The maximum liveliness needed to remove an electron form potassium metal is 3.70 × 10-19 J.?
If photons of frequency 9.16 × 1014 s-1 are shone on the surface of the metal, what is the kinetic energy (J) of the ejected electrons ?
Answers:
This is the basis of ESCA or XPS spectroscopy. The binding force of the ejected electrons are governed by the the energy of the xray and the kinetic dynamism of the ejected electrons by the equation:
E(binding) = E(photon) - E(kinetic) - φ,
where φ = the "work function" of the XPS spectrometer. We'll assume φ = 0 since you don't mention it. The energy of the photon is:
E = hν
E = (6.626x10^-34 J s) (9.16x10^14 /s)
E = 6.07x10^-19 J.
Substituting into the above equation give:
3.70x10^-19 J = 6.07x10^-19 J - E(kinetic);
E(kinetic) = 2.37x10^-19 J.
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Answers:
This is the basis of ESCA or XPS spectroscopy. The binding force of the ejected electrons are governed by the the energy of the xray and the kinetic dynamism of the ejected electrons by the equation:
E(binding) = E(photon) - E(kinetic) - φ,
where φ = the "work function" of the XPS spectrometer. We'll assume φ = 0 since you don't mention it. The energy of the photon is:
E = hν
E = (6.626x10^-34 J s) (9.16x10^14 /s)
E = 6.07x10^-19 J.
Substituting into the above equation give:
3.70x10^-19 J = 6.07x10^-19 J - E(kinetic);
E(kinetic) = 2.37x10^-19 J.
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