# Physics

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When light of wavelength 391 nm falls on a
potassium surface, electrons are emitted that
have a maximum kinetic energy of 1.72 eV.
1) What is the cutoﬀ wavelength of potassium?
2)What is the threshold frequency for potassium?

The speed of light is 3 × 10^8 m/s and Planck’s constant is 6.63 × 10 ^−34 J · s .

• Physics -

KE = 1.72 eV =1.72•1.6•10^-19 =2.76•10^-19 J.
ε = h•c/λ =
=6.63•10^-34•3•10^8 /391•10^-9 = =5.08•10^-19 J
Einstein's photoelectric equation:
ε =W + KE,
Work function is
W = ε – KE =
= 5.08•10^-19 - 2.76•10^-19 =
=2.32•10^-19 J.
W = h•c/λₒ.
λₒ = h•c/W =
= 6.63•10^-34•3•10^8 /2.32•10^-19=
=8.56•10^-7 m.
W =h•fₒ.
fₒ =W/h =2.32•10^-19/6.63•10^-34 = 3.5•10^14 Hz.

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