1. A coupon of very pure Gold and highly clean surface is exposed to an Argon Fl

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?

Explanation / Answer

a) What is the energy of a photon in the ArF laser beam expressed in electron-Volts (eV)?

E = h = hc/

h = 6.625 x 10-34 Js

c = 3x108 ms-1

= 193.4 nm = 193.4 x 10-9 m

E = h = hc/

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;

h = + Ek

where

h = Plank constant (6.63 x 10-34 J s)

= the frequency of the incident light in hertz (Hz) or s-1

= 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 h = 6.37eV

= ? (unknown)

Ek = 1.3 eV

6.37 = + 1.3

so = 6.37-1.3 = 5.07 eV

So, the Work Function of this Gold surface in 5.07 eV

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