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Chapter 15: Problem 38

\(\mathrm{MnF}_{6}{ }^{2-}\) has a crystal field splitting energy, \(\Delta_{0}\) of \(2.60 \times 10^{2} \mathrm{~kJ} / \mathrm{mol}\). What is the wavelength responsible for this energy?

Short Answer

Expert verified
Answer: The wavelength responsible for the crystal field splitting energy of MnF₆²⁻ is approximately 459 nm.

Step by step solution

01

Write down the given information

The information given is as follows: - Crystal field splitting energy, \(\Delta_{0}= 2.60\times10^{2}\mathrm{~kJ/mol}\) - Planck's constant, \(h = 6.626\times 10^{-34}\mathrm{~Js}\) - Speed of light, \(c = 3.00\times10^{8}\mathrm{~m/s}\)
02

Convert the energy to J/photon

First, we must convert the given energy from \(\mathrm{kJ/mol}\) to \(\mathrm{J/photon}\). We can do this using Avogadro's constant (\(N_{A} = 6.022\times10^{23} \mathrm{mol^{-1}}\)): $$ \Delta E = \frac{2.60\times10^{2}\times10^3}{6.022\times10^{23}}\mathrm{~J/photon} $$
03

Calculate the energy (E) of the photon in Joules

Now, calculate the energy \(\Delta E\) in Joules: $$ \Delta E \approx 4.32\times10^{-19}\mathrm{~J/photon} $$
04

Use the Planck-Einstein equation to solve for the wavelength

The Planck-Einstein equation is given by: $$ E = h\times c/\lambda $$ Where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon. We will now rearrange the equation to solve for \(\lambda\): $$ \lambda = \frac{h\times c}{E} $$ Now plug in the values for \(h\), \(c\), and \(\Delta E\) to find out the wavelength: $$ \lambda\approx\frac{6.626\times 10^{-34}\mathrm{~Js} \times 3.00\times10^{8}\mathrm{~m/s}}{4.32\times10^{-19}\mathrm{~J/photon}} $$
05

Calculate the wavelength responsible for the energy

Calculate the wavelength \(\lambda\): $$ \lambda\approx 4.59\times10^{-7}\mathrm{~m} $$ Or, the wavelength responsible for the crystal field splitting energy of \(\mathrm{MnF}_{6}{ }^{2-}\) is approximately \(459\mathrm{~nm}\).

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