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Photoelectron spectroscopy applies the principle of the photoelectric effect to study...
Photoelectron spectroscopy applies the principle of the photoelectric effect to study orbital energies of atoms and molecules. High-energy radiation (usually UV or x-ray) is absorbed by a sample and an electron is ejected. The orbital enegy can be calculated from the known energy of the radiation and the measured energy of the electron lost. The following energy differences were determined for several electrons transitions:
change of energy from 2 to 1= 4.098x10^-17 J
change of energy from 3 to 1= 4.854x10^-17 J
change of energy from 5 to 1= 5.254x10^-17 J
change of energy from 4 to 2 = 1.024x10^-17 J
Calculate the energy change and the wavelength of a photon emitted in the following transition: level 3 to 2.
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We are looking for the energy transition from level 3 to 2 and finding the wavelength of the associated photon released. We are given several energy transitions but the two that are relevant here are the first two given. We see the energy for going from 3 to 1 and also the energy for going from 2 to 1. If we subtract the 2 to 1 number from the 3 to 1 number, we are left with the energy associated with going from 3 to 2.
4.854 x 10^-17 J - 4.098 x 10^-17 J = 0.756 x 10^-17 J
Now that we know the energy for the transition, we can calculate the wavelength for the photon emitted via the equation below:
E = hc/lambda
where E is the energy, h is Plank's constant, c is the speed of light, and lambda is the wavelength. Solving for lambda:
lambda = hc/E
We can look up that the value of hc (remember that both numbers are constants) is 1.986 x 10^-25 Jm. Now divide by the energy to get the wavelength:
1.986 x 10^-25 Jm / 0.756 x 10^-17 J = 2.63 x 10^-8 m
This can be converted to nm:
2.63 x 10^-8 m * (10^9 nm/1 m) = 26.3 nm
Posted by ncchemist on July 1, 2013 at 4:23 AM (Answer #1)
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