Achieve Question 12- Electron Affinity

[Anna Turk 1D]

The electron affinity of thulium has been measured by a technique known as laser photodetachment electron spectroscopy. In this technique, a gaseous beam of the anions of an element is bombarded with photons from a laser. Electrons from the anion are then ejected and their energies are detected. The incident radiation had a wavelength of 1064 nm, and the ejected electrons were found to have an energy of 0.137 eV. The electron affinity is the difference in energy between the incident photons and the energy of the ejected electrons. Determine the electron affinity of thulium in units of electron volts per atom.
I was confused on how to calculate electron affinity after using to find the photon energy. Thanks for any help!

[Wesley Zhou 1F]

Hello,

The problem states that electron affinity is the difference in energy between the incident photons and the energy of the ejected electrons.

We know that the energy of the ejected electrons is 0.137 eV, as stated in the problem.

We also have the wavelength of the incident radiation (photons): 1064 nm.

Using the wavelength, we would find the energy of the incident photons using c= λv and E=hv.

Using that energy, subtract from the energy of ejected electrons, and that is your electron affinity.

Hope this helps! Check your units!!

[Rose Arcallana 2B]

So there are lots of ways to go through this question, this is my way:
1. Fine your photon energy by using E = h*v and c= lambda*v.
2. Convert your Joules to eV by dividing it by 1.602 x 10^-19
3. Find the difference between the given eV and calculated photon eV

You're done!! :)