When photons of wavelength `lambda_(1) = 2920 Å` strike the surface of metal A, the ejected photoelectron have maximum kinetic energy of `k_(1) eV` and the smallest de-Broglie wavelength of `lambda`. When photons of wavelength `lambda_(2) = 2640 Å` strike the surface of metal B the ejected photoelectrons have kinetic energy ranging from zero to `k_(2) = (k_(1) – 1.5) eV`. The smallest de-Broglie wavelength of electrons emitted from metal B is `2lambda`. Find
(a) Work functions of metal A and B.
(b) `k_(1)`
Take `hc = 12410 eV Å`

Answer

(a) `phi_(A) = 2.25 eV, phi_(B) = 4.2 eV`
(b) `2e V`