Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 326.5 nm.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to calculate the wavelength of a photon that has the same energy as the silicon-silicon bond strength, which is given as 3.80 eV. We will use the energy-wavelength relationship in physics that relates the energy of a photon to its wavelength.
02

Convert Energy from eV to Joules

The energy given is in electronvolts (eV). To use it with the wavelength formula, we first need to convert it to joules since 1 eV = 1.602 x 10^-19 J. Thus, the energy in joules is calculated as follows:\[ E = 3.80 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 6.088 \times 10^{-19} \, \text{J} \]
03

Use the Photon Energy Formula

The energy of a photon can be expressed using the formula: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength. We need to solve for \( \lambda \).
04

Rearrange the Formula to Solve for Wavelength

Rearrange the energy formula to find \( \lambda \): \[ \lambda = \frac{hc}{E} \] Substitute the values for \( h \), \( c \), and \( E \) into the formula to find \( \lambda \).
05

Substitute Values and Calculate Wavelength

Substitute the known values into the rearranged formula to find the wavelength:\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js} \times 3.00 \times 10^8 \, \text{m/s}}{6.088 \times 10^{-19} \, \text{J}} \] \[ \lambda = \frac{1.9878 \times 10^{-25}}{6.088 \times 10^{-19}} \, \text{m} = 3.265 \times 10^{-7} \, \text{m} \]Convert the answer to nanometers (1 m = 10^9 nm):\[ \lambda = 326.5 \, \text{nm} \]
06

Conclusion: Interpret the Result

The wavelength of photon required to disintegrate a silicon-silicon single bond with an energy of 3.80 eV is approximately 326.5 nm. This wavelength falls in the ultraviolet range of the electromagnetic spectrum.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Wavelength Calculation
Calculating the wavelength of a photon involves understanding the relationship between energy and wavelength. This relationship can be described using the formula:
\[ E = \frac{hc}{\lambda} \]where:
  • \( E \) is the energy of the photon (in joules).
  • \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \text{Js} \).
  • \( c \) is the speed of light, which is approximately \( 3.00 \times 10^8 \text{m/s} \).
  • \( \lambda \) is the wavelength that we need to find (in meters).
To find the wavelength with the given photon energy, rearrange the equation to solve for \( \lambda \):
\[ \lambda = \frac{hc}{E} \]By placing the correct values into this formula, we can calculate the wavelength of a photon based on its energy.
Silicon-Silicon Bond
Silicon is known to form a wide array of compounds, prominently getting into bonds with other silicon atoms. The silicon-silicon bond is significant in various chemical reactions and structures.
In the given problem, the strength of a silicon-silicon single bond is given as 3.80 eV. An important aspect to note is that bond strength refers to the energy required to break a bond between atoms.
Silicon bonds are integral to the structure of many materials, most notably in semiconductors and synthetic materials. The energy value indicates how much energy, typically in the form of photons, is needed to disrupt this connection.
Understanding this helps comprehend the various technological applications of silicon, including integrated circuits and other electronic devices.
Ultraviolet Light
Ultraviolet (UV) light is part of the electromagnetic spectrum with wavelengths shorter than visible light.
Typically ranging from about 10 nm to 400 nm, UV light lies just beyond the violet end of visible light and is divided into several subcategories based on their wavelength.
  • UVA (320-400 nm): This range is closest to visible light and can penetrate deeper into materials.
  • UVB (290-320 nm): It is more energetic than UVA and is responsible for causing sunburn.
  • UVC (100-290 nm): This type is mostly absorbed by the Earth's atmosphere and is not commonly encountered in nature.
A photon with a wavelength of 326.5 nm, as calculated in the example, falls into the UVA range of ultraviolet light, indicating its capability to sever strong molecular bonds like those in silicon-silicon bonds. Understanding this can help harness UV light for material applications where specific reactions or bond disruptions are desirable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

(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?

An alpha particle (\(m = 6.64 \times 10^{-27} kg\)) emitted in the radioactive decay of uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?

Determine \(\lambda_m\) , the wavelength at the peak of the Planck distribution, and the corresponding frequency \(f\), at these temperatures: (a) 3.00 K; (b) 300 K; (c) 3000 K.

Why is it easier to use helium ions rather than neutral helium atoms in such a microscope? (a) Helium atoms are not electrically charged, and only electrically charged particles have wave properties. (b) Helium atoms form molecules, which are too large to have wave properties. (c) Neutral helium atoms are more difficult to focus with electric and magnetic fields. (d) Helium atoms have much larger mass than helium ions do and thus are more difficult to accelerate.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free