) This lowest possible energy, which is not zero, is called the zero-point energy. (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

A certain atom has an energy level 3.50 eV above the ground state. When excited to this state, it remains \(4.0 \mu s,\) on the average, before emitting a photon and returning to the ground average, before emitting a photon and returning to the ground state. (a) What is the energy of the photon? What is its wavelength? (b) What is the smallest possible uncertainty in energy of the photon?

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and \(v=0.960 c ?\) (Hint: Use the correct relativistic expression for linear momentum if necessary.)
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