Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

The maximum kinetic energy (Kmax) for an electron is a function of its work function (φ) as well as the energy of the incident radiation.


Let Kmax of the electron be K1 when incident light has wavelength 600 nm and K2 when incident light has wavelength 400 nm. Note 1 nm = 10-9 m




According to the question, K2 is two times K1





To convert joule to eV, we can simply divide the value by 1.6 10-19. Thus, work function in electron volts can be determined by



1