A rubidium surface has a work function...

Question

A rubidium surface has a work function of 2.16eV. (a)What is the maximum kinetic energy of ejected electrons if the incident radiation is of wavelength 413 nm? (b) What is the treshold wavelength for this surface?

Answer: a) 0.84 eV b) 574 nm

This is what i did

work function = 2.16eV

lambda = 413 nm

Kmax=?

hc = [(6.626x10^-34 Js)(3.00x10^8)]/[1.6x10^-19]

= 12.42x10^-7

Kmax = (hc)/lambda - workfunction

Kmax= [(12.42x10^-7)/413x10^-9] - 2.16eV

= (3.008 - 2.16) eV

= 0.85 eV

b) I don't what to do here.

Can someone help me out here? I would appreciate it.

Answer 1

The threshold wavelength is the lowest energy light that will liberate photoelectrons. That energy is the work function.

frequency = energy / Planck's constant

= 2.16 eV / 4.14 X 10^-15 eV/sec = 5.22 X 10^14 Hz

wavelength = speed of light / frequency

= 3.0 x 10^8 m/s / 5.22 x 10^14 Hz = 5.74 x 10^-7 m