Homework Help: Physics

Posted by Jason on Friday, May 11, 2012 at 10:02pm.

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?
Answer in units of nm
2)What is the threshold frequency for potassium?
Answer in units of Hz

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

Physics - Elena, Saturday, May 12, 2012 at 5:29am
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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