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Chapter 7: Problem 137

Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the \(\mathrm{AgCl}\) particles, the following reaction occurs: $$ \mathrm{AgCl} \stackrel{\mathrm{hv}}{\longrightarrow} \mathrm{Ag}+\mathrm{Cl} $$ The silver metal that is formed causes the lenses to darken. The enthalpy change for this reaction is \(3.10 \times 10^{2} \mathrm{~kJ} / \mathrm{mol}\). Assuming all this energy must be supplied by light, what is the maximum wavelength of light that can cause this reaction?

Short Answer

Expert verified
The maximum wavelength of light that can cause the reaction is \(385\, \mathrm{nm}\).

Step by step solution

01

Write down the Planck's equation for energy

The Planck's equation shows the relationship between the energy (E) of a photon and its wavelength (λ). The equation is given by: $$ E = \dfrac{h c}{λ} $$ where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\, \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8}\, \mathrm{m/s}\)), and \(λ\) is the wavelength of the light.
02

Convert the enthalpy change to energy per photon

The enthalpy change is given in \(\mathrm{kJ/mol}\), which we need to convert to energy per photon. To do this, we'll perform the following conversions: 1. Convert the enthalpy change from \(\mathrm{kJ/mol}\) to \(\mathrm{J/mol}\). 2. Divide by Avogadro's number to convert from \(\mathrm{J/mol}\) to \(\mathrm{J/photon}\). The enthalpy change is \(3.10 \times 10^{2}\, \mathrm{kJ/mol} \times \dfrac{10^3\, \mathrm{J}}{1\, \mathrm{kJ}} = 3.10 \times 10^{5}\, \mathrm{J/mol}\). Now, divide by Avogadro's number (\(6.022 \times 10^{23}\, \mathrm{mol^{-1}}\)): $$ E_{photon} = \dfrac{3.10 \times 10^{5}\, \mathrm{J/mol}}{6.022 \times 10^{23}\, \mathrm{mol^{-1}}} = 5.15 \times 10^{-19}\, \mathrm{J/photon} $$
03

Solve for the wavelength using Planck's equation

Now that we have the energy per photon, we can use Planck's equation to solve for the wavelength of the light. Rearrange the equation to solve for λ: $$ λ = \dfrac{h c}{E_{photon}} $$ Insert the values for Planck's constant, speed of light, and energy of the photon: $$ λ = \dfrac{(6.626 \times 10^{-34}\, \mathrm{Js})(2.998 \times 10^{8}\, \mathrm{m/s})}{5.15 \times 10^{-19}\, \mathrm{J/photon}} $$
04

Calculate the maximum wavelength

Now, perform the calculation to find the maximum wavelength: $$ λ = \dfrac{(6.626 \times 10^{-34}\, \mathrm{Js})(2.998 \times 10^{8}\, \mathrm{m/s})}{5.15 \times 10^{-19}\, \mathrm{J/photon}} = 3.85 \times 10^{-7}\, \mathrm{m} $$ To express the wavelength in nanometers, multiply by \(10^{9}\, \mathrm{nm/m}\): $$ λ = 3.85 \times 10^{-7}\, \mathrm{m} \times \dfrac{10^9\, \mathrm{nm}}{1\, \mathrm{m}} = 385\, \mathrm{nm} $$ The maximum wavelength of light that can cause the reaction is \(385\, \mathrm{nm}\).

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Most popular questions from this chapter

One of the emission spectral lines for \(\mathrm{Be}^{3+}\) has a wavelength of \(253.4 \mathrm{~nm}\) for an electronic transition that begins in the state with \(n=5 .\) What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one- electron ions. Don't forget the \(Z\) factor: \(Z=\) nuclear charge \(=\) atomic number. \()\)

The ionization energy for a \(1 s\) electron in a silver atom is \(2.462 \times 10^{6} \mathrm{~kJ} / \mathrm{mol}\) a. Determine an approximate value for \(Z_{\text {eff }}\) for the Ag \(1 s\) electron. Assume the Bohr model applies to the \(1 s\) electron. \(Z_{\text {eff }}\) is the apparent nuclear charge experienced by the electrons. b. How does \(Z_{\text {eff }}\) from part a compare to \(Z\) for Ag? Rationalize the relative numbers.

Identify the following three elements. a. The ground-state electron configuration is \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\). b. The ground-state electron configuration is \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2} .\) c. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{1}\).

How many electrons in an atom can have the designation \(1 p\), \(6 d_{x^{2}-y^{2}}, 4 f, 7 p_{y}, 2 s, n=3 ?\)

For each of the following pairs of elements \((\mathrm{C}\) and \(\mathrm{N})\) \((\mathrm{Ar}\) and \(\mathrm{Br})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.
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