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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.

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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.

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Most popular questions from this chapter

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops 6.50 \(\times\) 10\(^{-14}\) m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 \(\times\) 10\(^{-27}\) kg. (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

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.

(a) A particle with mass \(m\) has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (\(Hint\): You must use the relativistic expressions for momentum and kinetic energy: \(E^2 = (pc^2) + (mc^2)^2\) and \(K = E - mc^2\).) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

What must be the temperature of an ideal blackbody so that photons of its radiated light having the peak-intensity wavelength can excite the electron in the Bohr-model hydrogen atom from the ground level to the \(n\) = 4 energy level?

An electron has a de Broglie wavelength of 2.80 \(\times\) 10\(^{-10}\) m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

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