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Chapter 7: Problem 48

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?

Short Answer

Expert verified
After converting the given light wavelength (225 nm) to energy (530.7 kJ/mol) using the speed of light, Planck's constant, and Avogadro's number, we can compare this value to the ionization energy of gold (890.1 kJ/mol). Since the calculated photon energy (530.7 kJ/mol) is less than the ionization energy of gold (890.1 kJ/mol), light with a wavelength of 225 nm is not capable of ionizing a gold atom in the gas phase.

Step by step solution

01

Gather necessary constants

Before beginning, let's list the required constants to solve the problem: 1. Speed of light, c = \(3.00 \times 10^8 m/s \) 2. Planck's constant, h = \(6.626 \times 10^{-34} J \cdot s\) 3. Avogadro's number, N_A = \(6.022 \times 10^{23} mol^{-1}\)
02

Determine the energy of the given light wavelength

To find the energy of a single photon from a light wave, we'll use the formula, E = h × c / λ, where E is the energy of the single photon, and λ is the wavelength of the light. Given λ = \(225 \times 10^{-9} m \) E = \(\frac{(6.626 \times 10^{-34} J \cdot s) (3.00 \times 10^8 m/s)}{225 \times 10^{-9} m}\) E = \(8.82 \times 10^{-19} J\)
03

Convert the energy of the photon to kJ/mol

Now, we'll convert the energy of one photon to kJ/mol. We will multiply the photon energy with Avogadro's number and then divide by \(10^3\) to convert Joules to kiloJoules. Photon energy in kJ/mol = \( (8.82 \times 10^{-19} J) \times (6.022 \times 10^{23} mol^{-1}) \div 10^3 \) Photon energy in kJ/mol = \(530.7 kJ/mol\)
04

Compare the photon energy with ionization energy

Now, let's compare the photon energy (530.7 kJ/mol) to the ionization energy of gold (890.1 kJ/mol). Since the photon energy (530.7 kJ/mol) is less than the ionization energy of gold (890.1 kJ/mol), light with a wavelength of 225 nm is not capable of ionizing a gold atom in the gas phase.

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Most popular questions from this chapter

In the ground state of element 115, Uup, a. how many electrons have \(n=5\) as one of their quantum numbers? b. how many electrons have \(\ell=3\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=1\) as one of their quantum numbers? d. how many electrons have \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers \(p, q, r\), and \(s .\) The rules for these quantum numbers are as follows: \(p=1,2,3,4,5, \ldots\) \(q\) takes on positive odd integers and \(q \leq p\) \(r\) takes on all even integer values from \(-q\) to \(+q\). (Zero is considered an even number.) \(s=+\frac{1}{2}\) or \(-\frac{1}{2}\) a. Sketch what the first four periods of the periodic table will look like in this universe. b. Wh?t are the atomic numbers of the first four elements you would expect to be least reactive? c. Give an example, using elements in the forst four rows, of ionic compounds with the formulas \(\mathrm{XY}, \mathrm{XY}_{2}, \mathrm{X}_{2} \mathrm{Y}, \mathrm{XY}_{3}\), and \(\mathrm{X}_{2} \mathrm{Y}_{3}\) d. How many electrons can have \(p=4, q=3 ?\) e. How many electrons can have \(p=3, q=0, r=0 ?\) f. How many electrons can have \(p=6\) ?

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