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.
Let the maximum energies of emitted electrons are K1 and K2 when 600 nm and 400 nm visible light are used according to question K2=2K1 Kmax=hν−ϕ=λhc−ϕ K1=λ2hc−ϕ=2K1 λ2hc−ϕ=2[λ1hc−ϕ]=λ12hc−2ϕ ϕ=hc[λ12−λ21] (∴ hc=1240 ev nm) Work function ϕ=66.2=1.03 eV.
