Photogray lenses incorporate small amounts of silver chloride in the glass of the lens. When light hits the AgCl particles, the following reaction occurs: $$ \operatorname{AgCl} \stackrel{h v}{\longrightarrow} \mathrm{Ag}+\mathrm{Cl} $$ The silver metal that is formed causes the lenses to darken. The energy 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?

We are given the energy change for the reaction as \(3.10 \times 10^{2} \mathrm{kJ/mol}\). We will now convert it to energy per photon by using Avogadro's constant, which is \(N_A = 6.022\times10^{23}\, \text{particles/mol}\). Energy per photon: \(E = \frac{3.10 \times 10^{2} \mathrm{kJ/mol}}{6.022\times10^{23}\, \text{particles/mol}}\) Now, convert the energy from kJ to J: \(E = \frac{3.10 \times 10^{2} \mathrm{ kJ/mol} \times 1000 \mathrm{\, J/ kJ}}{6.022\times10^{23}\, \text{particles/mol}}\) E = \(5.15 \times 10^{-19} \mathrm{J}\)

Now, we will use the equation \(E = \frac{h\cdot c}{\lambda}\) to find the maximum wavelength of light. Here, E is the energy per photon, h is Planck's constant (\(6.63\times10^{-34}\, \text{Js}\)), and c is the speed of light (\(3.00\times10^{8}\, \text{m/s}\)). Wavelength: \(\lambda = \frac{h \cdot c}{E}\) \(\lambda = \frac{6.63\times10^{-34}\, \text{Js} \cdot 3.00\times10^{8}\, \text{m/s}}{5.15 \times 10^{-19} \mathrm{J}}\) \(\lambda = 3.85 \times 10^{-7}\) m To convert the wavelength to nanometers, multiply by \(1\times10^{9}\, \text{nm/m}\): \(\lambda = 3.85 \times 10^{-7} \mathrm{m} \times 1\times10^{9}\, \text{nm/m}\) \(\lambda = 385 \mathrm{nm}\) So, the maximum wavelength of light that can cause the reaction is 385 nm.