/*! 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 19 If the energy of the \(\mathrm{H... [FREE SOLUTION] | 91Ó°ÊÓ

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If the energy of the \(\mathrm{H}_{2}\) covalent bond is \(-4.48 \mathrm{eV},\) what wavelength of light is needed to break that molecule apart? In what part of the electromagnetic spectrum does this light lie?

Short Answer

Expert verified
The required wavelength is 276 nm, in the ultraviolet region.

Step by step solution

01

Understand the Problem

We need to find the wavelength of light required to break an \(\text{H}_{2}\) covalent bond, given the energy of the bond is \(-4.48 \,\text{eV}\). The question also asks for the region of the electromagnetic spectrum that this wavelength lies in.
02

Convert Energy to Joules

First, we need to convert the energy from electron volts (eV) to joules (J). We use the conversion factor \(1 \, \text{eV} = 1.60219 \, \times 10^{-19} \, \text{J}\). Hence, \(-4.48 \, \text{eV} = -4.48 \, \times 1.60219 \, \times 10^{-19} = -7.18 \, \times 10^{-19} \, \text{J}\).
03

Calculate Wavelength

Next, use the energy-wavelength relation: \([c = 3.00 \, \times 10^{8} \, \text{m/s}], [h = 6.626 \, \times 10^{-34} \, \text{J·s}]\), to find the wavelength: \( E = \frac{hc}{\lambda} \). Rearranging gives \( \lambda = \frac{hc}{E} = \frac{6.626 \, \times 10^{-34} imes 3.00 \, \times 10^{8}}{7.18 \, \times 10^{-19}} \, \text{m}\). Calculating this yields \(\lambda \approx 2.76 \, \times 10^{-7} \, \text{m}\).
04

Determine Electromagnetic Spectrum Region

Convert the wavelength from meters to nanometers (n = m \times 10^{9}\ for simplicity in interpretation). \(2.76 \, \times 10^{-7} = 276 \, \text{nm}\). Light with this wavelength falls within the ultraviolet (UV) region of the electromagnetic spectrum, which ranges from approximately \(100 \, \text{nm} \) to \(400 \, \text{nm}\).

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Key Concepts

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

Covalent Bond Energy
Covalent bonds are chemical connections formed when atoms share pairs of electrons. These bonds hold the atoms together, forming molecules. In our example, the energy of the covalent bond in a hydrogen molecule (\(\text{H}_2\) is \(-4.48 \, \text{eV}\). This energy value indicates the amount of energy required to break the bond and separate the two atoms completely.The energy value is essential in calculating the wavelength of light needed to break this bond, as well as understanding its stability. A larger bond energy often signifies a stronger bond, making it more challenging to break. Converting the bond energy from electron volts (eV) to joules (J) is a crucial step in physics, using the factor \(1 \, \text{eV} = 1.60219 \, \times 10^{-19} \, \text{J}\). This conversion is vital in practical calculations of photon energies and interactions with matter.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from very short wavelengths (like gamma rays) to very long wavelengths (like radio waves). Each type of electromagnetic radiation has distinct properties and applications.
  • Gamma rays: less than 0.01 nm
  • X-rays: 0.01 nm to 10 nm
  • Ultraviolet (UV) light: 10 nm to 400 nm
  • Visible light: 400 nm to 700 nm
  • Infrared (IR): 700 nm to 1 mm
  • Microwave: 1 mm to 1 m
  • Radio waves: greater than 1 m
When you calculate a wavelength, like the one required to break the hydrogen covalent bond, it is helpful to know where this wavelength lies on the electromagnetic spectrum. For example, a wavelength of \(276 \, \text{nm}\) falls in the ultraviolet region. This segmentation helps scientists and engineers to choose the right type of radiation for various applications, such as imaging or communication.
Ultraviolet Light
Ultraviolet (UV) light is a type of electromagnetic radiation with a wavelength ranging from about 10 nm to 400 nm. It is found between visible light and X-rays on the electromagnetic spectrum. UV light is divided into several subtypes:
  • UVA: 315 nm to 400 nm, can penetrate the skin and contribute to aging
  • UVB: 280 nm to 315 nm, causes sunburn and can lead to skin cancer
  • UVC: 100 nm to 280 nm, most dangerous but largely absorbed by the Earth's atmosphere
In our problem, the wavelength calculated for breaking the covalent bond falls into the ultraviolet range, specifically around 276 nm, which is in the UVB region. Such energy levels can break molecular bonds, explaining their potentially hazardous effects yet useful in disinfection and sterilization processes. Understanding the behavior of UV light is crucial in fields like environmental sciences and healthcare.
Photon Energy Conversion
Photon energy conversion involves converting the energy of a photon, determined by its frequency or wavelength, into a usable form. The energy of a photon is given by the equation:\[ E = \frac{hc}{\lambda} \]where:- \(E\) is the energy in joules,- \(h\) is Planck's constant (\(6.626 \, \times 10^{-34} \, \text{J·s}\)),- \(c\) is the speed of light (\(3.00 \, \times 10^{8} \, \text{m/s}\)),- \(\lambda\) is the wavelength in meters.In the exercise, converting photon energy from electron volts to joules is the initial step. With known values for Planck's constant and the speed of light, you can determine the precise wavelength of the light required. Knowing this allows for precise targeting within the electromagnetic spectrum for various applications, from breaking chemical bonds to powering solar cells.

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Most popular questions from this chapter

The gap between valence and conduction bands in diamond is \(5.47 \mathrm{eV}\). (a) What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie? (b) Explain why pure diamond is transparent and colorless. (Hint: Will photons of visible light that strike a diamond be absorbed or transmitted?) (c) Most gem diamonds have a yellow color. Explain how impurities in the diamond can cause this color.

(a) Write out the electron configuration \(\left(1 s^{2} 2 s^{2},\right.\) and so on \()\) for \(L i\) and Na. (b) How many electrons does each of these atoms have in its outer shell?

Atoms of unusual size. In photosynthesis in plants, light is absorbed in light-harvesting complexes consisting of protein and pigment molecules. The energy absorbed is then transported to a specialized complex called the reaction center. Scientists believe that quantum mechanical effects may play an important role in this energy transfer. In a recent experiment, researchers used rubidium atoms cooled to a very low temperature to study a similar energy-transfer process in the laboratory. Laser light was used to excite an electron in each atom to a very highly excited state (large \(n\) ). The excited electron behaves very much like the single electron in a hydrogen atom, with an effective \(Z=1,\) but because \(n\) is so large, the excited electron is quite far from the atomic nucleus, with an orbital radius of approximately \(1 \mu \mathrm{m},\) and is weakly bound. Using these so-called Rydberg atoms, the researchers were able to study how energy is transported from one atom to the next in a process that may be a helpful model in understanding energy transport in photosynthesis. The size of a highly excited atom can be taken to be the diameter of the orbit of the excited electron. If the researchers want to do the experiment with the rubidium atoms in a gas, with atoms separated by 10 times their size, the density of the atoms should be about A. \(10^{5}\) atoms \(/ \mathrm{cm}^{3}\). B. \(10^{8}\) atoms \(/ \mathrm{cm}^{3}\). C. \(10^{11}\) atoms \(/ \mathrm{cm}^{3}\). D. \(10^{21}\) atoms \(/ \mathrm{cm}^{3}\).

What is the ratio of the number of different \(3 d\) states of the hydrogen atom to the number of \(5 d\) states of the atom?

An electron is in the hydrogen atom with \(n=5\). (a) Find the possible values of \(L\) and \(L_{z}\) for this electron, in units of \(\hbar\). (b) For each value of \(L\), find all the possible angles between \(L\) and the \(z\) axis.

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