Question

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?

Answer 1

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 + E_k

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

E_k = 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)

E_k = 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