# Chapter 7 Lecture Problems

## 7.2 The Nature of Light

1. What is the frequency of light at 120 nm (a wavelength I frequently used for experiments in graduate school)
2. What is the wavelength of the radiation for proton NMR in our instrument (300 MHz)
3. What wavelength of light corresponds to the energy of an O=O bond, 498 kJ/mol.
1. 498 kJ/mol = 8.27 * 10-19 J/molecule
2. This is the amount of energy that 1 photon must have to break the bond in an O=O atom. Since the energy of a photon is related to it's frequency as:
1. E = h v (v is the greek nu and h is Planck's constant)
2. v = 1.25 * 1015 Hz (Hz is frequency or s-1)
3. c = [lambda] * v
4. [lambda] = 2.40 * 10-7 meters or 240 nm

4. What energy (kJ/mol) corresponds to the wavelength of light required to remove an electron from a hydrogen atom (90 nm)
1. 90 nm = 90*10-9 m
2. Since c = [lambda] * v
1. v = (3*108 m/sec)/(90*10-9 m)
2. v = 3.33*1015 Hz
3. Since E=h v
4. E = (33.3*1015 Hz) * (6.626*10-34 J sec/photon)
5. E = 2.21*10-18 J/photon
6. Multiply by Avogadro's number to get J/mol
1. E = (2.21*10-18 J/photon) * (6.02*1023 photon/mol)
2. E = 1.33*106 J/mol
3. E = 1330 kJ/mol