Question

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 .

Answer 1

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.

Answer 2

this did not help at all. please come up with shorter answer. Thanks :)