The maximum force needed to remove an electron form potassium metal is 3.70 × 10-19 J.?
Part 2 of the question: If photons of frequency 13.54 × 1014 s-1 are shone on the surface of the metal, what is the kinetic energy (J) of the ejected electrons ?
Answers:
I devise the given enenergy is the minmum energy or minimum work needed to remove an electron.
The kinetic energy of the incoming photon is equal to the work required to remove an electron plus its kinetic vigour:
E_photon = W + E_kin
<=>
h·f = W + m·v^2/2
(h Planck's constant, f frequency of the photon, m mass of electron, v speed of electron)
Hence:
v = √ [ 2·(h·f - W) / m ]
= √ [ 2 · ((6.626×10^-34Js · 13.54×10^14s^-1) - 3.7×10^19J ) / 9.109×10^-31kg ]
= 1.08×10^6m/s Source(s): http://en.wikipedia.org/wiki/Photoelectr…
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Answers:
I devise the given enenergy is the minmum energy or minimum work needed to remove an electron.
The kinetic energy of the incoming photon is equal to the work required to remove an electron plus its kinetic vigour:
E_photon = W + E_kin
<=>
h·f = W + m·v^2/2
(h Planck's constant, f frequency of the photon, m mass of electron, v speed of electron)
Hence:
v = √ [ 2·(h·f - W) / m ]
= √ [ 2 · ((6.626×10^-34Js · 13.54×10^14s^-1) - 3.7×10^19J ) / 9.109×10^-31kg ]
= 1.08×10^6m/s Source(s): http://en.wikipedia.org/wiki/Photoelectr…
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