Question:
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.
Solution:
Let K1 and K2 be the maximum energy emitted by the electrons when 600 and 400 nm wavelength is used.
K2 = 2K1
Therefore, the work function, ϕ = 1.03 eV