contestada

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.

Respuesta :

Answer:

ΔE = 1.031 eV

Explanation:

For this exercise let's calculate the energy of the photons using Planck's equation

          E = h f

wavelength and frequency are related

         c = λ f

         f = c /λ

let's substitute

         E = h c /λ

let's calculate

         E = 6.63 10⁻³⁴ 3 10⁸/1064 10⁻⁹

         E = 1.869 10⁻¹⁹ J

let's reduce to eV

         E = 1.869 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

        E = 1.168 eV

therefore the electron affinity is

         ΔE = E - 0.137

         ΔE = 1.168 - 0.137

         ΔE = 1.031 eV

The electron affinity of thulium in units of electron volts per atom is; ΔE ≈ 1.031 eV

From Planck's equation, we can find the energy of the photons when given wavelength as;

E = hc/λ

Where;

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

c is speed of light = 3 × 10⁸ m/s

λ is wavelength

We are given;

wavelength; λ = 1064 nm = 1064 × 10⁻⁹ m

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1064 × 10⁻⁹)

E2 = 1.868 × 10⁻¹⁹ J

Converting to eV gives;

E2 = (1.868 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)

E2 = 1.1675 eV

We are given E1 = 0.137 eV.

Now, electron affinity is simply change in energy. Thus;

ΔE = 1.1675 - 0.137

ΔE ≈ 1.031 eV

Read more at; https://brainly.com/question/14004834

ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas