Asked by genius
Electrons are ejected from a metallic surface with speeds ranging up to 4.50x10^5 m/s when light with a wavelength of 615nm is used. (a) What is the work function of the metal? (b) What is the cutoff frequency for this surface?
Answers
Answered by
Elena
Einstein's photoelectric equation:
ε =W + KE,
h•c/λ = W +m•v²/2,
W = h•c/λ - m•v²/2 =
= 6.63•10^-34•3•10^8/615•10^-9 - 9.1•10^-31•(4.5•10^5)/2 =
= 2.3•10^-19 J = 1.44 eV.
W = h•f,
f = W/h = 2.3•10^-19/6.63•10^-34 = 3.48•10^14 Hz.
ε =W + KE,
h•c/λ = W +m•v²/2,
W = h•c/λ - m•v²/2 =
= 6.63•10^-34•3•10^8/615•10^-9 - 9.1•10^-31•(4.5•10^5)/2 =
= 2.3•10^-19 J = 1.44 eV.
W = h•f,
f = W/h = 2.3•10^-19/6.63•10^-34 = 3.48•10^14 Hz.
Answered by
kevin
good answer
Answered by
kevin u down bad
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