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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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Most popular questions from this chapter

(a) For an He+ ion, do the \(2 s\) and \(2 p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of $9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}$ What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Microwave ovens use microwave radiation to heat food. The energy of the microwaves is absorbed by water molecules in food and then transferred to other components of the food. (a) Suppose that the microwave radiation has a wavelength of $10 \mathrm{~cm} .\( How many photons are required to heat \)200 \mathrm{~mL}$ of water from 25 to \(75^{\circ} \mathrm{C} ?\) (b) Suppose the microwave's power is $1000 \mathrm{~W}\( ( 1 watt \)=1\( joule-second \)) .$ How long would you have to heat the water in part (a)?

For each element, indicate the number of valence electrons, core electrons, and unpaired electrons in the ground state: (a) sodium, (b) sulfur, (c) fluorine.

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).
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