Light of frequency 7×1015 Hz falls on a metal
surface.
If the energy of the resulting photoelec-
trons is 1.7 eV, what is the work func-
tion of the metal? Planck’s constant is
6.62607 × 10−34 J · s .
Answer in units of eV
surface.
If the energy of the resulting photoelec-
trons is 1.7 eV, what is the work func-
tion of the metal? Planck’s constant is
6.62607 × 10−34 J · s .
Answer in units of eV
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The light brings an energy hf to the electron. Energy φ (the work function) is required for the electrons to leave the surface, so the remaining energy of the electons is
E = hf - φ
Solve for φ
φ = hf - E
Although you've been given h in J∙s, I'm going to use 4.136×10^-15 eV∙s to save on unit conversions.
φ = (4.136×10^-15 eV∙s)(7×10^15 Hz) - 1.7 eV = 27.3 eV
which seems like an awful lot. Are you sure you have the frequency right?
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Well, if that's the frequency, then that would be the answer. Still seems odd to me, but I'm glad to help.
E = hf - φ
Solve for φ
φ = hf - E
Although you've been given h in J∙s, I'm going to use 4.136×10^-15 eV∙s to save on unit conversions.
φ = (4.136×10^-15 eV∙s)(7×10^15 Hz) - 1.7 eV = 27.3 eV
which seems like an awful lot. Are you sure you have the frequency right?
****************
Well, if that's the frequency, then that would be the answer. Still seems odd to me, but I'm glad to help.
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You're welcome.
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