In: Chemistry
1. A coupon of very pure Gold and highly clean surface is exposed to an Argon Fluoride (ArF) excimer laser beam of wavelength 193.4±0.2 nm inside an ESCA (Electron Spectroscopy for Chemical Analysis) ultrahigh vacuum spectrometer. The coupon surface faces the entrance aperture of a hemispherical electrostatic energy analyzer and detector. The coupon and the analyzer entrance aperture are electrically connected together at ground potential (zero Volts). A fraction of the photo-emitted electrons enters the analyzer and their energy distribution is measured. a) (2 points) What is the energy of a photon in the ArF laser beam expressed in electron-Volts (eV)? b) (2 points) The highest energy of the detected electrons was 1.3±0.1 eV. What was the Work Function of this Gold surface in eV?
a) What is the energy of a photon in the ArF laser beam expressed in electron-Volts (eV)?
E = hν = hc/λ
h = 6.625 x 10-34 Js
c = 3x108 ms-1
λ = 193.4 nm = 193.4 x 10-9 m
E = hν = hc/λ
E = 6.625 x 10-34 Js x 3x108ms-1/193.4 x 10-9 m = 0.102x10-17 J = 1.02 x 10-18 J
1eV = 1.6 x 10-19 J
so 1.02 x 10-18 J corresponds to 1.02x10-18/1.6x10-19 = 6.37 eV
b)The highest energy of the detected electrons was 1.3±0.1 eV. What was the Work Function of this Gold surface in eV?
The photoelectric equation involves;
hν = φ + Ek
where
h = Plank constant (6.63 x 10-34 J s)
ν= the frequency of the incident light in hertz (Hz) or s-1
φ = the work function in joules (J) or eV
Ek = the maximum kinetic energy of the emitted electrons in joules (J) or eV
we have already calculated the value of hν = 6.37eV
φ = ? (unknown)
Ek = 1.3 eV
6.37 = φ + 1.3
so φ = 6.37-1.3 = 5.07 eV
So, the Work Function of this Gold surface in 5.07 eV