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Stells [14]
3 years ago
8

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.
Physics
1 answer:
antoniya [11.8K]3 years ago
3 0

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

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we know that;

shear stress τ = 16T/πd³

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τ_{max} = 16T_{max} /πd³

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(b) Determine the angle of twist between the ends of the bar.

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now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

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