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Chapter 2: Problem 136

A carbon-oxygen double bond in a certain organic molecule absorbs radiation that has a frequency of \(6.0 \times 10^{13} \mathrm{s}^{-1}\). a. What is the wavelength of this radiation? b. To what region of the spectrum does this radiation belong? c. What is the energy of this radiation per photon? d. A carbon-oxygen bond in a different molecule absorbs radiation with frequency equal to \(5.4 \times 10^{13} \mathrm{s}^{-1} .\) Is this radiation more or less energetic?

Short Answer

Expert verified
a. The wavelength of the radiation is \(5.00\ \mu m\). b. The radiation belongs to the infrared region of the spectrum. c. The energy of the radiation per photon is \(3.98 \times 10^{-20}\ J\). d. The radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) is more energetic compared to the one with frequency \(5.4 \times 10^{13}\ s^{-1}\).

Step by step solution

01

Calculate the wavelength of the radiation

Given the frequency \(\nu = 6.0 \times 10^{13}\ s^{-1}\), we can find the wavelength \(\lambda\) using the equation \(c = \lambda \nu\), where \(c\) is the speed of light, which is approximately \(3.00 \times 10^8\ m/s\). To find \(\lambda\), we can rearrange the equation to get \(\lambda = \frac{c}{\nu}\), and then plug in the given values: \(\lambda = \frac{3.00 \times 10^8\ m/s}{6.0 \times 10^{13}\ s^{-1}} = 5.00 \times 10^{-6}\ m = 5.00\ \mu m\) The wavelength of the radiation is \(5.00\ \mu m\).
02

Determine the region of the spectrum

The wavelength \(5.00\ \mu m\) falls within the infrared (IR) region of the electromagnetic spectrum. So, the radiation belongs to the infrared region.
03

Calculate the energy per photon

To find the energy per photon, we can use the energy equation \(E = h\nu\), where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\ Js\)) and \(\nu\) is the frequency. \(E = (6.626 \times 10^{-34}\ Js)(6.0 \times 10^{13}\ s^{-1}) = 3.98 \times 10^{-20}\ J\) The energy of the radiation per photon is \(3.98 \times 10^{-20}\ J\).
04

Compare the energies of the two radiations

To compare the energies of the two radiations, we can calculate the energy of the radiation with the frequency of \(5.4 \times 10^{13}\ s^{-1}\) using the same energy equation: \(E = (6.626 \times 10^{-34}\ Js)(5.4 \times 10^{13}\ s^{-1}) = 3.58 \times 10^{-20}\ J\) Since the energy of the radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) (\(3.98 \times 10^{-20}\ J\)) is larger than the energy of the radiation with frequency \(5.4 \times 10^{13}\ s^{-1}\) (\(3.58 \times 10^{-20}\ J\)), the radiation with frequency \(6.0 \times 10^{13}\ s^{-1}\) is more energetic.

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Most popular questions from this chapter

Write the expected ground-state electron configuration for the following. a. the element with one unpaired \(5 p\) electron that forms a covalent compound with fluorine b. the (as yet undiscovered) alkaline earth metal after radium c. the noble gas with electrons occupying \(4 f\) orbitals d. the first-row transition metal with the most unpaired electrons

The work function of an element is the energy required to remove an electron from the surface of the solid element. The work function for lithium is \(279.7 \mathrm{kJ} / \mathrm{mol}\) (that is, it takes \(279.7 \mathrm{kJ}\) of energy to remove 1 mole of electrons from 1 mole of Li atoms on the surface of Li metal; 1 mol \(L i=6.022 \times\) \(10^{23}\) atoms Li). What is the maximum wavelength of light that can remove an electron from an atom on the surface of lithium metal?

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is \(890.1 \mathrm{kJ} / \mathrm{mol} .\) Is light with a wavelength of \(225 \mathrm{nm}\) capable of ionizing a gold atom (removing an electron) in the gas phase? ( 1 mol gold \(=6.022 \times 10^{23}\) atoms gold.)

Which of the following statements is(are) true? a. F has a larger first ionization energy than does Li. b. Cations are larger than their parent atoms. c. The removal of the first electron from a lithium atom (electron configuration is \(1 s^{2} 2 s^{1}\) ) is exothermic - that is, removing this electron gives off energy. d. The He atom is larger than the \(\mathrm{H}^{+}\) ion. e. The Al atom is smaller than the Li atom.

In defining the sizes of orbitals, why must we use an arbitrary value, such as \(90 \%\) of the probability of finding an electron in that region?
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