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Chapter 6: Problem 30

The energy from radiation can be used to rupture chemical bonds. A minimum energy of \(192 \mathrm{~kJ} / \mathrm{mol}\) is required to break the bromine- bromine bond in \(\mathrm{Br}_{2}\). What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

Short Answer

Expert verified
The longest wavelength of radiation that possesses enough energy to break the bromine-bromine bond is approximately 621.8 nm, which falls into the range of visible light, specifically in the orange region.

Step by step solution

01

Convert energy from kJ/mol to J/photon

First, we need to convert the energy given in kJ/mol to J/photon, using Avogadro's number (N): \(6.022 \times 10^{23} \mathrm{mol^{-1}}\). \(192 \mathrm{kJ/mol} \times \dfrac{1,000 \mathrm{J}}{1 \mathrm{kJ}} \times \dfrac{1 \mathrm{mol}}{6.022 \times 10^{23} \mathrm{particles}} = 3.19 \times 10^{-19} \mathrm{J/photon}\)
02

Use Planck's equation to solve for the wavelength

Now that we have the energy in J/photon, we can use Planck's equation to find the wavelength of the radiation: \(3.19 \times 10^{-19} \mathrm{J} = 6.626 \times 10^{-34} \mathrm{J\cdot s} \times \dfrac{3.0 \times 10^8 \mathrm{m/s}}{\lambda}\) Rearrange the equation to solve for \(\lambda\): \(\lambda = \dfrac{6.626 \times 10^{-34} \mathrm{J\cdot s} \times 3.0 \times 10^8 \mathrm{m/s}}{3.19 \times 10^{-19} \mathrm{J}}\) \(\lambda \approx 6.218 \times 10^{-7} \mathrm{m}\)
03

Convert the wavelength to nanometers and identify the type of radiation

We can convert the wavelength from meters to nanometers by multiplying by \(10^9\): \(6.218 \times 10^{-7} \mathrm{m} \times \dfrac{10^9 \mathrm{nm}}{1 \mathrm{m}} = 621.8 \mathrm{nm}\) The longest wavelength of radiation that possesses enough energy to break the bromine-bromine bond is approximately 621.8 nm. This wavelength falls into the range of visible light, specifically in the orange region. Therefore, the type of electromagnetic radiation is visible light.

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Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum encompasses a wide range of electromagnetic waves characterized by their wavelengths and frequencies. It includes various types of radiation, ranging from radio waves with long wavelengths to gamma rays with very short wavelengths.
When it comes to breaking chemical bonds, we are typically interested in the regions of the spectrum where energies are high enough to affect molecular structures. These include ultraviolet (UV), visible, and sometimes infrared (IR) light.
  • Radio waves: Longest wavelengths, used for communication
  • Microwaves: Heat food, used in microwave ovens
  • Infrared: Heat radiation, felt as warmth
  • Visible light: The spectrum of colors visible to the human eye
  • Ultraviolet: Can cause sunburn, used in disinfection
  • X-rays: Used for medical imaging
  • Gamma rays: Shortest wavelengths, used in cancer treatment
Knowing the properties of each type helps us understand how different electromagnetic waves can interact with matter. In this exercise, the focus is on finding electromagnetic radiation that can break a chemical bond.
Planck's Equation
Planck's Equation forms the backbone for understanding the relationship between energy and wavelength in electromagnetic radiation. It states that the energy (\( E \)) of a photon is directly proportional to its frequency (\( u \)), and inversely proportional to its wavelength (\( \lambda \)).
This equation is often written as:\[ E = h u = \frac{hc}{\lambda} \]
  • \( h \) is Planck's constant (\(6.626 \times 10^{-34} \; \mathrm{J \cdot s} \))
  • \( c \) is the speed of light in a vacuum (\(3.0 \times 10^8 \; \mathrm{m/s} \))
  • \( u \) is the frequency of the radiation
  • \( \lambda \) is the wavelength of the radiation
Utilizing Planck's equation allows us to calculate that longer wavelengths correspond to lower energy photons. In practical applications like breaking chemical bonds, knowing the exact wavelength that can transfer sufficient energy is crucial.
Chemical Bond Energy
Chemical bond energy refers to the amount of energy required to break a bond between two atoms in a molecule. It is commonly expressed in kilojoules per mole (kJ/mol).
This energy depends on the type of bond and the atoms involved.
  • Stronger bonds need more energy to break.
  • Single bonds are typically weaker than double or triple bonds of the same elements.
  • In the exercise, breaking a Bromine-Bromine (\(\mathrm{Br}_2\)) bond requires \(192 \; \mathrm{kJ/mol}\).
Understanding bond energy is vital when analyzing chemical reactions and stability. In photochemical processes, determining the precise energy needed to break specific bonds helps identify the suitable wavelength (and type) of light to use. For Bromine, this energy corresponds to visible light radiation, particularly in the orange spectrum, as calculated in the problem solution.

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Most popular questions from this chapter

Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X rays, ultraviolet light.

(a) Using Equation 6.5 , calculate the energy of an electron in the hydrogen atom when \(n=3\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=3 .(\mathbf{b})\) Is this line in the visible region of the electromagnetic spectrum?

What is the maximum number of electrons in an atom that can have the following quantum numbers? (a) \(n=3, m_{l}=-1 ;\) (b) \(n=4, l=2\); (c) \(n=4, l=3, m_{l}=-2 ;\) (d) \(n=5, l=2\), \(m_{l}=0 .\)

The speed of sound in dry air at \(20^{\circ} \mathrm{C}\) is \(343 \mathrm{~m} / \mathrm{s}\) and the lowest frequency sound wave that the human ear can detect is approximately \(20 \mathrm{~Hz}\) (a) What is the wavelength of such a sound wave? (b) What would be the frequency of electromagnetic radiation with the same wavelength? (c) What type of electromagnetic radiation would that correspond to? [Section 6.1]

The visible emission lines observed by Balmer all involved \(n_{\mathrm{f}}=2 .\) (a) Which of the following is the best explanation of why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum: (i) Transitions to \(n_{\mathrm{f}}=3\) are not allowed to happen, (ii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum, (iii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to \(n_{\mathrm{f}}=3\) emit photons that are at exactly the same wavelengths as those to \(n_{\mathrm{f}}=2 .\) (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which \(n_{1}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11
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