Question

(a) Photoelectrons with a maximum kinetic energy of 7.6eV are emitted from a metal when it is illuminated by ultraviolet radiation with a wavelength of 1.3×10^2 nm.
(i) What is the energy of the incident photons in electronvolts?
(ii) What is the wavelength of the radiation (in nanometres) corresponding to the lowest energy photons that can free electrons from the metal?
(iii) How do you explain the fact that, when infrared radiation is shone on this metal, no photoelectrons are emitted?
(b) What is the de Broglie wavelength (in nanometres) of an electron with a kinetic energy of 7.6eV?

Answer 1

https://physics.info/photoelectric/

See above link or Google Albert Einstein.
(i) get energy of incident photon from wavelength. E = h f but
f = c/wavelength
(ii)Ke emmitted = incident energy - energy needed to shake an electron loose
(iii) infrared is low frequency(energy)
http://calistry.org/calculate/deBroglieEquation